Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Today’s Outline - August 31, 2015

• Time-dependent solutions

• Problem 1.5

• Dirac bra-ket notation

• Solutions of the infinite square well

• Wavefunction orthogonality

• Wavefunction completeness

Office Hours: Monday, Wednesday, Friday 14:00-15:00

Reading Assignment: Chapter 2.3

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, September 02, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 1 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Constructing the time-dependent solution

Each spatial solutions, the station-ary states, will have a specific en-ergy and there are an infinite num-ber of such solutions.

Thus, a full time-dependent solu-tion can be obtained by a linearcombination of time-independentwave functions.

Ψ1(x , t) = ψ1(x)e−iE1t/~

Ψ2(x , t) = ψ2(x)e−iE2t/~

...

Ψn(x , t) = ψn(x)e−iEnt/~

Ψ(x , t) =∞∑n=1

cnψne−iEnt/~

The general solution does not have time-independent expectation values,nor does it have zero variance for the expectation value of the energy

We generally leave the time-dependent portion as an implicit part of thesolution.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 2 / 15

Problem 1.5

Consider the wave function

Ψ(x , t) = Ae−λ|x |e−iωt ,

where A, λ, and ω are positive real constants.

(a) Normalize Ψ.

(b) Determine the expectation values of x and x2.

(c) Find the standard deviation of x and the probabilityof finding the particle outside of this range.

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 3 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx

= A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |e iωte−λ|x |e−iωt dx

= A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx

= A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)

break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx

+

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞

− 1

2λe−2λx

∣∣∣∞0

)

= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞− 1

2λe−2λx

∣∣∣∞0

)

= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)

=A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (a)

The normalization integral is

1 =

∫ψ∗ψ dx = A2

∫ ∞−∞

e−λ|x |���e iωte−λ|x |���e−iωt dx = A2

∫ ∞−∞

e−2λ|x | dx

= A2

(∫ 0

−∞e+2λx dx +

∫ ∞0

e−2λx dx

)break this into two integrals

= A2

(1

2λe2λx

∣∣∣0−∞− 1

2λe−2λx

∣∣∣∞0

)= A2

(1

2λ+

1

2λ

)=

A2

λ

A =√λ

So the normalized wave function is

Ψ(x , t) =√λe−λ|x |e−iωt

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 4 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx

= 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx

= 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx

= 0

this vanishes due to an odd integrand.

The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand.

The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is

⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)

= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2,

du = 2x dx ,

dv = e−2λxdx ,

v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx ,

v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(− x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)

= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−�

����xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)

= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(− xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)

= +1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)

= +1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx

=1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (b)

The expectation value of x is

〈x〉 = λ

∫ ∞−∞

xe−2λ|x | dx = 0

this vanishes due to an odd integrand. The expectation value of x2 is⟨x2⟩

= λ

(∫ 0

−∞x2e+2λx dx +

∫ ∞0

x2e−2λx dx

)= 2λ

∫ ∞0

x2e−2λx dx

this can be solved applying integration by parts twice, first using

u = x2, du = 2x dx , dv = e−2λxdx , v = − 1

2λe−2λx

⟨x2⟩

= 2λ

(−��

����x2e−2λx

2λ

∣∣∣∣∞0

+1

λ

∫ ∞0

xe−2λx dx

)= +2

∫ ∞0

x e−2λx dx

and with u = x , du = dx for the second

⟨x2⟩

= 2

(−��

���xe−2λx

2λ

∣∣∣∣∞0

+1

2λ

∫ ∞0

e−2λx dx

)= +

1

λ

∫ ∞0

e−2λx dx =1

2λ2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 5 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2

=1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx

= 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ

= e−√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Problem 1.5 - part (c)

The variance is now given by

σ2 =⟨x2⟩− 〈x〉2 =

1

2λ2− 0 =

1

2λ2

σ =1√2λ

λ

-2λ -σ 0 +σ +2λ

|Ψ|2

x

and the probability that the particle is outside the interval −σ < x < σ is

Pout = λ

∫ −σ−∞

e2λx dx + λ

∫ ∞σ

e−2λx dx = 2λ

∫ ∞σ

e−2λx dx

= 2λ

(−e−2λx

2λ

)∣∣∣∣∞σ

= e−2λσ

= e−2λ/√2λ = e−

√2 = 0.2431

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 6 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Dirac bra-ket notation

Paul Dirac developed a formalism for quantum mechanics which iscommonly used. We will see it in detail in Chapter 3, however one part ofthis formalism is a compact notation which simplifies writing expectationvalue integrals. We will start using this “bra-ket” notation immediately.

integral bra-ket

bra ψ(x)∗ 〈ψ|complex conjugate isimplicit

ket ψ(x) |ψ〉

normalization∫ψ∗(x)ψ(x)dx = 1 〈ψ |ψ〉 = 1

expectationvalue

∫ψ∗Qψdx 〈ψ |Q ψ〉 operator is applied

to the right

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 7 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0.

Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging

and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging

and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ

= −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ

= −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well

Consider a potential

V (x) =

{0, if 0 ≤ x ≤ a.

∞, otherwise

Outside the well the solution mustbe ψ(x) ≡ 0. Inside the well, theSchrodinger equation is

− ~2

2m

d2ψ

dx2= Eψ

This is the classical standing waveboundary value problem with gen-eral solution

V(x)

xa0

rearranging and substituting

k =√

2mE/~

d2ψ

dx2= −2mE

~2ψ = −k2ψ

ψ(x) = A sin(kx) + B cos(kx)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 8 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) + B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

ψ(x) = A sin(kx) + B cos(kx)

ψ(0) = ψ(a) = 0

for ψ(0) = 0

0 = ����A sin(0) + B cos(0)

B = 0

ψn(x) = A sin(nπ

ax)

En =⟨n∣∣Hn

⟩

Starting with the general solution,apply boundary conditions

for an infinite potential, the deriva-tive must be discontinuous

for ψ(a) = 0

0 = A sin(ka) +�����B cos(ka)

ka = nπ, n = 1, 2, 3, · · ·

k =nπ

an = 1, 2, 3, · · ·

all that remains is to calculate En

and normalize ψn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 9 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩

=n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2

=~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Infinite square well - solution

Normalization

1 = 〈n | n〉

=

∫ a

0|A|2 sin2(knx)dx

= |A|2 a2

A =

√2

a

ψn(x) =

√2

asin(nπ

ax)

State Energy

∣∣Hn⟩

= − ~2

2m

d2

dx2A sin

(nπax)

= +n2π2~2

2ma2A sin

(nπax)

=n2π2~2

2ma2|n〉

En =⟨n∣∣Hn

⟩=

n2π2~2

2ma2〈n|n〉

=n2π2~2

2ma2=

~2k2n2m

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 10 / 15

Wavefunction properties

0 a

x

n=1

n=2

n=3

The first three wavefunctions areshown offset in the y direction

The states are alternating even andodd about the center of the infinitewell x = a/2

The energy and momentum arenaturally quantized

The states are mutually orthogonal

〈ψm |ψn〉 = δmn

The states form a complete basisset

f (x) =∞∑n=1

cnψn(x)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15

Wavefunction properties

0 a

x

n=1

n=2

n=3

The first three wavefunctions areshown offset in the y direction

The states are alternating even andodd about the center of the infinitewell x = a/2

The energy and momentum arenaturally quantized

The states are mutually orthogonal

〈ψm |ψn〉 = δmn

The states form a complete basisset

f (x) =∞∑n=1

cnψn(x)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15

Wavefunction properties

0 a

x

n=1

n=2

n=3

The first three wavefunctions areshown offset in the y direction

The states are alternating even andodd about the center of the infinitewell x = a/2

The energy and momentum arenaturally quantized

The states are mutually orthogonal

〈ψm |ψn〉 = δmn

The states form a complete basisset

f (x) =∞∑n=1

cnψn(x)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15

Wavefunction properties

0 a

x

n=1

n=2

n=3

The first three wavefunctions areshown offset in the y direction

The states are alternating even andodd about the center of the infinitewell x = a/2

The energy and momentum arenaturally quantized

The states are mutually orthogonal

〈ψm |ψn〉 = δmn

The states form a complete basisset

f (x) =∞∑n=1

cnψn(x)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15

Wavefunction properties

0 a

x

n=1

n=2

n=3

The first three wavefunctions areshown offset in the y direction

The states are alternating even andodd about the center of the infinitewell x = a/2

The energy and momentum arenaturally quantized

The states are mutually orthogonal

〈ψm |ψn〉 = δmn

The states form a complete basisset

f (x) =∞∑n=1

cnψn(x)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 11 / 15

Orthogonality

∫ψ∗mψndx

=2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β)

= (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (cosα cosβ + sinα sinβ)

− (cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

cos(α± β) = cosα cosβ ∓ sinα sinβ

cos(α− β)− cos(α + β) = (������cosα cosβ + sinα sinβ)

− (������cosα cosβ − sinα sinβ)

= 2 sinα sinβ

sinα sinβ =1

2cos(α− β)− 1

2cos(α + β)

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral

∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]

= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .

. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0.

We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.

∫ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Orthogonality

∫ψ∗mψndx =

2

a

∫ a

0sin(mπ

ax)

sin(nπ

ax)dx

=1

a

∫ a

0

[cos

(m − n

aπx

)− cos

(m + n

aπx

)]dx

now proceed to evaluate the integral∫ψ∗mψndx =

[1

(m − n)πsin

(m − n

aπx

)− 1

(m + n)πsin

(m + n

aπx

)∣∣∣∣a0

=1

π

[sin[(m − n)π]

m − n− sin[(m + n)π]

m + n

]= 0 for m 6= n

If m = n, the above expression is not valid because . . .. . . the first term becomes 0/0. We know, however that m = n is simplythe normalization condition. Thus, orthogonality is demonstrated.∫

ψ∗mψndx = δmn

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 12 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn

= cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Completeness

Because this set of functions iscomplete, an arbitrary functionf (x) may be constructed as asum of the stationary wavefunc-tions.

The coefficient cm may be ob-tained by taking the inner prod-uct with ψ∗m.

f (x) =∞∑n=1

cnψn(x)

〈ψm|f (x)〉 =

∫ψ∗m(x)f (x)dx

=∞∑n=1

cn

∫ψ∗m(x)ψn(x)dx

=∞∑n=1

cnδmn = cm

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 13 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:

Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:

Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx

=

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx

=∞∑

m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn

=∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

The full time-dependentsolution is thus the sum:Where the coefficients,cn, are determined bythe initial state of thewavefunction at t = 0,Ψ(x , 0).

The condition of normal-ization for Ψ(x , 0) leadsto a constraint on the co-efficients

Ψ(x , t) =∞∑n=1

cn

√2

asin(nπ

ax)e−i(n

2π2~/2ma2)t

Ψ(x , 0) =∞∑n=1

cnψn(x)

cn =

√2

a

∫ a

0sin(nπ

ax)

Ψ(x , 0)dx

1 =

∫|Ψ(x , 0)|2 dx =

∫ ( ∞∑m=1

cmψm(x)

)∗( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcn

∫ψ∗m(x)ψn(x)dx =

∞∑m=1

∞∑n=1

c∗mcnδmn =∞∑n=1

|cn|2

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 14 / 15

Arbitrary time-dependent wavefunction

Since the coefficients, cn, are independent of time, the expectation valueof the energy is also

⟨H⟩

=

∫Ψ∗HΨdx =

∫ ( ∞∑m=1

cmψm(x)

)∗H

( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcnEn

∫ψ∗mψndx =

∞∑n=1

|cn|2 En

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15

Arbitrary time-dependent wavefunction

Since the coefficients, cn, are independent of time, the expectation valueof the energy is also

⟨H⟩

=

∫Ψ∗HΨdx

=

∫ ( ∞∑m=1

cmψm(x)

)∗H

( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcnEn

∫ψ∗mψndx =

∞∑n=1

|cn|2 En

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15

Arbitrary time-dependent wavefunction

Since the coefficients, cn, are independent of time, the expectation valueof the energy is also

⟨H⟩

=

∫Ψ∗HΨdx =

∫ ( ∞∑m=1

cmψm(x)

)∗H

( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcnEn

∫ψ∗mψndx =

∞∑n=1

|cn|2 En

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15

Arbitrary time-dependent wavefunction

Since the coefficients, cn, are independent of time, the expectation valueof the energy is also

⟨H⟩

=

∫Ψ∗HΨdx =

∫ ( ∞∑m=1

cmψm(x)

)∗H

( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcnEn

∫ψ∗mψndx

=∞∑n=1

|cn|2 En

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15

Arbitrary time-dependent wavefunction

Since the coefficients, cn, are independent of time, the expectation valueof the energy is also

⟨H⟩

=

∫Ψ∗HΨdx =

∫ ( ∞∑m=1

cmψm(x)

)∗H

( ∞∑n=1

cnψn(x)

)dx

=∞∑

m=1

∞∑n=1

c∗mcnEn

∫ψ∗mψndx =

∞∑n=1

|cn|2 En

C. Segre (IIT) PHYS 405 - Fall 2015 August 31, 2015 15 / 15