Solutions6.pdf

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Quantum Mechanics I

Solutions 6.

HS 2012Prof. Ch. Anastasiou

Exercise 1. Harmonic Oscillator: coherent states

A coherent state of a one-dimensional harmonic oscillator is defined as an eigenstate of the

(non-Hermitian) annihilation operator a:

a | = | , (1)

where is, in general, a complex number. We recall that the annihilation operator a and creationoperator a act on harmonic oscillator number states |n, i.e., Fock states, in the followingmanner:

a |n = n |n 1 and a |n = n + 1 |n + 1 . (2)The states {|n} are a complete set of eigenstates of the number operator N = aa and of theharmonic oscillator Hamiltonian H = (N + 12) according to:

N|n = n |n and H|n = (n + 12

) |n with n|m = nm and n

|n n| = I (3)

a) Prove that

| = e||2/2ea |0 (4)is a normalized coherent state.

b) Prove the minimum uncertainty relation for such a state.

c) Write | as

| =n=0

f(n) |n . (5)

Show that the distribution of |f(n)|2 with respect to n is of the Poisson form. Find themost probable value of n, hence of E.

d) Show that the basis of coherent states {|} is overcomplete, i.e.,

| = 0 , and

d() d()

| | = I . (6)

Hint. To calculate the scalar product, make use of the Baker-Campbell-Hausdorff formula,valid for two operators A, B whose commutator [A, B]

C,

eA+B = eA eB e[A,B]/2 = eB eA e[A,B]/2 . (7)

To prove the completeness, write the integral in polar coordinates. You will maybe needthe Euler integral:

(n) = (n 1)! =+0

dt tn1 et . (8)

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Solution. Scalar product: From Baker-Campbell-Hausdorff

eA+B = eA eB e[A,B]/2 = eB eA e[A,B]/2

so thateA

eB = eB eA e[A,B]

and setting A = a and B = a, [A, B] = so that:

| = e ||2+||2

2 0|eaa |0 = e ||2+||2

2 e 0|ea |0 = e||2/2 (S.1)

Completeness:

d | | =n,m

1n!

m!|n m|

d() d()

n()me||

2

=

=n,m

1n!

m!|n m|

20

d

ei(nm)

+0

dn+m

e2 =

=n,m

1n!

m!|n m| 2nm

+0

dn+m

e2 =

n

1

n!|n n| 2

+0

d2n

e2 =

=n

1

n! |n n| +

0 dt t

n

et

=n|n n| = I

(S.2)

where we initially wrote = ei and then made the substitution 2 = t. The integration mea-

sure reads d = d() d(). There also is a cooler way of showing it by proving that the operatord() d()

| | commutes with all operators belonging to the algebra generated by a, a, therefore ithas to be proportional to identity by Schurs lemma, and since the expectation value over the vacuum is 1

then the proportionality constant is 1.

e) Prove that a coherent state |0 remains coherent when time evolution is given by theharmonic oscillator Hamiltonian H, i.e.,

|(t) = eiHt/ |0 = |0eit , (9)up to a time-dependent phase factor.

Solution.

|(t) = eiHt/ |0 = e||2/2eiHt/n

nn!|n = e||2/2

n

nn!

eiHt/ |n

= eit/2e||2/2n

nn!

eint |n = eit/2e||2/2

n

(eit)nn!

|n = eit/2 |(t) |0eit

(S.3)

f) Show that a coherent state can also be obtained by applying the translation (finite-displacement) operator eipl/ (where p is the momentum operator and l is the displace-ment distance) to the ground state.

Solution. The solution to this exercise is attached on page 5.

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Exercise 2. Delta-function potential

A particle of mass m in one dimension is bound to a fixed center by an attractive -functionpotential:

V(x) = (x) , ( > 0) .

Att

= 0, the potential is suddenly switched off (that is,V

= 0 fort >

0). Find the wavefunction for t > 0. (Write your answer using the propagator kernel.)

Hint.

exp

1

2iax2 + iJ x

=

2i

aexp

iJ22a

Solution. The solution to this exercise is attached on page 10.

Exercise 3. Schrodinger vs. Heisenberg pictures

In the lecture, you have seen that a spin- 12 system in the presence of a magnetic field can bedescribed by the Hamiltonian operator H = Sz. The energy eigenstates are the Sz eigenstates| and in the Schrodinger picture, a general state | evolves like

|(t) = C+eit2 |+ + Ce

it2 | , for |(t = 0) = C+ |+ + C | .

Here, we would like to describe the same system in the Heisenberg picture.

(a) Solve the Heisenberg equation of motion, idAdt = [A, H], for Sx,y,z(t) with initial condition

Sx(t = 0) =

2 (

|+

|+

|+

|)

S0x and analogously for Sy,z(t = 0).

Solution. [Sz, H] = 0, obviously, so the Sz operator is, as expected, a constant of motion, Sz(t) =Sz(0) = S

0z .

For Sx,y, we find, via the commutation relations of the Pauli matrices ([i, j ] = iijkk), the coupledsystem of equations

dSx

dt= Sy , dSy

dt= Sx . (S.4)

This can be solved e.g. by writing the system as a 2x2 matrix and diagonalising it, or faster by noting that

d

dt(Sx(t) iSy(t)) = i(Sx(t) iSy(t)) Sx(t) iSy(t) = (S0x iS0y)eit . (S.5)

One way or the other, one finds the solutions

Sx(t) = cos(t)S0

x sin(t)S0

y, Sy(t) = cos(t)S

0

y+ sin(t)S0

x, (S.6)

which obviously satisfy the initial condition, too.

(b) Check that you get the same result in the two pictures for the expectation valuesSx (t), Sy (t), Sz (t) when taking it with respect to i) the state |+, ii) the state |Sx, +.

Solution. i) Since (c.f. Sheet 5, exercise 1), S0z = 2 and S0x,y = 0 in this case, the time-dependentexpectation values are actually constant,

Sz (t) = 2

, Sx,y (t) = 0 . (S.7)

This is in accordance with the result from the Schrodinger picture, where the eigenstate

|+

just oscillates

with a phase that drops out in the expection values.

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ii) In this more interesting case, we rediscover the spin-precession. Similarly as before, S0x = 2 andS0y,z = 0. So with the results from part a), we find

Sx (t) = 2

cos(t) , Sy (t) = 2

sin(t) , Sz (t) = 0 , (S.8)

which coincides with the result in the Schrodinger picture (C= = C = 12 ).

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Solution to Exercise 1.

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