Quantum Mechanics I, Correction Sheet 7, Spring 2013

Responsible for this sheet: J. Guillod (julien.guillod@unige.ch), office 212, Sciences I

April 24, 2013 (Ecole de Physique, Auditoire Stückelberg)

Prof. D. van der Marel (dirk.vandermarel@unige.ch)

Tutorials: J. Guillod (julien.guillod@unige.ch), O. E. Peil (oleg.peil@unige.ch)

I. (*) FUNCTIONAL CALCULUS AND GENERALIZED COMMUTATORS

Consider an analytic function F : C → C so that

∞
X
F (x) = fn xn .
n=0

The function F (Â) of an operator  is then defined as

∞
X
F (Â) = fn Ân .
n=0

1. Let |ψi be an eigenvector of  with eigenvalue a. Therefore, for all n ∈ N, we have,

Ân |ψi = an |ψi .
so by definition of F (Â),
∞
X ∞
X
F (Â)|ψi = fn Ân |ψi = fn an |ψi = F (a)|ψi ,
n=0 n=0

and |ψi is an eigenvector of F (Â) with eigenvalue F (a).

  
2. If B̂, Â , Â = 0, we prove by induction that

B̂, Ân = n B̂, Â Ân−1 .

   

The base case is n = 1. For the inductive step, by assuming the relation true of n, we have
B̂, Ân+1 = B̂ Ân  − Ân ÂB̂ = B̂ Ân  − Ân B̂  + Ân B̂  − Ân ÂB̂
 

= B̂, Ân  + Ân B̂,  = n B̂,  Ân + Ân B̂, Â

       

= (n + 1) B̂, Â Ân ,
 

 
since B̂, Â commutes with Â. For a general function, we obtain
∞ ∞ ∞
 X  X X
n n
nfn Ân−1 = B̂, Â F 0 (Â) ,
      
B̂, F (Â) = B̂, fn  = fn B̂,  = B̂, Â
n=0 n=0 n=0

because by definition,
∞
X
0
F (x) = nfn xn−1 .
n=0

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3. If X̂, P̂ = i~, then we deduce

X̂, T (P̂ ) = X̂, P̂ T 0 (P̂ ) = i~T 0 (P̂ ) ,

   

and

P̂ , V (X̂) = P̂ , X̂ V 0 (X̂) = −i~V 0 (X̂) .

   

These two relations will be used in the next exercise.

II. EHRENFEST THEOREM AND HAMILTON’S EQUATIONS

In this exercise we consider a particle in three dimensions in a potential V , and the aim is to link
and see the differences between classical and quantum mechanics. The classical Hamiltonian is

p2
H= + V (q) ,
2m
where p = (p1 , p2 , p3 ) and q = (q1 , q2 , q3 ) are the generalized coordinates. The quantum Hamilto-
nian is
p̂2
Ĥ = + V (q̂) ,
2m
where p̂ = (p̂1 , p̂2 , p̂3 ) and q̂ = (q̂1 , q̂2 , q̂3 ) are the momentum and position operators.

A. Poisson brackets and commutators

The Poisson bracket of two classical observables is defined as

3  
X ∂A ∂B ∂A ∂B
{A, B} = − ,
∂qi ∂pi ∂pi ∂qi
i=1

and the commutator between two quantum observables by

 
Â, B̂ = ÂB̂ − B̂ Â .

1. By definition of the Poisson bracket, we obtain

3   X 3
X ∂qi ∂qj ∂qi ∂qj
{qi , qj } = − = (δik 0 − 0δjk ) = 0 ,
∂qk ∂pk ∂pk ∂qk
k=1 k=1
3   X 3
X ∂pi ∂pj ∂pi ∂pj
{pi , pj } = − = (0δjk − δik 0) = 0 ,
∂qk ∂pk ∂pk ∂qk
k=1 k=1
3   X 3
X ∂qi ∂pj ∂qi ∂pj
{qi , pj } = − = (δik δjk − 0) = δij .
∂qk ∂pk ∂pk ∂qk
k=1 k=1

2. In the representation given by

∂
q̂i ψ(q) = qi ψ(q) , p̂i ψ(q) = −i~ ψ(q) ,
∂qi

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 we have
 
q̂i , q̂j ψ(q) = q̂i q̂j ψ(q) − q̂j q̂i ψ(q) = qi qj ψ(q) − qj qi (q) = 0 ,
∂ ∂ ∂ ∂
p̂i , p̂j ψ(q) = p̂i p̂j ψ(q) − p̂j p̂i ψ(q) = −~2 ψ(q) + ~2
 
ψ(q) = 0 ,
∂qi ∂qj ∂qj ∂qi
  ∂ ∂
q̂i , p̂j ψ(q) = q̂i p̂j ψ(q) − p̂j q̂i ψ(q) = −i~qi ψ(q) + i~ (qi ψ(q))
∂qj ∂qj
∂ ∂ ∂qi
= −i~qi ψ(q) + i~qi ψ(q) + i~ = i~δij .
∂qj ∂qj ∂qj

The commutation relations between position and momentum operators in quantum me-
chanics are the analog of the Poisson brackets between position and momentum in classical
mechanics.

B. Ehrenfest theorem

1. By using the chain rule, the evolution of a classical observable A = F (q, p, t) is given by
3   3  
dA ∂A X ∂A ∂A ∂A X ∂A ∂H ∂A ∂H ∂A
= + q̇i + ṗi = + − = + {A, H} ,
dt ∂t ∂qi ∂pi ∂t ∂qi ∂pi ∂pi ∂qi ∂t
i=1 i=1

where we used Hamilton equations,

∂H ∂H
q̇i = , ṗi = − .
∂pi ∂qi

2. The evolution of the expectation value of a quantum observable  evolving under the action
of the Hamiltonian Ĥ, is
   
d d ∂ Â d d
 = hψ(t)|Â|ψ(t)i = hψ(t)| |ψ(t)i + hψ(t)| Â|ψ(t)i + hψ(t)| |ψ(t)i
dt dt ∂t dt dt
∂ Â 1 1
= hψ(t)| |ψ(t)i − hψ(t)|Ĥ Â|ψ(t)i + hψ(t)|ÂĤ|ψ(t)i
* +∂t i~ i~
∂ Â 1 D E
= + Â, Ĥ ,
∂t i~

where we used the Schrödinger equation and its adjoint,

d d
i~ |ψi = Ĥ|ψ(t)i , −i~ hψ(t)| = hψ(t)|Ĥ .
dt dt

C. Hamilton’s equations

1. By applying Ehrenfest theorem and the conclusions of the first exercise, we have
d 1 D E 1 1
q̂, p̂2 =
 
hq̂i = q̂, Ĥ = hp̂i ,
dt i~ 2i~m m
d 1 
D E 1  
hp̂i = p̂, Ĥ = p̂, V (q̂) = − h∇V (q̂)i .
dt i~ i~

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 For the classical system, the Hamilton’s equations are given by

d 1 d
q = p, p = −∇V (q) .
dt m dt

2. For a quadratic potential

mω 2 2
V (q) = q ⇒ ∇V (q) = mω 2 q ,
2
and therefore the quantum-classical correspondence

hq̂i ↔ q , hp̂i ↔ p ,

provides an exact analogy. This correspondence is not true for a generic potential. For
example for a quartic one,

λ 4
V (q) = q ⇒ ∇V (q) = λq3 ,
4
and consequently

h∇V (q̂)i = λ q̂3 6= λ hq̂i3 ↔ λq3 .

III. EVOLUTION OPERATOR

The time-evolution of a quantum state |ψ(t)i ∈ E where E is an Hilbert space is given by the
Schrödinger equation

d
i~ |ψ(t)i = Ĥ(t)|ψ(t)i ,
dt

where Ĥ(t) = Ĥ(t)† is the Hamiltonian of the system.

1. The evolution operator Û (t) : E → E is defined as Û (t)|ψ(0)i = |ψ(t)i. This operator is

linear because if
d d
i~ |a(t)i = Ĥ(t)|a(t)i , i~ |b(t)i = Ĥ(t)|b(t)i ,
dt dt
then
d
i~ |ψ(t)i = Ĥ(t)|ψ(t)i , with |ψ(t)i = α|a(t)i + β|b(t)i ,
dt
which prove that

Û (t) (α|a(0)i + β|b(0)i) = Û (t)|ψ(0)i = |ψ(t)i

= α|a(t)i + β|b(t)i = αÛ (t)|a(0)i + β Û (t)|b(0)i .

2. By using the Schrödinger equation, we have

d d
i~ Û (t)|ψ(0)i = i~ |ψ(t)i = Ĥ(t)|ψ(t)i = Ĥ(t)Û (t)|ψ(0)i ,
dt dt

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 and also by definition,
Û (0)|ψ(0)i = |ψ(0)i .
Since the last two relations are valid for all |ψ(0)i ∈ E, the evolution operator satisfies the
following differential equation
d ˆ
i~ Û (t) = Ĥ(t)Û (t) , Û (0) = I.
dt
This equation also defines the evolution operator uniquely.

3. (*) By using the differential equation satisfied by the evolution operator and its conjugate,
we have
d    d  
d

i~ Û (t)† Û (t) = i~ Û (t)† Û (t) + Û (t)† i~ Û (t)
dt dt dt
 †
= − Ĥ(t)Û (t) Û (t) + Û (t)† Ĥ(t)Û (t) = 0 ,

and
Û (0)† Û (0) = Iˆ† Iˆ = Iˆ ,
which together prove that
Û (t)† Û (t) = Iˆ .
ˆ In
Strictly speaking, to prove that Û (t) is unitary, it remains to show that Û (t)Û (t)† = I.
fact this is automatically true by the fact that Û (t) is surjective: for all |ψ(t)i ∈ E there
exists |ψ(0)i ∈ E such that |ψ(t)i = Û (t)|ψ(0)i and therefore,
Û (t)Û (t)† |ψ(t)i = Û (t)Û (t)† Û (t)|ψ(0)i = Û (t)|ψ(0)i = |ψ(t)i .

4. If the Hamiltonian is time-independent, the evolution operator satisfies the linear differential
equation
d
i~ Û (t) = Ĥ Û (t) , Û (0) = Iˆ .
dt
Since this differential equation is linear it is sufficient to check that

Û (t) = e−iĤt/~ ,
is a solution:
d d
i~ Û (t) = i~ e−iĤt/~ = Ĥe−iĤt/~ = Ĥ Û (t) , Û (0) = e−iĤ0/~ = Iˆ
dt dt
 
5. (*) If Ĥ(t), Ĥ(s) = 0, then the evolution operator is given as for an ordinary differential
equation by
 Z t 
−i
Û (t) = exp Ĥ(s) ds .
~ 0
However, if the Hamiltonian does not commute at different times, then
Z t 2 Z t Z t Z t
d
Ĥ(s) ds = Ĥ(t) Ĥ(s) ds + Ĥ(s) ds Ĥ(t) 6= 2Ĥ(t) Ĥ(s) ds ,
dt 0 0 0 0

and the evolution operator does not satisfies the differential equation.

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