the quantum theory of atoms schrodinger equation 2009
TRANSCRIPT
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The Quantum Theory of
Atoms and Molecules
The Schrdinger equation and
how to use wavefunctions
Dr Grant Ritchie
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An equation for matter waves?
De Broglie postulated that every particles has an associated wave of wavelength:
ph/
Wave nature of matter confirmed by electron diffraction studies etc (see earlier).If matter has wave-like properties then there must be a mathematical function that is the
solution to a differential equation that describes electrons, atoms and molecules.
The differential equation is called the Schrdinger equation and its solution is called the
wavefunction,.
What is the form of the Schrdinger equation ?
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The classical wave equation
2
2
22
21
tvx
We have seen previously that the wave equation in 1d is:
Where v is the speed of the wave. Can this be used for matter waves in free space?
Try a solution: e.g.)(
),(tkxi
etx
Not correct! For a free particle we know thatE=p2/2m.
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An alternative.
tx
2
2
)(
),(tkxi
etx
ti
xm
2
22
2
Try a modified wave equation of the following type:
(is a constant)
Now try same solution as before: e.g.
Hence, the equation for matter waves infree space is:
For )(),(tkxi
etx ),(),(
2
22
txtxm
k
then we have
which has the form: (KE) wavefunction = (Total energy) wavefunction
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The time-dependent Schrdinger equation
),(2
2
txVm
pE For a particle in a potential V(x,t) then
and we have (KE + PE) wavefunction = (Total energy) wavefunction
titxV
xm
),(
2 2
22
TDSE Points of note:
1. The TDSE is one of the postulates of quantum mechanics. Though the SE
cannot be derived, it has been shown to be consistent with all experiments.2. SE is first order with respect to time (cf. classical wave equation).
3. SE involves the complex number i and so its solutions are essentially complex.
This is different from classical waves where complex numbers are used imply
for conveniencesee later.
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The Hamiltonian operator
HtxV
xmtxV
xm),(
2),(
2 2
22
2
22
)(2
)(2
2
2
22
xVm
pxV
xmH x
xipx
LHS of TDSE can
be written as:
whereis called theHamiltonian operator which is the differential operator that
represents thetotal energy of the particle.
Thus
where themomentum operator is
Thus shorthand for TDSE is:t
iH
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Solving the TDSEAaargh!
)(
2 2
22
xVxmt
i
)()(),( tTxtx
Suppose the potential is independent of
time i.e. V(x, t) = V(x) then TDSE is:
LHS involves variation of with twhile RHS involves variation of withx. Hence
look for a separated solution:
t
TiTxV
xT
m
)(
2 2
22then
Now divide byT:t
T
TixV
xm
1)(
1
2 2
22
LHS depends only uponx, RHS only on t. True for allx and tso both sides must equal a
constant,E(E= separation constant).
Thus we have:
ExVxm
Et
T
Ti
)(1
2
1
2
22
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Time-independent Schrdinger equation
Solving the time equation:
/)(
1 iEtAetTdt
iE
T
dTE
dt
dT
Ti
This is exactly like a wave e-it withE = . Therefore T(t) depends upon the energyE.
To find out what the energy actually is we must solve the space part of the problem....
The space equation becomes
EHExVxm
)(
2 2
22
or
This is the time independentSchrdinger equation(TISE) .
** All Prelims problems are concerned with solving TISE rather than the
TDSE!
The TISE can often be very difficult to solveit depends upon V(x)!
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Eigenvalue equations
The Schrdinger Equation is the form of anEigenvalue Equation: EH
where is the Hamiltonian operator,
is the wavefunction and is an eigenfunctionof;
Eis the total energy (T+ V) and an eigenvalue of.Eis just a constant!
)(2
2
22
xVdx
d
mVTH
Later in the course we will see that the eigenvalues of an operator give the possible
results that can be obtained when the corresponding physical quantity is measured.
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TISE for a free-particle
Exm
2
22
2
)/()()(),(
EtkxietTxtx
For a free particle V(x) = 0 and TISE is:
and has solutions
Thus the full solution to the full TDSE is:
m
kEwhereeore
ikxikx
2
22
Corresponds to waves travelling in either x direction with:
(i) an angular frequency, =E/ E= !
(ii) a wavevector, k= (2mE)1/2 / = p /p = h /!
WAVE-PARTICLE DUALITY!
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Interpretation of(x,t)
As mentioned previously the TDSE has solutions that are inherently complex (x,t)
cannotbe a physical wave (e.g. electromagnetic waves). Therefore how can (x,t)
relate to real physical measurements on a system?
The Born Interpretation
dxtxPdxtxdxtxtx ),(),(),(),(2*
* is real as required for a probability distribution and is the probabilityper unitlength (or volume in 3d).
The Born interpretation therefore calls theprobability amplitude, * (= P(x,t) )
theprobability density and * dx theprobability.
Probability of finding a particle in a small length dx at position x and time t is equal to
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Expectation values
xxPxxi
ii )(
dxtxxdxxxPxx2
),()(
Thus if we know (x, t) (a solution of TDSE), then knowledge of* dx allows the
average position to be calculated:
In the limit that x 0 then the summation becomes:
dxtxxdxxPxx2222
),()(Similarly
The average is also know as the expectation value and are very important in quantum
mechanics as they provide us with the average values of physical properties because inmany cases precise values cannot, even in principle, be determinedsee later.
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Normalisation
1),()(2dxtxdxxP
Total probability of finding a particle anywhere must be 1:
This requirement is known as theNormalisation condition. (This condition arises
because the SE is linear in and therefore if is a solution of TDSE then so is c
where c is a constant.)
Hence if original unnormalised wavefunction is (x,t), then the normalisation integral is:
dxtxN22
),(
And the (re-scaled) normalised wavefunction norm= (1/N) .
Example 1: What value ofNnormalises the functionN x (x L) of 0 xL?
Example 2: Find the probability that a system described by the function 21/2sin (x) where
0 x 1 is found anywhere in the interval 0 x 0.25.
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Boundary conditions for
In order for to be a solution of the Schrdinger equation to represent a physicallyobservable system, must satisfy certain constraints:
1. Must be a single-valued function ofx and t;
2. Must be normalisable; This implies that the 0 asx;
3. (x) must be a continuous function ofx;
4. The slope ofmust be continuous, specifically d(x)/dx must be
continuous (except at points where potential is infinite).
(x)
x
(x)
x
(x)
x
(x)
x
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Stationary states
/)()()(),(
iEtextTxtx
Earlier in the lecture we saw that even when the potential is independent of time the
wavefunction still oscillates in time:
Solution to the full TDSE is:
But probability distribution is static:
2//2)()()(*),(),( xexextxtxP
iEtiEt
Thus a solution of the TISE is known as a Stationary State.
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What other information can you get from ?
(and how!)
)(
2
2
2
2
xV
dx
d
m
H
.xi
px
nnnnnnnnn
nnn
ExExExH
EH
ddd
We have seen how we can use the probability distribution to calculate the
average position of a particle. What happens if we want to calculate the average
energy or momentum because they are represented by the following differential
operators:
Do the operators work on , or on , or on alone?
Take TISE and multiply from
left by and integrate:
NB is normalised.
Suggest that in order to calculate the
average value of the physical quantity
associated with the QM operator we
carry out the following integration:
xnn d
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Momentum and energy expectation values
dxtxxi
txpx ),(),(*
where .
xipx
is the operator for thex component of momentum.
Example: Derive an expression for the average
energy of a free particle. m
pE
2
2
then
m
pE
2
2
Since V= 0 the expectation value for energy for a particle moving in one dimension is
dxtxxm
txE ),(2
),(*2
22
The expectation value ofmomentum involves the representation of momentum as a
quantum mechanical operator:
Our definition of the expectation value is one of the postulates of QMsee later
lectures.
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Summary
titxV
xm
),(2 2
22
dxtxPdxtxdxtxtx ),(),(),(),(2*
1),()(2dxtxdxxP
TDSE:
Born interpretation:
Normalisation:
TISE:
EHExVxm
)(
2 2
22
or
/)()()(),(
iEtextTxtx
Boundary conditions on : single-valued, continuous, normalisable, continuous first derivative.
Expectation value of operator:
dxtxtx ),(),(
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Its never as bad as it seems.