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Algebraic Method of Solving the Linear Harmonic
Oscillator
Lecture by Gable Rhodes
PHYS 773: Quantum MechanicsFebruary 6th, 2012
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Gable Rhodes, February 6th, 2012 2
Simple Harmonic Oscillator• Many physical problems can be modeled as small
oscillations around a stable equilibrium• Potential is described as parabolic around the
minimum energy
• The Hamiltonian is formulated in the usual way for canonical variables q and p.
22
21)( qmqV
)(2
2
qVmpH
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Raising & Lowering Operators - Defined
• After substituting in our potential, we can rearrange the constants slightly to give the Hamiltonian a more suggestive appearance
• If we “factor” the operators, we get
• Recognizing that the last term is the commutator of the canonical variables, we get
22222222
21
21 pqm
mqmp
mH
qppqimipqmipqmm
H 21
pqimipqmipqmm
H ,21
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Gable Rhodes, February 6th, 2012 4
pqimipqmipqmm
H ,21
Raising & Lowering Operators - Defined
• We can now define the operator a
• Upon substitution
• And Simplifying
mpiqma
2
pqimaamm
H ,221 †
mpiqma
2†
pqiaaH ,
2†
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Gable Rhodes, February 6th, 2012 5
Raising & Lowering Operators - Defined
• If we reverse the order of the operators, a similar expression is obtained
• Subtracting the two forms yields the commutator
• Simplifying to
0,2
,2
††
pqiaapqiaaHH
0,†† pqiaaaa
pqiaa ,, †
pqiaaH ,
2†
pqiaaH ,
2†
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Raising & Lowering Operators - Defined
• It is important to note that the a, a† operators are defined in such a way that as long as the state variables follow the canonical commutation relation, the a, a† commutator will be 1.
• And the Hamiltonian can be written as a linear function of a, a†
21†aaH
pqiaa ,, †
1, †
iiaa
21†aaH
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Properties of a, a†
• If a wave function has the property that it is an eigenfunction of the Hamiltonian
• We can introduce the equivalent statement for the operator aa† with generic eigenvalue, λ
• We then apply the a† operator to a†ψ and test if the result is the same eigenvector.
)(2
2
qVmpH EH
aa†
†††††††† 111 aaaaaaaaaaa
Use commutator
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Properties of a, a†
• And the equivalent method for a
• Using the relationship of Hamiltonian, we can then relate eigenvalues
aa†
aaaaaaaaaaa 1111 †††
EH aa†
21
21 †† aaaaH
21†aaEH
21
21† aaE
21E
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Raising & Lowering Operators -Properties
• a† is called the raising (or creation) operator.
• And a is the lowering (or annihilating) operator.
• The rungs of the ladder are all evenly separated.
• No degeneracy.
ψ
a† ψ
aψ
aaψ
a† a† ψ
λ
λ+ 1
λ+2
λ-2
λ- 1W
ave
vect
ors
Eige
nval
ues
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What is the Significance?• If we have any solution, we can find infinitely more
solutions by repeated application of the raising and lowering operators
• But, importantly, although there are infinite solutions we know that the are are no solutions with negative energy (both kinetic and potential components for the Hamiltonian are always positive)
• Therefore, a ground state must exist. (this is also a property of Sturm-Liouville PDE)
• Applying the lowering operator to the ground state will result in a null vector (trivial state).
• Eq.10-77 0a
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Raising & Lowering Operators -Eigenvectors
• Once we have a ground state, repeated application of the raising operator will result in an infinite set of eigenvectors with distinct (non-degenerate) eigenvalues.
• And introducing an arbitrary starting point λ0.
• But for the special case of the ground state
• Must be zero on the right side (by our definition), so λ0 is exactly 0.
0† n
n a
0†
00†††
nnn anaaaaa
000† 0 aa
00 00 nn 0
...2,1,0n
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Determine the Energy Levels• Using the previous equation relating the Hamiltonian
and aa†, we can relate energy to λ (n).
• And since we started at the ground state, we can relate the energy level to the eigenvectors
• If we use a normalization constant
• Where An can be found by direct integration at each step or by algebraic tricks (later)
21
21 nE
0† n
n a
21nEn
0† n
nn aA
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What Are These Operators Good For Anyway?
• We found Energy exactly and wavevectors in abstract form.
• What else can we do with them?• What about expectation values? In chapter 5,
problem 2, we were asked to find <V>. This required direct integration with the (explicitly known) wavefunction.
• Can this be done without knowing the wavefunction?
22
21)( qmqV
22
21 qmV
2*2
21 qmV
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Expectation value of Potential• If we use the definition of a†, a and rewrite q and
p operators in terms of a and a† we get
• Now these can be substituted into <V>
mpiqma
2†
mpiqma
2
22 †aam
q
22 †aaimm
p
2†*22*2
221
21 aa
mmqmV
)(22
1 ††††* aaaaaaaaV
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Expectation value of Potential
• Of the four terms in the integral, we see that two of them vanish due to the orthogonality of the wavevectors.
• And the other two are known to us from the previous work.
2,2**
nnnnaa 02,2*††* nnnnaa
111 ,*†* nnnaa nnnn
nnnaa nnnn ,*†*
)(22
1 ††††* aaaaaaaaV
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Expectation value of Potential• This makes the result pretty straightforward
• And the this result agrees with problem 5.2, and the virial theorem
• But, we did not need to know the explicit form of the wavefunction.
10022
1 nnV
nEnV
21
212
21
)(22
1 ††††* aaaaaaaaV
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<q>, <p>, <q2>, <p2>• Other key expectation values can be easily
obtained.
22 †aam
q
22 †aaimm
p
02
†**
aam
02
†**
aaimm
pp
212 n
mq
2†*222*2
2aam
mpp
aaaaaaaamp ††††*2
2
21100
22 nmnnmp
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Variance and Uncertainty?• The variance is therefore
• And this agrees with the uncertainty principle
21222 n
mqqq
21222 nmppp
221
npq
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What is left then?• The Normalization constant, which can also be
determined algebraically.
• Square both sides and integrate
• After integration by parts and throwing out the boundary terms.
• So that gives us our wavefunction in terms of the ground state and raising operator
1†
nn Ba
1,12
1*
1†*†
nnnnnn BBBaa
2*† Baa nn nnnn nnB ,*2 11
1† 1 nn na
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Normalization• Lets try ψ1.
• Next try ψ2.
• We can now write the general equation
0†
!1
nn a
n
1† 1 nn na
0†
1010 a 0†
1 a
1†
1111 a 0
2†2 !2
1 a
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And what is ψ0? • Starting with our condition for the ground state
• And using the definition of the operator
• We get a first order ODE
00 a
mpiqma
2
02 00
mpiqma
00
mpiq
qip
00 qm
dqd
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And what is ψ0?• The equation is separable and easily solved.
• After normalization we get
• Which is consistent with the solutions in chapter 5.
00 qm
dqd
2
0 2exp qmconst
2
41
0 2exp qmm
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Finding ψ1. • Here we can use the raising operator to generate
further solutions
• Substitute in the operator
• After rearranging, we get the desired result.
0†
!1
nn a
n
01†
1 !11 a
01 2
mpiqm
01 22
qm
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Finding ψn. • We find that continuing up the ladder is in fact a
generating algorithm for the Hermite polynomials, and the general equation then identical to eq. 5.39 0
†
!1
nn a
n
0!2
1
qmH
nnnn
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Matrix representation of a, a†
• We can show that the lowering operator relates neighboring wave vectors with the normalization factor.
• Adding the bra.
• This gives the matrix elements of a as shown.
1 nn na
nna nnnn 111
00400
300200
10
a
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Matrix representation of a, a†
• As an example, lowering n=2,nna nnnn 111
2020
010
100
00300
20010
010
221 a
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Matrix representation of a, a†
• The equivalent representation of the raising operator is derived from the expression
• Adding the bra.
1† 1 nn na
03
002001
00
†a
11 11†
1 nna nnnn
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Matrix representation of a, a†
• With the lowering and raising operators in matrix form, we can then solve for the q and p operators in terms of a, a†.
• For position we get,
22 †aam
q
03302
20110
222 †
maa
mq
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Matrix representation of a, a†
• And the equivalent expression for momentum is,
• The complex constant insures that the anti-symmetric matrix is Hermitian.
22 †aaimm
p
03302
20110
222 †
miaaimm
p