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Homework 9 Solutions
April 30, 2016
1a) The expectation value of position is given by ∫ (^) L
0
Ψ∗xΨdx =
∫ L
0
L
sin
( (^) nπx L
eiωtx
L
sin
( (^) nπx L
e−iωtdx (1)
∫ L
0
L
sin^2 (nπx/L)xdx (2)
The time dependent parts cancel out in this case, as the eigenstates are station- ary states. We can write the integral as
〈x〉 =
L
∫ L
0
sin^2 (ax)xdx where a = nπ/L (3)
and going to an online integrator we find ∫ sin^2 (ax)xdx =
x^2 4
cos(2ax) 8 a^2
x sin(2ax) 4 a
Now we must evaluate the integral at the limits 0 and L. The cosine term is equal to 1 at both the x = 0 and x = L and so cancels, while the sine term is zero at both x = 0, L. So the only remaining non-zero term is the first term evaluated at L, which gives L^2 /4. So
〈x〉 =
L
L^2
L
or midway in the box, which is what we expect for the average just by looking at the pictures.
1b) The superposition of the first two states is
Ψ = C
[√
L
sin
( (^) πx L
eiω^1 t^ +
L
sin
2 πx L
eiω^2 t
]
Multiplying this by its complex conjugate gives the probability distribution
|Ψ|^2 = C^2
L
[
sin^2
( (^) πx L
2 πx L
( (^) πx L
sin
2 πx L
ei(ω^1 −ω^2 )t^ + e−i(ω^1 −ω^2 )t
)]
we see there is a cross term with time dependence that does not go away. In fact this cross-term is just a cosine!
|Ψ|^2 = C^2
L
[
sin^2
( (^) πx L
2 πx L
( (^) πx L
sin
2 πx L
cos(∆ωt)
]
where ∆ω = ω 1 − ω2. This looks a lot like when we added intensities for EM waves – there was a cross-term. We have to choose the constant C so that the probability distribution is normalized, that is
∫ (^) L
0
|Ψ|^2 dx = 1 (9)
We already know from the eigenstates that that the integrals of the first two terms in |Ψ|^2 are each L/2, and the third term is zero. So we find
∫ (^) L
0
|Ψ|^2 dx = C^2
L
[
L
L
]
from this we see that C^2 = 1/2 or C =
1d) To get the expectation value we just need
〈x〉 =
∫ L
0
Ψ∗xΨdx =
∫ L
0
|Ψ^2 |xdx (11)
where the |Ψ^2 | is given above. We know from part 1a) that the integral of the first two terms each just give L/2 so we just need to do the last integral. Doing so we find
〈x〉 =
L
16 L
9 π^2
cos(∆ωt) (12)
3a) Looking up the spherical Laplacian, and setting all theta and phi derivatives to zero, and V = 0 the time-independent Schrodinger equation is
ℏ^2
2 m
r^2
∂r
r^2 ∂ψ ∂r^2
= Eψ (13)
making the suggested substitution u = ψ/r and carrying out the chain rule we find ∂^2 u ∂r^2
= k^2 ψ (14)
with k = (2mE/ℏ^2 )^1 /^2. The solution is just
u(r) = A sin(kr) + B cos(kr) (15)
