A system is defined by the wavefunction: −L L 2πx for ≤x≤ ψ(x) = A cos L 4 4 (a) Determine the normalization constant A. (b) What is the probability that the particle will be found between x = 0 L and x = ? 8
1.1 1.1.1
Solution Part (a) Z
+∞
ψ ∗ (x)ψ(x)dx
1= −∞
Z
−L 4
1=
L 4
Z 0dx +
A2 cos2
−L 4
−∞
1 = A2
L 4
Z
cos2
−L 4
1 = A2 A2 =
1.2
2πx L
2πx L
Z
L )= 8
Z
L 8
dx
L 4
4 L
ψ(x)∗ ψ(x)dx
0 L 8
2πx 4 cos2 dx L L 0 L 4 (2 + π)L P (0, ) = 8 L 16π
P (0,
L )= 8
P (0,
Z
L 2+π )= ≈ 0.41 8 4π
1
0dx+ L 4
Part (b) P (0,
∞
dx +
2
Problem 6.5
A particle with zero energy has a wavefunction x2
ψ(x) = Axe− L2 Find and sketch V(x).
2.1
Solution
Start with the Schrodinger Equation. −~2 d2 ψ(x) + V (x)ψ(x) = Eψ(x) 2m dx2 Notice that the particle has zero energy. ~2 d2 ψ(x) = V (x)ψ(x) (1) 2m dx2 Now to figure out the second derivative of the wavefunction. Make sure to use the chain rule! x2 dψ(x) −2x − x22 = A(1)e− L2 + Ax e L dx L2 x2 dψ(x) 2x2 = A 1 − 2 e− L2 dx L −4x − x22 2x2 −2x − x22 d2 ψ(x) L = A e + A 1 − e L dx2 L2 L2 L2 3 x2 d2 ψ(x) 4x 6x = A − e− L2 2 4 2 dx L L Now we can plug this and the original wavefunction back into Equation 1. 3 x2 x2 ~2 4x 6x A − e− L2 = V (x)Axe− L2 4 2 2m L L Now we can cancel a bunch of terms out to get ~2 4x2 6 V (x) = − 2m L4 L2 This is an oscillator’s potential.
2
3
Problem 6.6
The wavefunction of a particle is given as ψ(x) = A cos(kx) + B sin(kx) Show that it is a solution is the Schrodinger Equation with V(x)= 0 and find the energy.
3.1
Solution
Start with the Schrodinger Equation. −~2 d2 ψ(x) + V (x)ψ(x) = Eψ(x) 2m dx2 There’s no potential so −~2 d2 ψ(x) = Eψ(x) 2m dx2 Calculate the second derivative of the wavefunction
(2)
dψ(x) = −Ak sin(kx) + Bk cos(kx) dx d2 ψ(x) = −k 2 (A cos(kx) + B sin(kx)) dx2 Plug this back into Equation 2 ~2 k 2 [A cos(kx) + B sin(kx)] = E [A cos(kx) + B sin(kx)] 2m So this is a solution to the Schrodinger Equation provided we have ~2 k 2 p2 = 2m 2m Which is exactly what we would expect for a free particle. E=
4
Problem 6.11
L −L A particle is confined to a box from ≤x≤ 2 2 What are the wavefunctions and probability densities for the n = 1, 2, and 3 states? Sketch them.
3
4.1
Solution
The general solution for a particle in a box is still: ψ(x) = A sin(kx) + B cos(kx) But now the boundary conditions are L L ψ(− ) = 0; ψ( ) = 0 2 2 The first of these tells us 0 = −A sin(kL/2) + B cos(kL/2) The second of these tells us 0 = A sin(kL/2) + B cos(kL/2) The difference in sign is due to cos(x) being symmetric and sin(x) being antisymmetric. Equating the two gives A sin(kL/2) = −A sin(kL/2) But this only makes sense if A = 0, so the wavefunction is ψ(x) = B cos(kx) Returning to the boundary conditions we have 0 = B cos(kL/2) This means that 1 kL = (n + )π 2 2 So k=
π (2n + 1) L
This makes the wavefunction ψn (x) = A cos
xπ L
(2n + 1)
Normalizing will still yield A=
p
2/L
So ψn (x) =
p
2/L cos
4
xπ L
(2n + 1)
This means the probability density will be Pn (x) =
5
xπ 2 cos2 (2n + 1) L L
Problem 6.24
A solution of Schrodinger’s equation for an oscillator is 2
ψ(x) = Cxe−αx
(a) Express α in terms of m and ω. What is the energy of this state? (b) Normalize it.
5.1 5.1.1
Solution Part (a)
As always, start with the Schrodinger equation: −~2 d2 ψ(x) 1 2 + kx ψ(x) = Eψ(x) 2m dx2 2 Notice that this wavefunction is essentially the same as the one from problem 6.5 except we replaced L12 with α. This means it’s second derivative will be: 2 d2 ψ(x) = C 4α2 x3 − 6αx e−αx 2 dx If we plug this into the Schrodinger Equation we get 2 2 2 −~2 1 C 4α2 x3 − 6αx e−αx + kx2 Cxe−αx = ECxe−αx 2m 2 This simplifies to 1 −~2 4α2 x2 − 6α + kx2 = E 2m 2 We can see that the x2 terms must cancel each other out, giving ~2 1 4α2 x2 = kx2 2m 2 Solving for alpha gives r α=
km 4~2
Remember that r ω= 5
k m
(3)
Plugging this in gives mω 2~ The other consequence of Equation 3 is α=
−~2 (−6α) = E 2m This simplifies to 3~2 α m
E= Plugging in alpha gives
3 ~ω 2 Notice that this is the energy of a quantum harmonic oscillator in the n=1 state! E=
5.1.2
Part (b)
We have the standard rule for normalization Z +∞ 1= ψ ∗ (x)ψ(x)dx −∞
Z
+∞
2
C 2 x2 e−2αx dx
1= −∞
r 1 π −3/2 α 1=C 4 2 r r 2 mω 3/2 2 3/2 2 mω 3/2 C2 = 4 =√ α =4 π π 2~ ~ π 2
6
Problem 6.29
An electron is bound to x > 0 with the wavefunction ψ(x) = Ce−x 1 − e−x (a) Normalize the wavefunction. (b) What is the most probable value of x? (c) What is the expectation value of x?
6
6.1 6.1.1
Solution Part (a)
Again start with Z
+∞
ψ ∗ (x)ψ(x)dx
1= −∞
Z 1=
+∞
C 2 e−2x 1 − e−x
2
0
1 = C2
1 12
C 2 = 12
6.2
Part (b)
We find the most probable value by finding where the derivative of the probability is 0: 2 d 12 e−2x 1 − e−x dx We can throw away the normalization constant, but must make sure to use the chain rule 0=
0 = (−2)e−2x (1 − e−x )2 + e−2x (2)(1 − e−x )1 (e−x ) Canceling terms and factoring gives 0 = (2e−x − 1)(1 − e−x ) The term in the right parentheses is only 0 when x = 0, but that is outside the allowed range. This means 0 = 2e−x − 1 Which has the solution x = ln 2 ≈ 0.69
7
6.3
Part (c)
To find the expectation value, we use Z +∞ ˆ ˆ hAi = ψ ∗ (x)Aψ(x)dx −∞
In our case this is: Z
+∞
hxi = 12
xe−2x 1 − e−x
2
dx
0
This works out to be 13 ≈ 1.1 12 This means that the most probable value is NOT the expectation value. This is because the expectation value is just the average value. The most probable value is just the highest point on the plot of P (x) vs x. If you have an asymmetric curve, these two are not necessarily the same. hxi =
7
Problem 6.35
Which of the following are eigenfunctions of the momentum operator pˆ = i~
d dx
(a) ψ(x) = A sin(kx) (b) ψ(x) = A sin(kx) − A cos(kx) (c) ψ(x) = A cos(kx) + iA sin(kx) (d) ψ(x) = Aeik(x−a)
7.1
Solution
The generic eigenvalue problem for an operator Aˆ is ˆ = cψ Aψ ˆ In our case Aˆ = pˆ = i~ d Where c is a constant and an eigenvalue of A. dx
7.2
Part (a)
d (A sin(kx)) = i~Ak cos kx 6= c(A sin(kx)) dx So it is NOT an eigenfunction. i~
8
7.3 i~
d (A sin(kx)−A cos(kx)) = i~k(A sin(kx)+A cos(kx)) 6= c(A sin(kx)−A cos(kx)) dx So it is NOT an eigenfunction.
7.4 i~
Part (b)
Part (c)
d (A cos(kx)+iA sin(kx)) = i~(−Ak sin(kx)+ikA cos(kx)) = −~k(A cos(kx)+iA sin(kx)) dx So this IS an eigenvector with eigenvalue c = −~k = −p
7.5
Part (d)
d ik(x−a) Ae = i~(ik)Aeik(x−a) = −~k Aeik(x−a) dx So this IS an eigenvector with eigenvalue i~
c = −~k = −p
8
Additional Problem
8.1
Solution
The wavefunction is: 1/2 2 2 mω 3/2 xe−αx ψ(x) = √ ~ π
With α= 8.1.1
mω 2~
Part (a) Z
+∞
ψ ∗ (x)ˆ xψ(x)dx
hxi = −∞
Z 2 mω 3/2 ∞ 3 −2αx2 x e dx hxi = √ ~ π −∞ hxi = 0 2
Since x3 is odd (antisymmetric) and e−2αx is even (symmetric), their product is odd. Integrating an odd function over an eve interval will give zero 9
8.1.2
Part (b) Z
2
+∞
ψ ∗ (x)ˆ x2 ψ(x)dx
hx i = −∞
Z 2 mω 3/2 ∞ 4 −2αx2 2 x e dx hx i = √ ~ π −∞ r 2 mω 3/2 3 π −5/2 2 hx i = √ α ~ 16 2 π hx2 i = 8.1.3
3 ~ 2 mω
Part (c) ∆x =
p
hx2 i − hxi2 = r
∆x = 8.1.4
p hx2 i
3 ~ 2 mω
Part(d) Z
+∞
ψ ∗ (x)ˆ pψ(x)dx
hpi = −∞
Z 2 2 mω 3/2 ∞ −αx2 d hpi = √ xe−αx dx xe −i~ ~ dx π −∞ Z ∞ 3/2 2 2 2 mω hpi = √ xe−αx (−i~) 1 − 2αx2 e−αx dx ~ π −∞ These will both lead to odd functions being integrated over an even interval, so we know it will go to zero. hpi = 0 8.1.5
Part (e) hp2 i =
Z
+∞
ψ ∗ (x)ˆ p2 ψ(x)dx
−∞
2 Z 2 2 mω 3/2 ∞ −αx2 d hp i = √ xe −i~ xe−αx dx ~ dx π −∞ 2
We already calculated the second derivative of this wavefunction, so we can just plug it in Z 2 2~2 mω 3/2 ∞ −αx2 hp2 i = − √ xe 4α2 x3 − 6αx e−αx dx ~ π −∞ 10
Z ∞ Z ∞ 2~2 mω 3/2 2 4 −2αx2 2 −2αx2 x e dx + (−6α) x e dx 4α hp i = − √ ~ π −∞ −∞ 2
r r 2~2 mω 3/2 3 π −5/2 1 π −3/2 2 hp i = − √ 4α α − 6α α ~ 16 2 4 2 π 2
r 3 6 2~2 mω 3/2 −1/2 π hp2 i = − √ α − ~ 2 4 4 π hp2 i = 8.1.6
3 mω~ 2
Part (f ) ∆p =
p p hp2 i − hpi2 = hp2 i r ∆p =
8.1.7
3 mω~ 2
Part (g) r ∆x∆p =
3 ~ 2 mω
∆x∆p =
r
3 mω~ 2
3 ~ 2
~ This obeys the Heisenberg uncertainty principle, which says that ∆x∆p ≥ . 2 If you repeat this for an n = 0 quantum harmonic ocillator, you will actually ~ get the minimum value of ∆x∆p = 2
