PHYSICS 621 - Fall Semester 2013 - ODU
Graduate Quantum Mechanics - Problem Set 5 - Solution
Problem 1)
An operator A, corresponding to a physical observable, has two normalized eigenstates φ1 and φ2
with non-degenerate eigenvalues a1 and a2, respectively. A second operator B, corresponding to a
different physical observable, has normalized eigenstates χ1 and χ 2 , with eigenvalues b1 and b2,
respectively. The two sets of eigenstates are related by:
φ1 =
2 χ1 + 3 χ 2
13
and
φ2 =
3 χ1 − 2 χ 2
13
.
The physical observable corresponding to A is measured and the value a1 is obtained. Immediately
afterwards, the physical observable corresponding to B is measured, and again immediately after that,
the one corresponding to A is remeasured. What is the probability of obtaining a1 a second time?
Answ.:
After the first measurement of A, we know that the state must have “collapsed” into an eigenstate of
A with eigenvalue a1, i.e., φ1 =
2 χ1 + 3 χ 2
. If we now measure B, the results can be either b1 or b2,
13
with the corresponding projection operators P1 = χ1 χ1 or P2 = χ 2 χ 2 . The corresponding
2
4
9
and Prob(b2 ) = P2 φ1 = . Accordingly, the wave function
13
13
after this measurement will be either χ1 (with probability 4/13) or φ2 (with probability 9/13). Either
probabilities are Prob(b1 ) = P1 φ1
2
=
way, the probability to measure A again and find the result a1 is given by φ1 φ1 χ1
case and φ1 φ1 χ 2
2
=
2
=
4
in the first
13
9
in the second case. So the total probability for measuring a1 a second time
13
4 4 9 9 97
+
=
, slightly larger than ½. This shows that measuring B destroys the eigenstate
13 13 13 13 169
created before and therefore makes the next measurement of A unpredictable – the outcome could be
is
a2 with a probability of 72/169.
Problem 2)
The ammonia molecule NH3 has two different possible configurations: One (which we will call 1 )
where the nitrogen atom is located above the plane spanned by the three H atoms, and the other one
(which we will call 2 ) where it is below. (These two states span the Hilbert space in our simple
example). In both states, the expectation value of the energy n H n is the same, E (n = 1,2). On the
other hand, the two states are not eigenstates of the Hamiltonian; in fact, we have
2 H 1 = 1 H 2 = −V (where V is some positive value).
PHYSICS 621 - Fall Semester 2013 - ODU
[Comment: V tells us how strongly “coupled” the two states are – as you can see by looking at the
Schrödinger equation, if you start out in either basis state, over time you will “tunnel” into the other one
– and back…]
1) Write down the Hamiltonian in matrix form
2) Find both eigenvalues and (normalized) eigenstates of the Hamiltonian
3) The latter are the stable configurations of the system. Interpret their physical meaning. How can
they be distinguished, and what do you think will be the ground state of ammonia?
4) Comment on the behavior of these two eigenstates under the Parity operation. Are they
eigenstates? With what eigenvalues (called “parity”)?
Answ.:
" E
1) H = $
# −V
−V %
'
E &
2
2) The characteristic equation is ( E − λ ) = V 2 ; λ1,2 = E ±V . The corresponding eigenstates are
ϕ1 =
1 " 1 %
1 " 1 %
$
'; ϕ 2 =
$ '
2 # −1 &
2# 1 &
3) These two states are linear superpositions of the two eigenstates to the operator measuring the
position of the N atom. In fact, it is clear that for both of them, the probability to find the N either above
or below the H3 plane is 50% - it is in “both positions at once”. Obviously, they are distinguished by
their energy eigenvalues – what you would measure if you measured the energy of the system. The 2nd
eigenstate, with energy E-V, is clearly lower in energy and therefore should be the ground state of the
system.
4) Under the parity operation, all coordinates change sign, interchanging the meaning of “up” and
“down”. This means that 1 → 2 and 2 → 1 under this operation. Applying this to the two energy
eigenstates, we find that the ground state is invariant (it is an eigenstate of the parity operator with
eigenvalue +1 – “even parity”) and the 1st eigenstate acquires a minus sign (it is also an eigenstate
under parity, but with eigenvalue -1 – “odd parity”).
Problem 3)
Consider the following three operators (representing physical observables) on the 2-dimensional Hilbert
space of complex-valued column vectors (we will learn shortly which physical system they represent):
! 0 1 $
! 0 −i $
! 1 0 $
Lx = #
& ; Ly = #
& ; Lz = #
&
" 1 0 %
" i 0 %
" 0 −1 %
1) What are the possible values one can obtain when Lz is measured?
2) Assume Lz is measured and one finds the value -1. Afterwards, what are Lx , Lx2 and ΔLx ?
3) Find the normalized eigenstates and eigenvalues of Lx.
PHYSICS 621 - Fall Semester 2013 - ODU
4) What are the possible values one could measure for Lx and what are their probabilities if you
start out with the same state as in 2)? Show that you get the same answer as in 2) if you
calculate the “classical (probabilistic) expectation value” and “RMS” for the probability
distribution of possible measurement results.
5) Calculate the commutators between any two of the three operators above (all 3). Do you see a
pattern? Is it possible to find simultaneous eigenstates for any two of the operators? Therefore,
is it possible to prepare a state of the system with well-defined values for all three?
Answ.:
! 1 $
! 0 $
1) Obviously, the basis vectors #
& and #
& of this Hilbert space are eigenvectors for Lz with
" 0 %
" 1 %
eigenvalues +1 and -1. These are then the only two possible values a measurement of Lz can yield.
! 0 1 $! 0 $
! 1 0 $! 0 $
2) Lx = 0 1 #
&#
&#
& = 0; Lx2 = 0 1 #
& = 1 and therefore
" 1 0 %" 1 %
" 0 1 %" 1 %
(
ΔLx =
)
Lx2 − Lx
(
2
)
=1.
3) It’s easy to see that the eigenstates of Lx are
1 ! 1 $
1 " 1 %
# & with eigenvalue +1 and
$
' with
2" 1 %
2 # −1 &
eigenvalue -1.
4) Clearly, one can only measure +1 or -1. The probability to measure +1 if you start out with the 2nd
2
! 0 $
1
1
1 1 #
eigenvector of Lz is P1 =
& = and the same for the probability to measure -1. The
2
2
" 1 %
(
)
expectation values is therefore Lx = 12 (+1) + 12 (−1) and the RMS is given by
(ΔLx )
2
= 12 (+1− 0)2 + 12 (−1− 0)2 = 1 , just as we found before.
% 0 1 (% 0 −i ( % 0 −i (% 0 1 ( % i 0 (
5) !" Lx , Ly #$ = '
*'
*−'
*'
* = 2'
* = 2iLz . Similarly, !" Ly , Lz #$ = 2iLx
& 1 0 )& i 0 ) & i 0 )& 1 0 ) & 0 −i )
and [ Lz , Lx ] = 2iLy (this is the pattern you were supposed to see).
It is clearly not possible to find simultaneous eigenvectors to any two of them, since (as demonstrated
above), the product of any two (different ones) of them changes sign if you interchange them. So, if for
instance you found an eigenvector with Lx ψ = lx ψ and Ly ψ = ly ψ , then
Lx Ly ψ = lx ly ψ = −Ly Lx ψ = −lylx ψ which clearly is only possible if ψ = 0 , q.e.d.