Homework 4, Quantum Mechanics 501, Rutgers: October 28, 2016
Homework 4, Quantum Mechanics 501, Rutgers: October 28, 2016
Homework 4, Quantum Mechanics 501, Rutgers: October 28, 2016
1
2) A particle of mass m is in a one-dimensional potential of form V (x) = 1/2mω 2 x2 +mgx
with some real number g. (Think of this as an oscillator potential plus a constant force
mg in −x direction acting on the particle).
Without doing much heavy math, can you write down the lowest energy eigenstate
of this potential? (Think about the classical analog a weight hanging on a vertical
spring. How does gravity affect the equations and solution for the harmonic spring
potential energy?)
Ans.: The Hamiltonian can be cast into the following form
~2 d2 1 2 g 2 1 g2
H=− + mω (x + ) − m 2 (8)
2m dx2 2 ω2 2 ω
If we introduce a new variable xnew = x + ωg2 , the above Hamiltonian has a canonical
form of H.O. apart from the energy shift. The solutions are then solutions of H.O. The
ground state is
mω 1/4 mω 2 )2
ψ0 (x) = e− 2~ (x+g/ω (9)
π~
2
with energy E0 = ~ω/2 − 12 m ωg 2 .
What is the probability that a particle starting out in the ground state of the harmonic
oscillator potential only (first part of V (x)) ends up in the new ground state once the
force is ”switched on”?
Ans.: If we denote the unshifted oscilator wave function by ψ00 (x), then the probability
is
which is
mg 2
P = e− 2~ω3
∆E
and can also be written as P = e− ~ω where ∆E is the change of the energy due to
additional term in the Hamiltonian.
3) Find the eigenvalues and eigenstates of the one-dimensional Hamiltonian with potential
1
2
mω 2 x2 x < 0
V (x) = (11)
∞ x≥0
2
the corresponding Hermite polynomials have only odd powers of x. In other words,
the eigenstates are |ni, n = 1, 3, 5, ... and the eigenvalues are E = ~ω(n + 1/2).
Note that one also has to renormalize the wave functions (with an extra factor 2) since
the integral in the spatial variable x now only goes from −∞...0.
Axe−ax x ≥ 0
ψ(x) = (12)
0 x<0
~2 2 a
− (a − 2 )ψ(x) + V (x)ψ(x) = Eψ(x) (14)
2m x
2
For the equation to be satisfied, we need V (x) = − ~ma x1 + C. Without loss of
2 2
generality, we can set C = 0, which gives E = − ~2ma . Note that this holds only
for x > 0, as the wave function vanishes at x = 0 (with finite derivative) and
potential is therefore infinite at x < 0.
b) Find the potential energy expectation value hV i for this state
Ans.:
Z ∞ Z ∞ 2
2 2 2 −2ax ~a ~2 A2
hV i = V (x)ψ(x) = A dxx e − =− (15)
0 0 mx m 4a
and ∞ ∞
A2
Z Z
1= 2
dxψ(x) = A 2
dxx2 e−2ax = (16)
0 0 4a3
2 2
hence hV i = − ~ ma .
c) Find the expectation value of the kinetic energy for this state.
2 2
Ans.: hKi = E − hV i = ~2ma
5 The eigenstates, which are accesible to a single electron, have energies ε0 , ε1 and
ε2 and their states are |0i, |1i and |2i. When two electrons are introduced in such
system, what are possible wave-functions of the system of two electrons, if we neglect
interaction between the two electrons?
3
a) How many possible states can you write down, which have correct statistics?
Write them down.
Ans.:
1
|ψ0 i = √ (|0i ⊗ |1i − |1i ⊗ |0i) (17)
2
1
|ψ1 i = √ (|0i ⊗ |2i − |2i ⊗ |0i) (18)
2
1
|ψ2 i = √ (|1i ⊗ |2i − |2i ⊗ |1i) (19)
2