PHYSICS 304 QUANTUM PHYSICS II (2005) Assignment 6 Solutions

PHYSICS 304 QUANTUM PHYSICS II (2005) Assignment 6 Solutions

1. Consider a particle confined to move in the region x > 0 in the presence of an attractive potential V(x) = ∞ for x < 0 and V(x) = −α ² /x for x > 0. In this case, the solution to the Schr¨odinger equation can be written (you are NOT required to show this)

ψ E (x) = e ^−kx

∞

X

n =0

c n (x/a) ^{n +1} x > 0

where a = ~ ² /2mα ² , E = −~ ² k ² /2m and the coefficients c n satisfy the recurrence relation c _n = 2akn − 1

n(n + 1) c _n−1 n ≥ 1.

(a) Given that, except in certain cases, the infinite series above behaves like e ^2kx , show how this recurrence relation leads to the result that the energy of the particle is quantised, with energies

E n = − mα ⁴ 2~ ² . 1

n ² .

(b) Show that the normalised wave function for the lowest allowed energy level is of the form ψ ₁ (x) = 2k ^3/2 xe ^−kx . Determine k and sketch the associated probability distribu- tion, indicating where this distribution has a maximum.

(c) Calculate the average distance of the particle from the origin when in this state.

You may need the integral Z ∞

0

x ⁿ e ^−2kx dx = n!

(2k) ^{n +1} . SOLUTION

(a) The information supplied tells us that if the recurrence relation continues to generate an infinite sequence of non-zero coefficients, then the resultant solution for the wave function will behave as exp(kx), i.e. it will diverge as x → ∞ and hence the wave function cannot be normalized to unity, and so would not be a physically acceptable solution to the Schr¨odinger equation. The physically acceptable solutions will only arise if the recurrence relation terminates after a finite number of terms, i.e. if the coef- ficient c n vanishes for some value of n, n = m say. In that case, all further coefficients c m +1 , c m +2 , . . . will also vanish, in which case the infinite series solution will terminate to form a finite series, a polynomial in x. The wave function would then behave like a polynomial× exp(−kx), it would vanish as x → ∞, and could be normalized to unity.

Thus we are searching for conditions under which the recurrence relation c _n = 2akn − 1

n(n + 1) c _n−1 n ≥ 1

for n = m say. In this case, we find that c m = 0. Using this value of c m to calculate c m +1

readily shows that c m +1 = 0, and repeating the procedure shows that all coefficients c n , for n ≥ m will vanish.

The required condition is then that (reverting to the index n):

k n = 1 2an

which translates into an expression for the allowed energies given by E n = − mα ⁴

2~ ² . 1 n ² .

(b) The lowest energy level occurs when n = 1 in which case k = k 1 = 1/2a. If we then substitute this value of k into the recurrence relation we get

c n = n − 1

n(n + 1) c _n−1 n ≥ 1.

Starting with a non-zero value for c 0 , we find that this recurrence relation tells us that c ₁ = 0 and hence all coefficients c n for n ≥ 1 will also be zero. Thus the wave function becomes

ψ ₁ (x) = c 0 e ^−kx (x/a).

To normalize this wave function we require Z ∞

0

|ψ ₁ (x)| ² dx = 1

where we note that the integral will cut off at x = 0 as the wave function will be zero for x < 0.

Writing out the integrand gives Z ∞

0

|ψ 1 (x)| ² dx = Z ∞

0

|A| ² x ² e ^−2kx dx = 1

where c 0 /a has been combined into one unknown factor A. The integrand can be readily performed to give

|A| ² 1 4k ³ = 1 and hence

A = 2k ^3/2 . The normalized wave function is then

ψ ₁ (x) = 2k ^3/2 xe ^−kx .

(c) The average distance of the particle from the origin when in state ψ ₁ (x) is given by hxi = Z ∞

0

x|ψ ₁ (x)| ² dx = Z ∞ 0

4k ³ x ³ e ^−2kx dx = 3/2k = 6a = 3~ ² /mα ² .

2. The Friedmann equation for the ‘radius’ R(t) of the universe takes the form (with the cos- mological constant set to zero)

R ˙ ² = 2GM R − kc ²

where G = 6.67 × 10 ⁻¹¹ m ³ kg ⁻¹ s ⁻² is the gravitational constant, k = 0, ±1, c is the speed of light and M ∼ 10 ⁶⁰ kg can be interpreted as the total mass of the universe.

(a) Show that this equation can be derived by Hamiltonian methods based on the Hamil- tonian

H = ¹ ₂ p ² − GM R [Hamilton’s equations are ˙p = − ∂H

∂R and ˙ R = ∂H

∂p . You will also need ˙R ¨R = ¹ ₂ d ˙ R ² dt .]

(b) Assuming the techniques of canonical quantisation can be applied to this system, write down Schr¨odinger’s equation for the wave function Ψ(R, t).

(c) Derive an expression for the lowest energy eigenvalue for the system, and also deter- mine the most probable value for R. [Use the results from Question 1.] What does this result suggest about the presence of quantum effects on a cosmological scale?

SOLUTION (a) From the Hamiltonian

H = ¹ ₂ p ² − GM R we obtain Hamilton’s equations

R ˙ = ∂H

∂p = p

˙p = − ∂H

∂R = − GM R ² From this follows

R ¨ = ˙p = − GM R ² . Multiplying by ˙ R gives

R ¨ ˙ R = ¹ ₂ d ˙ R ²

dt = − GM R ² R ˙ which when integrated wrt time gives

1 2 R ˙ ² = −

Z GM

R ² Rdt ˙ + C

where C is a constant of integration. The integral on the right hand side can be readily

carried out: Z

GM R ² Rdt ˙ =

Z GM

R ² dR = − GM R so that overall

= 2GM

+ 2C

(b) The Schr¨odinger equation will be

− ~ 2

∂Ψ(R, t)

∂R − GM

R ² Ψ(R, t) = i~ ∂Ψ(R, t)

∂t

(c) This is, mathematically, identical to the Schr¨odinger equation of Question 1 above, with the replacements m → 1 and α ² → 2GM so that a = ~ ² /4GM and E = −~ ² k ² /2.

With k restricted to the values k = 1/2an as before.

The lowest energy level will have an ‘energy’ given by E ₁ = − ~ ²

8a ² = −2  GM

~

 ²

≈ 10 ⁸³ J.

The wave function associated with this lowest energy state will be ψ ₁ (R) = 2k ^3/2 Re ^−kR

so that the probability distribution for R will be

|ψ ₁ (R)| ² = 4k ³ R ² e ^−2kR

which will have a maximum, obtained by the usual methods of differential calculus,

at R = 1/k = 2a = ~ ² /2GM ≈ 10 ⁻¹¹⁸ m. At that size, it would be expected that

quantum e ffects associated with the large scale structure of the universe would be

immeasureably tiny.

3. Magnetic resonance imagery makes use of magnetic fields to invert the spin of atomic nuclei in organic systems, such as people. The following problem is related to this procedure (though in practice time dependent fields are used).

The Hamiltonian of a spin half system with magnetic dipole moment ˆ µµµ in a magnetic field B is given by ˆ H = −ˆµµµ · B = −µ 

B _x S ˆ _x + B y S ˆ _y + B z S ˆ _z  .

(a) Show that, in the presence of a magnetic field B = B x i + B z k that this Hamiltonian takes the form

H  ˆ



 

 

 



h +| ˆH|+i h+| ˆH|−i h−| ˆ H| +i h−| ˆH|−i



 

 

 

 =



 

 

 



− −V

−V 



 

 

 



where |±i are the eigenstates of ˆ S z with eigenvalues ± ¹ ₂ ~. Obtain expressions for  and V .

(b) If at time t the state of the system is |ψ(t)i = c ₊ (t)|+i+c − (t)|−i, show that c ₊ (t) satisfies the equation

¨c ₊ + ~ ⁻²  ² + V ²
c ₊ = 0.

(c) The system is observed at t = 0 to be in the eigenstate of ˆS z with eigenvalue + ¹ ₂ ~.

What is the probability of observing it to be in this state as a function of time t?

The matrices representing the three components of spin of a spin half particle are, in the ˆ S _z -representation

S ˆ _x = ¹ ₂ ~



 

 

 

 0 1 1 0



 

 

 



S ˆ _y = ¹ ₂ ~



 

 

 

 0 −i i 0



 

 

 



S ˆ _z = ¹ ₂ ~



 

 

 



1 0

0 −1



 

 

 

 SOLUTION

(a) Using the matrices representing the three spin operators, the Hamiltonian for the sys- tem can be written

H ˆ = −µ 

B x S ˆ x + B z S ˆ z



 − ¹ ₂ ~



 

 

 



0 µB x

µB x 0



 

 

 



− ¹ ₂ ~



 

 

 



µB z 0 0 −µB z



 

 

 

 = ¹ ₂ ~µ



 

 

 



−B z −B x

−B x B z



 

 

 

 .

Thus we can make the identifications  = ¹ ₂ ~µB z and V = ¹ ₂ ~µB x . (b) The Schr¨odinger equation in operator form will be

H|ψ(t)i ˆ = i~ d|ψ(t)i dt

so by making the substitution |ψ(t)i = c + (t)| +i + c − (t)|−i, the Schr¨odinger equation in matrix form becomes



 

 

 



− −V

−V 



 

 

 





 

 

 

 c ₊ c −



 

 

 

 = i~



 

 

 



˙c ₊

˙c −



 

 

 

 . Written out as a pair of di fferential equations, this is

−c ₊ − Vc − = i~˙c + (1)

Substituting for ˙c − in Eq. (3) from Eq. (2) gives

− ˙c ₊ − V

i~ (−Vc ₊ + c − ) = i~¨c ₊ . Finally, substituting for c − from Eq. (1) gives

− ˙c ₊ + V ² i~ c ₊ + 

i~ (i~˙c ₊ + c ₊ ) = i~¨c ₊ which can be simplified to give

¨c ₊ + ~ ⁻²  ² + V ²
c ₊ = 0.

(c) The solution to the equation for c ₊ is of the general form c ₊ (t) = A sin ωt + B cos ωt where ω = √

 ² + V ² /~. The initial state is one for which c ₊ = 1 and c 0 = 0 from which we have

c ₊ (0) = 1 = B and hence

c ₊ (t) = A sin ωt + cos ωt.

We still need to determine A. This can be done by making use of, for instance, Eq. (1).

If we substitute our solution for c ₊ into this equation we get

i~(Aω cos ωt − ω sin ωt) = −(A sin ωt + cos ωt) − Vc ⁻ (t).

Using the second initial condition, i.e. that c − = 0 we get i~Aω = −

and hence A = i/~ω. Overall then c ₊ (t) = i

~ω sin ωt + cos ωt.

The probability of observiing the system in the state | +i at time t is then |c + (t)| ² given by

|c ₊ (t)| ² =  

~ω

 2

sin ² ωt + cos ² ωt.

4. The Hamiltonian for the harmonic oscillator can be written in the form H ˆ = ~ω 

ˆa ^† ˆa + ¹ ₂ 

where the operator ˆa is given, in terms of the position and momentum operators of the oscillator by

ˆa = r mω

2~ ˆx + i ˆp mω

! .

(a) Obtain an expression for ˆx in terms of ˆa and ˆa ^† .

(b) The coherent state |αi where α is a complex number is given by

| αi = e ^{− 1 2 |α|}

²

∞

X

n =0

α ⁿ

√ n! |ni and can be shown to be an eigenstate of ˆa with eigenvalue α.

Show that the time evolved state e ^{−i ˆ Ht/~} | αi = e ^−iωt/2 |αe ^−iωt i.

(c) Calculate the expectation value of the position of the particle as a function of time if the particle is initially in the coherent state |αi in the case in which α is a real number.

(d) By reinterpreting ˆx and ˆp as operators for the electromagnetic field in a single mode cavity (i.e. ˆp is proportional to the electric field and ˆx is proportional to the vector potential) calculate the mean electric field inside the cavity, that is, calculate the ex- pectation value of

E(z) ˆ = −i s

~ω

 ₀ V (ˆa ^† − ˆa) sin kz i when the field is in the coherent state |αe ^−iωt i.

SOLUTION (a) Since

ˆa = r mω

2~ ˆx + i ˆp mω

! then

ˆa ^† = r mω

2~ ˆx − i ˆp mω

! so that

ˆa + ˆa ^† =

r 2mω

~ ˆx and hence

ˆx = r

~ 2mω

 ˆa + ˆa ^†  . (b) The time evolved coherent state is given by:

| α(t)i = e ^{−i ˆ Ht/~} | αi = e ^{−iω( ˆ N + 1 2 )t} |αi.

Inserting the series expansion of the coherent state we have

| α(t)i = e ^−iωt/2 e ^{− 1 2 |α|}

²

∞

X

n =0

α ⁿ

√ n!

e ^{−iω ˆ Nt} |ni.

Since the number states |ni are eigenstates of the number operator ˆ N with eigenvalue n, this becomes

| α(t)i = e ^−iωt/2 e ^{− 1 2 |α|}

²

∞

X

n =0

α ⁿ

√ n!

e ^−iωnt |ni

= e ^−iωt/2 e ^{− 1 2 |α|}

²

∞

X (αe ^−iωt ) ⁿ

√ |ni

(c) The expectation value of the position of the particle is hx(t)i = hαe ^−iωt | ˆx|αe ^−iωt i =

r

~

2mω hαe ^−iωt |  ˆa + ˆa ^† 

|αe ^−iωt i where it is to be noted that the factor exp(−iωt/2 cancels out.

Using the fact that the coherent states are eigenstates of the annihilation operator, we have

ˆa|αe ^−iωt i = αe ^−iωt |αe ^−iωt i.

From this it also follows that

hαe ^−iωt |ˆa ^† = α ^∗ e ^iωt hαe ^−iωt |.

Combining this in the calculation of the expectation value of position gives hx(t)i =

r

~

2mω αe ^−iωt + α ^∗ e ^iωt  . As we are given that α is real, this reduces to

hx(t)i = r 2~

mω α cos ωt.

(d) The expectation value of the cavity electric field is given by

hE(z, t)i = hαe ^−iωt | ˆ E(z)|αe ^−iωt i = −i s

~ω

 ₀ V hαe ^−iωt |  ˆa ^† − ˆa 

|αe ^−iωt i sin kz i.

Assuming α is real, as before, this becomes

hE(z, t)i = 2 s

~ω

 ₀ sin ωt sin kzi

which is the expression for a standing wave inside the cavity.