Assignment 8, solutions

Problem 1:

A system has a wave function y(x,y,z) = N*(x + y + z)*exp(-r2/a2) with a real.   If Lz and L2 are measured, what are the probabilities of finding 0 and 2h2?

Solution:

Concepts, principles, relations that apply to the problem:
The spherical harmonics, the postulates of quantum mechanics.
Why do they apply?
If we want to find the probability of measuring the eigenvalue a of an observable A we must express the wavefunction of the system as a linear combination of eigenfunctions of A.  Then  P(a) = Si|<ai|y>|2.
How do they apply?
Express the wavefunction in terms of spherical harmonics.

.

.

.

.

                   f(r)                                  c(q,f)

   
for properly chosen N.
c(q,f) = Sl,mdlmYlm(q,f), dlm = 0 for l ¹ 1.
Let P(l,m) denote the probability of finding the eigenvalues l(l+1)h2 and mh .
P(l,m) = |dlm|2.  P(1,1) = 1/3,  P(1,0) = 1/3,  p(1,-1) = 1/3,  P(l,m)l¹1 = 0.

Details of the calculation:
None

Problem 2:

A measurement of L2 and Lz for a free particle yields the values l = 1 and m = 1.  Later a measurement of Ly is made.
(a)  What are the possible values of Ly?
(b)  Calculate the probabilities for each of the possible values in part (a).

Solution:

Concepts, principles, relations that apply to the problem:
The orbital angular momentum operator L2, the raising and lowering operators 
L+ = Lx + iLy and L- = Lx - iLy,  L±|k,l,m> = [l(l+1) - m(m±1)]1/2h|k,l,m±1>. 
Why do they apply?
The initial state is an eigenstate of Lz.  We can find the matrix of Ly in the eigenbasis of Lz by using
<k,l,m'| L±|k,l,m> = [l(l+1) - m(m±1)]1/2hdm',m±1.  We can then find the eigenvectors and expand the initial state in terms of these eigenvectors.
How do they apply?
(a)  Let the vectors {|k,1,1>, |k,1,0>, |k,l1,-1>} be the eigenvectors of Lz with eigenvalues h, 0, -h  respectively, which span the subspace E(k,l) with l = 1.  After the measurement of L2 and Lz the particle is in state |k,1,1>.  E(k,l) is invariant under L. The matrix of Ly in the above basis is

.

It can be constructed using  Ly = (-i/2)(L+ - L-).
|L±|k,l,m> = [l(l+1) - m(m±1)]1/2h|k,l,m±1>.
<k,l,m'|L±|k,l,m> = [l(l+1) - m(m±1)]1/2hdm',m±1.
The eigenvalues of Ly are hb/21/2,

where .

We have    -b3 + 2b = 0, b(2 - b2) = 0, b = 0 or ±21/2.

For the eigenvalue 0 we have
c2 = 0, c1 = c3, |y1> = 2-1/2(|k,1,1> + |k,1,-1>).
For the eigenvalue h we have
c1 = -i2-1/2c2, c3 = i2-1/2c2, |y2> = (1/2)(|k,1,1> + i21/2|k,1,0> - |k,1,-1>).
For the eigenvalue -h we have
c1 = i2-1/2c2, c3 = -i2-1/2c2, |y3> = (1/2)(|k,1,1> - i21/2|k,1,0> - |k,1,-1>).

(b)  P(Ly = 0) = |<y1|k,1,1>|2 = 1/2,
P(Ly = h) = |<y2|k,1,1>|2 = 1/4,
P(Ly = -h) = |<y3|k,1,1>|2 = 1/4.

Note: k may denote the energy of the free particle.  If the particle does not have a well defined energy then the state of the particle after the measurement of L2 and Lz is
Skak|k,1,1>  with Sk|ak|2 = 1.

Details of the calculation:
None

Problem 3:

The quantum numbers l1 and l2 of the orbital momenta of particle A and particle B are 1 and 2, respectively.  Find the 15 possible ‘kets’ in the coupled representation (notation |l1,l2;L,ML>) where L represents the quantum number of the total orbital momentum.

Solution:
Concepts, principles, relations that apply to the problem:
Addition of angular momentum
Why do they apply?
We are supposed to add two angular momenta l1 and l2  and find the possible values for L and M.

How do they apply?
 l1 = 1 and l2 =2.  The possible values for L are L = 3, 2, 1.
L = 1:  Possible values M are 1, 0, -1.
Corresponding kets: |1,2; 1,1>, |1,2; 1,0>, |1,2; 1,-1>.
(notation: |l1,l2;L,m>)
L = 2:  Possible values M are 2, 1, 0, -1, -2.
Corresponding kets: |1,2; 2,2>, |1,2; 2,1>, |1,2; 2,0>, |1,2; 2,-1>, |1,2; 2,-2>.
L = 3:  Possible values M are 3, 2, 1, 0, -1, -2 -3.
Corresponding kets: |1,2; 3,3>, |1,2; 3,2>, |1,2; 3,1>, |1,2; 3,0>, |1,2; 3,-1>, 
|1,2; 3,-2>, |1,2; 3,-3>.
We have listed 15 kets. 

Details of the calculation:
None

Problem 4:

Consider two particles with angular momenta:
J = J1 + J2, Jx = J1x + J2x,  Jy = J1y + J2y,  Jz = J1z + J2z.
J1 and J2 are the angular momentum operators of particle 1 and 2.
Show that the commutators [J2,J12] and [J1z,J2] are zero and nonzero, respectively.  What does it mean in terms of measurements and Heisenberg’s uncertainty principle?

Solution:

Concepts, principles, relations that apply to the problem:
Addition of angular momenta, commutator algebra
Why do they apply?
We have to add two angular momenta and evaluate two commutators.
How do they apply?
J1 and J2 are angular momentum operators.  For any angular momentum operator we have
[Ji,Jj] = eijkJk and [J2,Ji] = 0.
The operator J2 = (J1 + J2)2 may be expressed as J2 = J12 + J22 + 2J1×J2, since [J1,J2] = 0.  J1 and J2 operate in different spaces.
We may write J1×J2 = J1xJ2x + J1yJ2y +J1zJ2z = (1/2)(J1+J2- + J1-J2+ ) +J1zJ2z.
J2 commutes with J12 and J22.
[J2,J12] = [J12 + J22 + 2J1×J2,J12] = 0
since [J12,J1i] = 0.
[J2,J1z] = [J12 + J22 + 2J1×J2,J1z] = 2[J1×J2,J1z] = 2[J1xJ2x + J1yJ2y,J1z]
= 2(-J1yJ2x + J1xJ2y) ¹ 0.
J2 and J12 are commuting observables and can be measured simultaneously, i.e. the measurement of one does not cause loss of information obtained in the measurement of the other.  After a measurement of J2 and J12 we know their exact values.
 J2 and J1z are non-commuting observables.  They cannot be measured simultaneously, i.e. the measurement of one does cause loss of information obtained in the measurement of the other.  The generalized uncertainty relation states that in any state of the system   DJ2DJ1z2 ³ (1/4)|<i[J2,J1z]>|.
Details of the calculation:
None

Problem 5:

For any two quantum-mechanical operators A and B, the uncertainty principle says that
<(DA)2><(DB)2>  ³ (1/4)|<[A,B]>|2.  Consider a spin ½ particle.  Show that for the spin operators Sx and Sy the eigenstate |+> of the Sz operator is a minimum uncertainty state.

Solution:

Concepts, principles, relations that apply to the problem:
The two dimensional state space of a spin 1/2 particle
Mean value <A> = <y|A|y> and root mean square deviation DA º (<(A - <A>2)>)1/2 = (<A2> - <A>2))1/2.
Why do they apply?
We are asked to evaluate <(DSx)2 ><(DSy)2> and show that it is equal to (1/4)|<[Sx,Sy]>|2
How do they apply?
The state space corresponding to the observable Sz of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sz by |+> and |->.  The matrices of Sz,  Sx, and Sy in the eigenbasis of Sz , {|+>, |->} ,

, , .

The matrices of  Sz2 , Sx2, and Sy2 are


[Sx,Sy] = ihSz.  These are the commutation relations for angular momentum.
The column matrices representing |+> and |-> are  and   respectively.
For the eigenstate |+> we have:

<Sz> = h/2,

<Sx> = 0,  <Sy> = 0,  <Sz2> = h2/4,  <Sx2> = h2/4,  <Sy2> = h2/4.

Therefore
(DSx)2 = <Sx2> - <Sx>2 = h2/4,  (DSy)2 = <Sy2> - <Sy>2 = h2/4.
<(DSx)2 ><(DSy)2> = h4/16.
<[Sx,Sy]> = < ihSz > = ih2/2
<(DSx)2 ><(DSy)2> = (1/4)|<[Sx,Sy]>|2

The equality holds, |+> is a minimum uncertainty state.

Details of the calculation:
None

Problem 6: 

Let Si, i = 1, 2 denote the spin vectors of two spin-1/2 particles. The interaction is given by

H = V0 (S1 · S2 − 3 S1zS2z).

Find the energy eigenstates and eigenvalues.

Solution:

Concepts, principles, relations that apply to the problem:
The state space of two spin-1/2 particles
Why do they apply?
We have to diagonalize the matrix of H in the state space of two spin 1/2 particles
How do they apply?
The eigenstates of S1 · S2 are the singlet and the triplet states.
{|S,Sz> = |0,0>, |1,1>, |1,0>, |1,-1>}
S1 · S2|S,Sz> = ½(S2 – S12 – S22) |S,Sz> = ½(S(S+1) – 3/2)h2.
The singlet and the triplet states are also eigenfunctions of S1zS2z.
S1zS2z |1,1> = S1zS2z |++> = (h2/4)|1,1>.
S1zS2z |1,0> = (1/2)1/2 S1zS2z (|+-> + |-+>) = -(h2/4)|1,0>.
S1zS2z |1,-1> = S1zS2z |--> = (h2/4)|1,-1>
S1zS2z |0,0> = (1/2)1/2 S1zS2z (|+-> - |-+>) = -(h2/4)|0,0>

The eigenstates of H are the the singlet and the triplet states.  In that bais H is diagonal.
H |1,1> = V0(1-3) (h2/4)|1,1> = -2V0(h2/4)|1,1>
H |1,0> = V0(1+3) (h2/4)|1,0> = 4V0(h2/4)|1,0>
H |1,-1> = V0(1-3) (h2/4|1,-1> = -2V0(h2/4)|1,1>
H |0,0> = V0(-3+3) (h2/4|0,0> = 0

Details of the calculation:
None

Problem 7: 

Some organic molecules have a triplet (S = 1) excited state that is located at an energy D above the singlet (S = 0) ground state.  Consider an ensemble of N such molecules where N is of the order of Avogadro’s number
(a)  Find the average magnetic moment <m> per molecule in the presence of a magnetic field B.  Assume Boltzmann statistics.  You may also assume that D is large compared to the field-induced level splittings.
(b) Show that the magnetic susceptibility c = N d<m>/dB is approximately independent of D when kBT >> D.

Solution:

Concepts, principles, relations that apply to the problem:
Angular momentum and magnetic dipole moment, the energy of a magnetic dipole in an external magnetic field, Boltzmann statistics
Why do they apply?
The magnetic moment of a molecule is proportional to its angular momentum. Molecules with S = 0 have no magnetic moment but molecules with S = 1 have a magnetic moment m proportional to S.  The energy of a magnetic moment in an external magnetic field is quantized.  The three possible different projections of the magnetic moment onto the direction of the magnetic field have different energies.  The relative probabilities for the different projections are found using Boltzmann statistics.  Once we have the relative probabilities, we can find the average magnetic moment.
How do they apply?
(a)  The ratio Nexcited/Nground of the number of molecules in the excited state to the number of atoms in the ground state is
Nexcited/Nground = (gexcited/gground)exp(-D/(kBT))
where g is the degeneracy. 
Here Nexcited/Nground = 3 exp(-D/(kBT))
with Nexcited + Nground = N we have Nexcited = N/(1 + exp(D/(kBT)))/3).

The magnetic moment m of a molecule is proportional to S, m = -(gmB/h)S.
The energy of a magnetic moment in a magnetic field is given by E = -m×B.
Let the magnetic field B point into the z-direction. Then
E = -mzB = g(mBh)BSz = gmBBm, with m = -1, 0, 1.

Let N1 denote the number of excited molecules with m = 1, N0 the number of excited molecules with m = 0, and N-1 denote the number of excited molecules with m = -1.  Then, using Boltzmann statistics,
N-1/N0 = exp(gmBB/(kBT)), N-1/N1 = exp(2gmBB/(kBT)), N1 + N0 + N-1 = Nexcited.
N-1(1 + exp(-gmBB/(kBT)) + exp(-2gmBB/(kBT))) = Nexcited.
N-1 = Nexcited/(1 + exp(-gmBB/(kBT)) + exp(-2gmBB/(kBT)))
N+1 = Nexcitedexp(-2gmBB/(kBT))/(1 + exp(-gmBB/(kBT)) + exp(-2gmBB/(kBT)))
<mz> = (N-1 gmB - N1 gmB)/N
= gmB(1 + (1/3)exp(D/(kBT))-1*
(1 - exp(-2gmBB/(kBT)))/(1 + exp(-gmBB/(kBT)) + exp(-2gmBB/(kBT)))
= average magnetic moment <mz > per molecule in the presence of a magnetic field B.

Details of the calculation:
(b)  Let kBT >> D >> 2gmBB.
Then <mz> ~ (1/2)B(gmB)2/(kBT).
c = N d<m z >/dB = (1/2)N(gmB)2/(kBT).