Quantum Mechanics - Homework Assignment 6 Alejandro G´ omez omez Espinosa∗ October 23, 2012
Shank Shankar Ex. 7.3.6 Consider a particle in a potential V (x) = 12 mw 2 x2 , x > 0 and V (x) = , x 0. What are the boundary boundary condition conditionss on the wave functions now?. Find the eigenvalues and eigenfunctions.
∞ ≤
This problem is the regular simple harmonic oscillator (SHO) with a boundary condition ψ (0) = 0. Then, from Shankar Shankar (7.3.22), we know that the eigenvalues eigenvalues for the SHO are: ψn (x) =
mw π 22n (n!)2
mwx2
− exp
mw
x (1) 2 where H n are the Hermite polynomials. Using the boundary condition, it is clear that the only way that ψ (0) vanishes is when the even terms of the Hermite polynomials vanish, therefore the solution should only include the odd terms. H n
In addition, for the eigenfunctions of the complete SHO, the have the normalization condition: ∞ ∞ ψn∗ ψn dx = 2 ψn∗ ψn dx = 1
0
−∞
Rename the eigenfunctions of this problem φn , this solutions must be normalized in the range of x given by the problem:
∞ ∗
φm φm dx = 1
0
where m are odd integers, according to the description above. Then,
∞ ∗
φm φm dx = 2
0
∞ 0
ψn∗ ψn dx
⇒ φm =
√ ψ 2
n
where n and m are dummy dummy indices. Thus, Thus, using this result with m odd, i.e. m = 2n + 1, we found: φn =
√
2ψn (x) =
√ 2mw π 22n+1 ((2n + 1)!)2
− mwx2
exp
2
H (2n (2n+1)
mw
x
(2)
with n = 0, 1, 2, 3,.... Finally, the eigenvalues of (1 (1) are E = (n + 12 ) w for all n. In our case, with n odd, we can rewrite this values as: E n =
1 2n + 1 + 2
again for n = 0, 1, 2, 3,.... ∗
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1
w
=
3 2n + 2
w
2) Using ladder-operator algebra, show that T = V in any eigenstate n of the harmonic oscillator (here T = 21m P 2 is the kinetic energy and V = 12 mw 2 x2 is the potential energy).
Using the kinetic energy operator T =
P 2
= = = = = = =
Then,
P 2 2m ,
and the definition of P , calculate P 2 :
n|P 2|n n| − mw2 (a† − a)2 |n
−
mw
2
† † † † | − − | † † | | | † | | † |
mw
2 2
mw
2
n
a a
n
n a a n + n aa n
(n n n + (n + 1) n n )
|
|
(2n + 1)
2
1 2m
mw
(2n + 1)
2
Now, using Now, using the potent potential ial energy energy operator operator V = 2 calculate X :
= = = = = = =
V =
=
w
(2n + 1)
4
mw2 X 2 2
(3)
and the definition for X ,
n|X 2|n n| 2mw (a† + a)2 |n
2mw
2mw
2mw
2mw
2mw
† † † † | | † † | | | † | | †| n
a a + a a + aa + aa
n
a a + aa
n
n
n a a n + n aa n
(n n n + (n + 1) n n )
|
|
(2n + 1)
mw2 X 2
= mw2
2
n
mw
2m
Then,
aa + aa
a a
a a + aa
n
mw
2 T = P =
X 2
|
2
2mw
From (3 (3) and (4 (4), we found that T = V .
2
(2n + 1)
=
w
4
(2n + 1)
(4)
3) Using ladder-operator algebra, compute 2 P 3 1 (the matrix element of P 3 between the first and second second excited oscillator eigenstates.)
| |
3
2|P |1
= = = = = =
3/2
| †− | | † † † − † † − † † † − † † − | † † †† †† | − | † †| | † †| | † † | √ √ √ − | √ − imw imw
3/2
2
imw
2
imw
2
a a
a aa + aa a + a a a
1
( 2 a aa 1 + 2 aa a 1 + 2 a a a 1 )
3/2
2 2+3 2+
2
imw
aa a + aaa
a aa
3/2
2
imw
a a a
a + a† aa − aa† a + aaa |1
3/2
2
2
a)3 1
2 (a
2
2
22
3/2
6 2
Shankar Ex, 7.4.6. Show that a(t) = e−iwt a(0) and that a (t) = eiwt a† (0) . Hint: There may be more than one way to do it, but a straightforward way is to write ψ (t = 0) = n cn n , write ψ (t) by inserting the appropriate time time-evolution factor into each term in the sum, and then compute ψ (t) a ψ (t) and ψ (t) a ψ (t) .
|
|
|
†
|| Let’s derivate with respect to time the expression a(t): d a(t) dt
= = =
d a(t) dt a(t)
= =
d ψ (t) a ψ (t) dt i [a, H ]
|| − − i wa −iwa e−iwt a(0)
Now, let’s try to show the second part:
a† (t)
= = =
3
a(t)∗ ∗ e−iwt a(0) eiwt a† (0)
| †|
5) Consider the three operators
0 1 , 1 0
σx =
σy =
− 0
i
0
i
σz =
,
1 0
0 1
−
(5)
(You may recognize these as the famous Pauli spin matrices.) a) Check that for each of these operators the eigenvalues are 1 and obtain normalized vectors. Let’s calculate the eigenvalues of σx :
±
det
−
λ
1
λ
then, the eigenvectors:
λ=
−1 ⇒ λ=1⇒
− 1
y=
⇒ λ2 − 1 = 0 ⇒ λ = −1, 1
=0
0 1 1 0
−x,
y = x,
x y
x y
=λ
−y ⇒ |σx, −1 = √ 12 (|1 − |2) x=y⇒ |σx, 1 = √ 12 (|1 + |2) x=
Then, the eigenvalues of σy : det
−
λ i
⇒ − ⇒ −
−i −λ
then, the eigenvectors:
λ2
=0
0 i
i
0
x y
1=0
=λ
λ=
−1, 1
x y
−1 ⇒ −iy = −x, ix = −y ⇒ |σy , −1 = √ 12 (|1 − i|2) −iy = x, ix = y ⇒ |σy , 1 = √ 12 (|1 + i|2) λ=1⇒ λ=
Then, the eigenvalues of σz : det
− 1
λ
0
then, the eigenvectors:
⇒ − − − 0
1
λ
1 0
λ2
=0
0 1
x y
=λ
1 =0
x y
⇒ λ = −1, 1
− λ = −1 ⇒ x = −x, −y = −y ⇒ |σz , −1 = |2 λ=1⇒ x = x, −y = y ⇒ |σz , 1 = |1
4
b) Check that the generalized uncertainty relation (∆A)(∆B )
≥ 12 |[A, B]|
(6)
is satisfied for A = σx , B = σy , and the system is in the +1 eigenvector of (That is, explicit explicitly ly compute ompute the varian variancce ∆A, the variance ∆B , and the σz . (That right-hand side.) Calculate the variance of A:
− σx2
∆A = ∆σx =
σx2
= =
2
σx
1 0
0 1 1 0
0 1 1 0
1 0
1 0 0 1
1 0
1 0
= 1
=
σx
1 0
0 1 1 0
1 0
0 1
=
= 0 Therefore, ∆σx =
− σx2
σx
1 0
2 = √ 1 − 0 = 1
(7)
Then, the same procedure for the variance of B :
− − − σy2
∆B = ∆σy =
σy2
= =
0
1 0
0
1 0 0 1
1 0
0
i
i
2
σy
i
i
0
1 0
1 0
= 1
σy
= =
−
1 0 1 0
= 0 Therefore, ∆σy =
− σy2
5
σy
0 i
i
0
1 0
0 i
2 = √ 1 − 0 = 1
(8)
Now the RHS: [A, B ] = [σx , σy ] = σxσy = = =
[A, B]
= =
− σ y σx
− − − − − − − −
0 1 1 0 i
0 2i 0
0
0
i
0
0 1 1 0
i
0
i
i 0 0 i
i
0 2i
1 0
2i 0
1 0
2i 0
= 2i
0
i
0 2i
1 0
|[A, B]| = 2
(9)
Finally, plug in (8 (8) and (9 (9) in (6 (6):
≥ 12 |[A, B]| ⇒ 1 = 1
(∆A)(∆B ) that proves the principle.
c) Repeat (same A and B) but now the system is in the +1 the +1 eigenvector σx . Calculate the variance of A: ∆A = ∆σx = σx2 σx 2
σx2
= = =
σx
−
1 1 1 2 1 1 1 2 1 (2) = 1 2
= = =
0 1 1 0
0 1 1 0
1 0 0 1
1 1
1 1 1 2 1 1 1 2 1 (2) = 1 2
Therefore, ∆σx =
− σx2
1 1
σx
0 1 1 0
1 1
1 1
2 = √ 1 − 1 = 0
(10)
Since ∆A = 0 we don’t need to calculate ∆B because the product (∆A)(∆B ) = 0. 6
Now the RHS: [A, B ] = [σx , σy ] =
[A, B]
− − −
2i 0
0 2i
1 1 1 2 1 = 1 1 2 = 0 =
2i 0
0 2i
1 1
2i 2i
|[A, B]| = 0 Finally in (6 (6): (∆A)(∆B )
≥ 12 |[A, B]| ⇒ 0 = 0
that proves the principle.
7
(11)
