FYSIKUM Stockholms Universitet

Tentamen, Kvantmekanik III Exam, Quantum Mechanics III Tuesday, January 13, 2015 Time: 09:00 – 14:00, Place: Room FR4 Allowed help: Physics Handbook (and the formulas and tables at the end of this exam) 1. Answers with correct and relevant comments (3p). Also well explained (1p*). Which of the following quantum mechanical statements or expressions are correct, or at least make sense, when interpreted in the “standard” way, and which are not? Give a short explanation for each one: (a) (U V )† = U † V † . (b) If a matrix element of an operator is time independent in the Schr¨odinger picture, the corresponding Heisenberg picture matrix element is time dependent. 00 00 (c) The propagator , t ; x0 , t0 ) evolves a wave function (x, t) by R R K(x 00 00 0 0 (x , t ) = dx dt K(x00 , t00 ; x0 , t0 ) (x0 , t0 ) for t00 > t0 .

(d) In quantum mechanics all particles behave equally in a fixed gravitational field, just as in classical mechanics (cf. Galileo). (e) For Hermitean operators A and B one can prove eiA eiB = eiB eiA (f) The sudden approximation and the standard WKB approximation are typically applicable under similar physical circumstances. Solution: (a) Wrong! The correct relation (U V )† = V † U † can be obtained from the definition of the Hermitean conjugate. (b) Wrong! Matrix elements are the same in both pictures. (c) Wrong! There should be no t0 integral. The propagator K expresses the matrix elements of the time evolution operator in position space.

(d) Wrong! In gravity-induced quantum interference experiments the interference e↵ects are sensitive to the mass. (e) Wrong! The inequality does not hold in general, as can be seen by Taylor expanding. (It can however be proven if A and B commute.) (f) Wrong! The standard version of WKB concerns time-independent equations, while the sudden approximation concerns time dependence. (In fact, one can envision cases where they can both be useful, but that would have to be specially motivated.)

2. Definitions (2p). Demonstration and discussion (2p*). Describe the probability interpretation of quantum mechanics and demonstrate that the expectation value of an observable defined in the formalism equals its average measured value. Discuss the role of Hermiticity. Solution: For the standard probability interpretation see Sakurai&Napolitano section 1.4. Crucial concepts are identically prepared systems, ensembles, and the probability for measuring a0 given by P (a0 ) = |ha0 |↵i|2 provided the ensemble is pure and characterised by the ket |↵i. Furthermore the measurement of the value a0 will prepare the ensemble to be characterised afterwards by the ket |a0 i.

The average measured value is the weighted sum of di↵erent possible values X X X X a0 P (a0 ) = a0 |ha0 |↵i|2 = a0 h↵|a0 iha0 |↵i = h↵|a0 ia0 ha0 |↵i = h↵|A|↵i, a0

a0

a0

a0

the last expression being the formal definition of the expectation value of an observable A. For all the equalities to hold it is necessary that the sum over all possible values a0 captures all the potential pure ensembles. This is guaranteed by the completeness of the space of eigenkets |a0 i which can be demonstrated for Hermitean operators A. Hermiticity also leads to real valued eigenvalues for observables which are always Hermitean.

3. Let |s1 i and |s2 i be two orthogonal states of a system: hs1 |s1 i = hs2 |s2 i = 1, hs1 |s2 i = 0. Consider the ensemble consisting of N >> 1

systems in the states |s i = a |s1 i+bei |s2 i, with a, b real and satisfying a2 + b2 = 1, and uniformly distributed on the interval (0, 2⇡). (a) Determine if the given ensemble is equivalent to the ensemble consisting of N a2 systems in the state |s1 i and N b2 systems in the state |s2 i. (3p)

(b) Calculate the entropies of the two ensembles. (1p*) Solution:

(a) The density operator describes all observable data of an ensemble. Thus, if the density operators of two ensembles agree, they are equivalent. (Alternatively, one could determine whether all possible transition probabilities or all possible ensemble averages agree.) The density operator of an individual system is ⇢ = |s i hs | = a2 |s1 i hs1 | + abei |s2 i hs1 | + abe i |s1 i hs2 | + b2 |s2 i hs2 | and a uniform distribution of gives Z 2⇡ 1 ⇢= d ⇢ = a2 |s1 i hs1 | + b2 |s2 i hs2 | , 2⇡ 0 which is the density operator of the ensemble to compare with. They are thus equivalent. (b) The entropy associated to the given ensemble is S=

k tr(⇢ log ⇢) =

ka2 log a2

k(1

a2 ) log(1

a2 ).

4. Consider a particle of spin 12 , and denote by |±i the eigenkets of z |±i = ±|±i.

z:

(a) Write down the operator U (ˆ n, ) that performs a counter-clockwise rotation by an angle around the axis n ˆ . Give its matrix form in the |±i basis. (3p)

(b) In the particular case that n ˆ is parallel to the z axis, calculate 0 i

⌘ U (ˆ z, ) i U (ˆ z, ) 1 .

and give its physical interpretation. (1p*)

Solution: i

(a) The rotation operator is e i J·ˆn/¯h = e 2 ·ˆn in the spin 12 representation. Recalling the Pauli matrices from the formula sheet we find ✓ ◆ nz nx iny ·n ˆ= nx + iny nz and ( · n ˆ )2 = I. This means

✓ ◆k 1 X 1 i U (ˆ n, ) = ·n ˆ k! 2 k=o ✓ ◆(2k) X ✓ ◆(2k+1) 1 1 X 1 i 1 i = + (2k)! 2 (2k + 1)! 2 k=o k=o = cos

2

I

i sin

·n ˆ

·n ˆ,

2

which is the matrix form of U in the |±i basis.

(b) Multiplying Pauli matrices we get 0 z

=

z,

0 x

= (I cos( /2) i z sin( /2)) x (I cos(( /2) + i z sin(( /2)) = ... = cos2 (( /2) sin2 ( /2)) x + 2 sin( /2) cos( /2) y = x cos( ) + y sin( ) and similarly 0 y

=

x

sin +

y

cos .

The interpretation of operator relations like A0 = U †AU = U 1 AU is always a rotation of the operator. In this case U is actually replaced by U 1 which means either that the system is rotated in the opposite clockwise sense or we think of the rotation being rotation of the coordinate system.

5. Consider a pair of spin 12 particles with opposite electric charges that have a magnetic-dipole-moment interaction ⌘ e2 g1 g2 ⇣ (1) (2) A S ·S 2m1 m2

H = A(µ1 · µ2 ) =

Ignore all degrees of freedom other than those due to spin. (a) Determine the energy eigenvalues and eigenstates. Express the results in terms of the parameter a = e2 g1 g2 A/4m1 m2 . (3p) (b) The system is initially (at t = 0) in the state |+i ⌦ | i. Calculate the probability of finding the system in the state | i ⌦ |+i at a later time t > 0. What is the maximum value of this probability and at what time is it attained? (1p*) Solution: The Hamiltonian of the system is H = (1)

(2)

2aSz Sz .

⇣ ⌘ (1) (2) (1) (2) a S+ S + S S +

(a) Acting on the products of the one-particle spin states, we get

H| i⌦| i =

h ¯2 |+i ⌦ |+i , 2 h ¯2 a | i ⌦ | i, 2

H|+i ⌦ | i =

a¯h2 | i ⌦ |+i + a

H |+i ⌦ |+i =

H| i ⌦ |+i =

a

h ¯2 |+i ⌦ | i, 2 h ¯2 a¯h2 |+i ⌦ | i + a | i ⌦ |+i. 2

We obtain the following eigenvectors and corresponding energy eigenvalues: |1, 1i ⌘ |+i ⌦ |+i,

E(1,1) =

a

h ¯2 2

1 1 |1, 0i ⌘ p |+i ⌦ | i + p | i ⌦ |+i, 2 2 h ¯2 |1, 1i ⌘ | i ⌦ | i, E(1, 1) = a 2 1 1 |0, 0i ⌘ p |+i ⌦ | i p | i ⌦ |+i, 2 2

h ¯2 2

E(1,0) =

a

E(0,0) = a

3¯h2 2

(b) The evolved state of the system is | (t)i = e 1 =p e 2 1 = e 2

iHt/¯ h

iE(1,0) t/¯ h

iE(1,0) t/¯ h

|+i ⌦ | i = e

1 |1, 0i + p e 2 +e

iE(0,0) t/¯ h

iHt/¯ h

✓

iE(0,0) t/¯ h

1 1 p |1, 0i + p |0, 0i 2 2

◆

|0, 0i

|+i ⌦ | i +

1 e 2

iE(1,0) t/¯ h

e

iE(0,0) t/¯ h

The probability for the flipped state can be read o↵ as " # E E t (0,0) (1,0) P = sin2 = sin2 (a¯ht) . 2¯h This probability reaches its maximum, P = 1 at t = ⇡(2n+1)/2¯ha for integer n.

6. A harmonic oscillator with Hamiltonian H=

p2 m! 2 x2 + 2m 2

is in its ground state. Then, for t 0 a perturbation of the form V (x, t) = m! 2 x2 e t/⌧ is applied, with and ⌧ real and positive. (a) Using first-order time dependent perturbation theory, compute the probability that the harmonic oscillator is not in the ground state when t ! 1. (3p)

(b) Calculate the rate of transition by turning on the potential slowly. You can take ✏ = 1/⌧ positive and small, and let the perturbation act from all times t0 ! 1 to t0 = t. Compare the result with the result of (a) in the limit !⌧ ! 1 and comment on the relation. (1p*) Solution: According to the formula for time-dependent perturbation theory, Z Z i t i t i!ni t0 (1) 0 0 cn (t) = hn|VI (t )|iidt = e Vni (t0 )dt0 h ¯ t0 h ¯ t0

| i ⌦ |+i

with h ¯ !n0 = En E0 = n¯h! for the harmonic oscillator. Here the t0 and x dependence can be separated, so that Vn0 (t0 ) = hn| m! 2 x2 e

Using the formula

x= we see that

r

t/⌧

|0i = m! 2 e

t/⌧

hn|x2 |0i.

h ¯ a† + a 2m!

h ¯ ! e t/⌧ 2 hn| a† + a |0i 2 ⇣ ⌘ p h ¯ ! e t/⌧ = hn| |0i + 2|2i 2 (a) We are only interested in the part that is not the ground state (n = 0) so only n = 2 contributes, with Z i 1 i2!t0 h ¯! i! 0 (1) p e t /⌧ dt0 = p lim c2 (t) = e t!1 h ¯ 0 2 2 2i! ⌧1 Vn0 (t0 ) = m! 2 e

t/⌧

hn|x2 |0i =

Thus the transition probability to lowest order is (1)

|c2 |2 =

!2⌧ 2 8! 2 ⌧ 2 + 2

2

as t ! 1. (b) For a slow turn on of the potential we get Z Z i t i (1) 0 0 cn (t) = hn|VI (t )|iidt = h ¯ 1 h ¯

t

0

ei!ni t Vni (t0 )dt0

1

and the transition contribution from n = 2: Z i t i2!t0 h ¯ ! ✏t0 0 i! e(✏+i2!)t (1) p e dt = p c2 (t) = e h ¯ 1 2 2 (2i! + ✏) implying that the rate goes to lim ✏!0

d (1) 2 h ¯ 2!2 2✏ |c2 (t)| = lim 2 ✏!0 (✏ + 4! 2 ) dt

which is proportional to (!). The vanishing of the rate for ! 6= 0 means that the result is consistent with the time derivative of the probability calculated in (a). (Actually, not only the rates, but also the probabilities obtained by the two methods agree.)

GOOD LUCK!

Some useful formulas Z

Z

Z

x2

2n

dxx e 1

=

e

x2

a2 t2 ibt

r

1

J± |j, mi = h ¯

e dt =

⇡(2n)!

1 22n+1 n! n+ 2

dxx2n+1 e

0

1

p

=

n! 2

⇡ e a2

n+1

b2 /(4a2 )

p (j ⌥ m)(j ± m + 1)|j, m ± 1i

For the harmonic oscillator: p a|ni = n|n 1i p a† |ni = n + 1|n + 1i [a, a† ] = 1

a† a|ni = n|ni r h ¯ x= a† + a 2m! r

m¯h! † p=i a a 2 r ⌘ p h ¯ ⇣p hl|ˆ x|ni = n l,n 1 + n + 1 l,n+1 2m! r ⌘ p m¯h! ⇣ p hl|ˆ p|ni = i n l,n 1 + n + 1 l,n+1 2 Ground state wave function for the one-dimensional harmonic oscillator: ⇣ m! ⌘ 14 m!x2 hx|0i = e 2¯h ⇡¯h Some spherical harmonics: 1 Y00 = p 4⇡

r

3 = cos ✓ 4⇡ r 3 ±1 Y1 = ⌥ sin ✓e±i 8⇡ r 5 Y20 (✓) = (3 cos2 ✓ 1) 16⇡ @ Lz = i¯h @ Y10 (✓)

Commutators:

@ F (p) @p @ [p, G(x)] = i¯h G(x) @x ~ = ¯h ~ Spin operator for spin-1/2 particles: S 2 Pauli matrices: ✓ ◆ ✓ ◆ ✓ ◆ 0 1 0 i 1 0 , 2= , 3= 1 = 1 0 i 0 0 1 [x, F (p)] = i¯h

Time independent perturbation, non-degenerate case: X

|ni = |n(0) i + n

= En

E

(0)

k6=n

|k (0) i 2

= Vnn +

Vkn (0)

c(0) n (t) = c(1) n (t)

=

i h ¯

Z

t t0

0

X

|Vnk |2

(0)

En

(0)

Ek

+ ...

ni

i h ¯

0

hn|VI (t )|iidt =

Fermi’s Golden Rule:

wi!n =

+ ...

Ek

k6=n

Time dependent perturbation theory:

(0)

En

2⇡ |Vni |2 (En h ¯

Z

t

0

ei!ni t Vni (t0 )dt0

t0

Ei )

for constant perturbation; wi!n =

2⇡ |Vni |2 (En h ¯

for harmonic perturbation.

Ei + h ¯ !) +

2⇡ † 2 |V | (En h ¯ ni

Ei

h ¯ !)