Download Quantum Theory. Problem Set 1 1. (a) Consider a particle in a box

Quantum Theory. Problem Set 1
1.
(a) Consider a particle in a box. The Scrödinger equation in Dirac formalism is
Ĥ|ni = En |ni,
where {|ni, En } are the nth eigenstate and eigenenergy respectively. We pass to coordinate representation by multiplying both sides of this equation by the position bra hx| and using the
continuous identity resolution
Z
ˆ
dx0 |x0 ihx0 | = I.
Then
hx|Ĥ|ni = hx|En |ni,
where
hx|En |ni = En hx|ni ≡ En ψn (x),
and where
ˆ
hx|Ĥ|ni = hx|Ĥ · I|ni
Z
=
dx0 hx|Ĥ|x0 ihx0 |ni
Z
=
dx0 H(x, x0 )δ(x − x0 )ψn (x0 )
= H(x)ψn (x),
where
~2 ∂ 2
H(x) = −
+ V (x).
2m ∂x2
Finally, for the wavefunction in x representation, the Scroedinger equation can be written as
¸
·
~2 ∂ 2
+ V (x) ψn (x) = En ψn (x).
−
2m ∂x2
a.1) Solve for V (x) = 0. Try
ψn (x) = An cos kn x + Bn sin kn x,
and obtain
En =
~2 kn2
.
2m
a.2) Boundary conditions at x = ±a : ψn (±a) = 0
cos kn a = 0 ⇒ kn =
(2n + 1)π
,
2a
sin kn a = 0 ⇒ kn =
nπ
(2n)π
=
,
a
2a
(n = 0, 1, 2, . . .),
(n = 1, 2, . . .).
a.3) Resumé:
µ
nπ
kn =
,
2a
En = ~
2
n2 π 2
8ma2
¶
,
(n = 0, 1, 2, . . .),

 An cos kn x (n ODD),
ψn (x) =

Bn sin kn x (n EVEN).
a.4) Normalization:
Z
An :
1 =
A2n
=
A2n
a
dx cos2 kn x
−a
a
Z
dx(1 + cos 2kn x)
2 −a
A2
1
= n (2a) ⇒ An = √ .
2
a
Z
Bn :
1 =
Bn2
a
dx sin2 kn x
−a
1
= Bn2 a ⇒ Bn = √ .
a
Finally


ψn (x) =

√1
a
cos kn x (n ODD),
√1
a
sin kn x (n EVEN).
(b) It is given the wavefunction

 N (a − x) x ≥ 0,
Φ(x) = N (a − |x|) =
 N (a + x) x < 0.
b.1) Normalization:
Z
N:
1 = N
2
a
dx(a − |x|)2
−a
·Z
= N
Z
0
2
dx(a + x) +
−a
¸
a
2
2
dx(a − x)
0
substituting x → −x in the second integral produces
Z
1 = 2N
2
0
2
dx(a + x) = 2N
3
2a
3
−a
µ
⇒N =
3
2a3
¶1/2
b.2) By definition
ˆ
hn|Φi = hn|I|Φi
Z
=
dx0 hn|x0 ihx0 |Φi
Z a
=
dx0 ψn∗ (x0 )Φ(x0 )
−a
where
ψn (x0 ) ≡ hx0 |ni ⇒ ψn∗ (x0 ) = (hx0 |ni)∗ = hn|x0 i,
and
Φ(x) ≡ hx|Φi.
Then
Z
a
hn|Φi =
−a
dx0 ψn∗ (x0 )Φ(x0 )
= 0,
(n EVEN),
since Φ(x) is symmetric while sin kn x is antisymmetric.
Z a
hn|Φi =
dx0 ψn∗ (x0 )Φ(x0 )
−a
1
= √
a
µ
µ
3
2a3
¶1/2 Z
¶1/2
a
dx0 cos(kn x0 )(a − |x|)
−a
1
3
2
= √
3
kn2
a 2a
r µ
¶
3
8
=
,
(n ODD)
2
2 n π2
.
b.3) By definition
µ
hx|Φi = Φ(x) =
3
2a3
¶1/2
(a − |x|).
b.4) By definition
hx|x0 i = δ(x − x0 ).
2. Using the rules of bra-ket algebra, prove or evaluate the following:
a) (XY )† = Y † X † . The Hermitian adjoint of XY is defined by the correspondence:
XY |ai ↔ ha|(XY )† .
However, we also have:
XY |ai = X(Y |ai) ↔ (ha|Y † )X † = ha|Y † X † .
Therefore, (XY )† = Y † X † .
b) tr(XY ) = tr(Y X).
tr(XY ) =
X
ha0 | XY |a0 i
a0
=
X
ha0 | XIY |a0 i
a0
=
X
ha0 | X
Ã
X
a0
=
XX
a0
=
=
X
a00
0
ha | X |a00 i ha00 | Y |a0 i
ha00 | Y |a0 i ha0 | X |a00 i
a00
ha00 | Y
a00
=
X
X
!
|a0 i ha0 | X |a00 i
ha | Y X |a00 i
= tr(Y X) ,
X
X
where we used I =
|a0 i ha0 | =
|a00 i ha00 |.
a00
Ã
a0
00
a00
a0
|a00 i ha00 | Y |a0 i
a00
XX
a0
!
c)
P
a0
Ψ∗a0 (x0 )Ψa0 (x00 ).
X
Ψ∗a0 (x0 )Ψa0 (x00 ) =
X
∗
hx0 | a0 i hx00 | a0 i
a0
a0
=
X
ha0 | x0 i hx00 | a0 i
a0
=
X
hx00 | a0 i ha0 | x0 i
a0
= hx00 |
Ã
X
!
|a0 i ha0 | |x0 i
a0
00
0
= hx | x i
= δ(x00 − x0 ) ,
where δ is the Dirac δ-function.
3. Since
h+|+i = 1 = h−|−i,
h+|−i = 0,
we associate
.
|+i =
µ ¶
1
,
0
.
|−i =
µ ¶
0
;
1
¢
. ¡
h+| = 1 0 ,
¢
. ¡
h−| = 0 1 .
Then
.
|+ih+| =
µ ¶
µ
¶
¡
¢
1
1 0
⊗ 1 0 =
,
0
0 0
.
|+ih−| =
µ ¶
µ
¶
¡
¢
1
0 1
⊗ 0 1 =
,
0
0 0
.
|−ih+| =
µ ¶
µ
¶
¡
¢
0
0 0
⊗ 1 0 =
,
1
1 0
.
|−ih−| =
µ ¶
µ
¶
¡
¢
0
0 0
⊗ 0 1 =
.
1
0 1
Then, for example
Ŝx =
.
=
.
=
~
(|+ih−| + |−ih+|)
2 ·µ
¶ µ
¶¸
~
0 1
0 0
+
0 0
1 0
2
µ
¶
~ 0 1
.
2 1 0
In the same way one calculate
. ~
Ŝy =
2
µ
0 −i
i 0
¶
,
. ~
Ŝz =
2
µ
1 0
0 −1
¶
.
Now that the matrix representation of the spin operators Ŝi , (i = x, y, z) is available, it is simple
to calculate the required commutators just by multiplying matrices. For example
[Ŝx , Ŝy ] = Ŝx Ŝy − Ŝy Ŝx
µ ¶2 µ
¶µ
¶ µ ¶2 µ
¶µ
¶
~
~
0 1
0 −i
0 −i
0 1
=
−
1 0
i 0
i 0
1 0
2
2
µ ¶2 ·µ
¶ µ
¶¸
~
i 0
−i 0
−
=
0 −i
0 i
2
µ ¶2 µ
¶
~
1 0
=
2i
0 −1
2
¶¸
µ ¶ ·µ ¶ µ
2
~
~
2
1 0
=
2i
0 −1
4
~
2
= i~Ŝz .
The other cases can be worked exactly in the same way. We illustrate a single example for the
anticommutator.
{Ŝx , Ŝy } = Ŝx Ŝy + Ŝy Ŝx
µ ¶2 µ
¶µ
¶ µ ¶2 µ
¶µ
¶
~
~
0 1
0 −i
0 −i
0 1
=
+
1 0
i 0
i 0
1 0
2
2
µ ¶2 ·µ
¶ µ
¶¸
~
i 0
−i 0
=
+
0 −i
0 i
2
= 0.
5.
a) Let us call the matrix Jz :


0 1 0
Jz =  1 0 1  .
0 1 0
To find eigenvalues, solve det|λI − Jz | = 0, so λ2 − λ = 0, ⇒ λ = 0, ±1. So, there is no
degeneracy. To find the eigenvectors, let us write the generic eigenvector as ~uλ = (x1 , x2 , x3 ).
Then
√
√
λ=0
⇒
x
=
0,
x
+
x
=
0
⇒
~
u
=
(1/
2,
0,
−1/
2),
2
1
3
0
√
√
√
λ = −1 ⇒
2x
+
x
=
0,
x
+
2x
=
0
⇒
~
u
=
(−1/2,
1/
2,
−1/2),
2
2
3
−
√1
√
√
λ=1
⇒
− 2x1 + x2 = 0, −x2 + 2x3 = 0 ⇒ ~u+ = (1/2, 1/ 2, 1/2).
b) The eigenvalues are 0, ±1 so we have a spin 1 system, and the matrix is probably related
to operators for spin 1. Later on, when we shall study angular momentum, this matrix
will re-appear and that is why we called it Jz as above. Spin 1 appears in many different
systems; for example in the hyperfine structure of the Hydrogen atom, in many elementary
particles (ρ, φ, W, Z), and in the isospin symmetry of pions.
6. Consider the following Hamiltonian for a 2-state system:
H = a(|1i h1| − |2i h2| + |1i h2| + |2i h1|) .
The matrix representation of H in the basis |1i and |2i is given by:
µ
¶
h1| H |1i h1| H |2i
H→
h2| H |1i h2| H |2i
Let us compute h1| H |1i:
h1| H |1i =
=
=
=
h1| [a (|1i h1| − |2i h2| + |1i h2| + |2i h1|)] |1i
a(h1| 1i h1| 1i − h1| 2i h2| 1i + h1| 1i h2| 1i + h1| 2i h1| 1i)
a(1 − 0 + 0 + 0)
a
Likewise,
h1| H |2i = a
h2| H |1i = a
h2| H |2i = −a
Therefore, we have
µ
H→a
1 1
1 −1
¶
The eigenvalues E of H are the solutions of det(H − EI) = 0, that is
µ
¶
a−E
a
det
= 0
a
−a − E
−(a − E)(a + E) − a2 = 0
−(a2 − E 2 ) − a2 = 0
E 2 = 2a2 = 0
√
⇒ E = ± 2a
Note that a general ket |αi can be decomposed on the basis |1i and |2i as
|αi = α1 |1i + α2 |2i ,
so that the vector representation of |αi on this basis is given by
¶
µ
α1
|αi →
α2
In particular, the vector representations of the eigenkets of H on this basis are the eigenvector of
the matrix representation of H on the same basis:
µ
¶µ ¶
µ ¶
1 1
x
x
a
=E
1 −1
y
y
(
x + y = Ea x
⇒
x − y = Ea y
µ
¶
E
⇒y=
−1 x
a
Therefore, the eigenvectors have the form
µ
N
E
a
1
−1
s
where N is a normalization factor: N = 1/ 1 +
µ
¶
¶2
E
−1 .
a
The eigenvalues and eigenkets of H are given by:
E=
√
2a
√
E = − 2a
h
³√
´ i
1
q
|1i
+
2
−
1
|2i
¡√
¢2
1+
2−1
h
³ √
´ i
1
q
|1i
+
−
2
−
1
|2i
¡ √
¢2
1+ − 2−1