Quantum Mechanics - Homework Assigment 2 Alejandro G´ omez omez Espinosa
∗
September 23, 2012
Shankar, Ex 1.8.10 By considering the commutator, show that the following Hermitian matrices may be simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitary transformation to this basis, both matrices are diagonalized. 1 0 1 2 1 1 Ω= 0 0 0 Λ= 1 0 1 , 1 0 1 1 1 2
−
−
Since Ω is degenerate and Λ is not, you must be prudent in deciding which matrix dictates the choice of basis. Considering the commutator [Ω, [Ω, Λ]: [Ω, [Ω, Λ] = ΩΛ ΛΩ 1 0 1 = 0 0 0 1 0 1 =
=
− 3 0 0 0 30 00 0 0
2 1 1 2 1 0 −1 − 1 1 −1 2 1 3 3 0 3 0 − 0 0 0 3 3 0 3 0 0
1 − 0
1 0 1
1 1 2
−
0 1 0 0 1 0 1
0 0 0
It means that Ω and Λ are two commuting Hermitian operators. Then, from Theorem 13, there exist (at least) a basis of common eigenvector eigenvectorss that diagonalizes diagonalizes them both. Now, Now, our task is to find the eigenvector eigenvectorss of Λ that diagonalizes diagonalizes both operators. For this: 0 = det[Λ
− λI ]
= = = =
∗
(2 − λ) det 1
1 1 λ 1 1 (2 λ)
− − 1 − − −λ(2 − λ) − 1 − 1 − (2 − λ) + λ − (2 − λ) −λ + 4λ 4λ − λ − 6 (λ + 1)(λ 1)(λ − 2)(λ 2)(λ − 3) 2
2
[email protected]
1
3
The eigenvalues are λ = 1, 2, 3. To find the corresponding eigenvectors, we have to compute: Λ v = λ v , then for λ = 1:
|
−
|
−
2 1 1
(1)2x (1)2x + y + z =
x x − y = −1 y
1 0 1
1 1 2
−
z
− z = −y; from (1) + (3): 4x 4x + 4z 4z = 0 ⇒ x = −z
for λ = 2:
−x;
z
2 1
(2)x
y= for λ = 3:
2 1
(2)x
−z
z
− z = 2y;
(3)x
x=
−y =0
−z
x x − y = 3 y 1 1 2
−
− x + y + z = 0;
and then: 2z 2z = y
z
x=y
1 0 1
1
(1)
1 1 2
−
(1)y (1)y + z = 0;
2z = −z − y + 2z
x x − y = 2 y
1 0 1
1
(3)x
(2)x
z
z
− 3y − z = 0;
y=0
(3)x
−y−z =0
x=y
Thus, the normalized eigenvectors are:
1 |v− = √ 16 −2 ;
1 |v = √ 13 1 ;
1
1 |v = √ 12 0
2
−1
3
−1
1
From the eigenvectors, we can build a unitary matrix that diagonalizes the operators: 1
√ U = − √ √ 6 2
6 1
−
6
√ 1 3 √ 1 3 − √ 1
3
√ 1
0 √ 2
1
2
Finally, we can verify that the matrices are diagonalized when we introduce this unitary matrix: 1
√ U † ΩU = √ √
6 1 3 1 2
1
√ U † ΛU = √ √
6 1 3 1 2
2
1
− √ − √ 1 √ − √ 0 6
3
3
√ 1
0
2
0 0 1 0 1
6 1
1
2
1
− √ − √ 2 √ − √ 1 6
1
6 1
3
0
3
√ 1
2
1
1
√ 1 6 3 2 √ 1 6 3 1 √ √ − −1
√ 1 0 − √
1 0 1
−
6
√ 1
2
1
3
1
√ 1 6 3 2 √ 1 6 3 − √ 1 − √ 1
√ 1 −1 − √ 2
6
0 0 = 0 √ 0
3
2
√ 1
−1 0 = 0 √ 0 2
1
2
0 0 0 0 0 2
0 0 2 0 0 3
Thus Thus we demostrate demostrate that the chose chose unitary matrix diagonalizes our initial operators. operators. 2
2) A two-dimensional space is spanned by the orthonormal basis vectors 1 and 2 . If operator B has ket-bra representation representation B = 1 2 2 1 , find the corr corresp esponding onding π explicit ket-bra representation of cosh( 3 B ). (Hint: You may find it easier to work in matrix notation, and then covert back to ket-bra notation at the end.)
|
| | − | |
|
The Taylor serie of cosh( π3 B ) is: 2
π 1 π 1 π π cosh B = 1+ B + B + ... = 1 + B + 2
3
2
4
3
24
2
3
18
π4 4 B + ... 1944
We can represent the orthonormal basis vectors explicitly as:
|1 =
1
|2 =
0
0 1
Therefore, Therefore, the operator B has this matrix notation: B = = = =
|112|− |2 1| 1 0 1 − 0 1 0 0 0 1 0 0 − 1 0 0 0 0 1 −1
0
Then, to replace the matrices B 2 and B 4 in the cosh serie, we calculate: B
2
0 1 0 1 −1 0 = −1 0 −1 0 = 0 −1 −1 0 −1 0 1 0
B4 =
0
−1
0
−1
=
0 1
Finally,
π cosh B 3
π2 1 0 = 1+ 1 18 0 π2 = 1 (1 1 + 2 18 1 π2 = 1 + 2! 32 π = cos I 3 1 = I 2
−
−
π4 1 0 + + ... 1944 0 1 π4 2 )+ ( 1 1 + 2 2 ) + ... 1944 1 π4 + ... ( 1 1 + 2 2 ) 4! 34
− | | | | −
3
| | | | | | | |
3) For operator Ω depending on scalar λ, show that d(Ω−1 ) = dλ
−Ω−1 dΩ Ω−1 dλ
Using the unitary operator, ΩΩ−1 = Ω−1 Ω = I , we found: d(ΩΩ−1 ) dλ d(Ω) −1 d(Ω−1 ) Ω +Ω dλ dλ d(Ω) −1 Ω dλ d(Ω) −1 Ω−1 Ω dλ d(Ω) −1 Ω−1 Ω dλ
−
=
d(I ) dλ
= 0 =
−1
−Ω d(Ωdλ ) d(Ω− ) − −Ω Ω dλ 1
= =
1
d(Ω−1 ) dλ
4) Referring to operators X and K and K in Shankar (1.10.41), show that [ that [X 2, K 2 ] = 2I + I + 4iXK iX K in two ways: a) By applying X 2 K 2 K 2 X 2 to an arbitraty f ( f (x) and seeing what happens. The two operators are:
−
X f
xf (x), | → xf (
f (x) | → −i df ( dx
K f
thus: 2
X f
2
f (x), | → x f (
2
K f
| →−
d2 f ( f (x) dx2
applying to an arbitrary function f ( f (x): [X 2 , K 2 ] f
|
= X 2 K 2 f K 2 X 2 f d2 f d2 2 2 = x + (x f ) f ) dx2 dx2 d2 f d df = x2 2 + 2xf + x2 dx dx dx d2 f df df d2 f 2 2 = x + 2f 2f + 2x 2x + 2x 2x +x dx2 dx dx dx2 df = 2f + 4x 4x dx = 2I + 4iX 4iXK K
| −
− −
|
−
4
b) By working solely with the operators and repeatedly applying commutator rules (1.5.10) and (1.5.11) with (1.10.41). Using commutator rules: [X 2 , K 2 ] = K [X 2 , K ] + [X [X 2 , K ]K = K (X [X, K ] + [X, [ X, K ]X ) + (X ( X [X, K ] + [X, [ X, K ]X )K = KX [ KX [X, K ] + K [X, K ]X + X + X [X, K ]K + K + [X, [X, K ]X K = KXiI + KXiI + KiIX + KiIX + XiIK + XiIK + iIXK = 2iIKX + iIKX + 2iIXK 2iIXK = 2iXK
2 iIXK − 2i[X, K ] + 2iIXK
= 4iXK + 2I 2I
Shankar, Ex 1.10.1 Show that δ (ax) ax) = δ (x)/ a . (Consi (Consider der that δ (x) = δ ( x).)
||
−
δ (ax) ax)d(ax) ax).
Rememb emember er
Using the hint:
∞
δ (ax) ax)dx =
−∞ = = =
∞ δ (|a|x)dx −∞ ∞ 1 δ (|a|x) d(|a|x) |a| −∞ ∞ 1 |a| −∞ δ (|a|x)d(|a|x) ∞ 1 |a|
using δ (x) = δ ( x)
−
δ (x)d(x)
−∞
Finally, we show that δ (ax) ax) = δ (x)/ a .
||
Shankar, Ex 1.10.2 Show that
δ (x − x) δ (f ( f (x)) = i
|df/dx | i
i
where xi are zeros of f ( f (x). Hint: Hint: wher where does does δ (f ( f (x)) blows up?. up?. Expand Expand f ( f (x) near such points in a Taylor series, keeping the first nonzero term. Expanding f ( f (x) in Taylor series around xi : )(x − x ) + ... − x ) + 12 f (x )(x if f ( f (x ) = 0, keeping the first nonzero term: f ( f (x) = f (x )(x )(x − x ). No Now, w, using using the 2
f ( f (x) = f ( f (xi ) + f (xi )(x )(x
i
i
i
i
i
i
results of Ex 1.10.1:
δ (f ( f (x)) = δ (f (xi)(x )(x
1 − x )) ⇒ δ (f ( f (x)) = |f (x )| δ (x − x ) i
i
i
To generalized our result, we can summ over all the nonzero terms:
δ (f ( f (x)) = i
5
δ (x xi ) df ( f (xi)/dxi
|
−
|
Shankar, Ex 1.10.3 Consider the theta function θ(x x ) which vanishes if x x is negative and equals 1 if x x is positive. Show that δ (x x ) = d/dxθ( d/dxθ(x x ).
−
−
Using a test well-beha well-behaved ved function g (x):
∞
d g (x) θ(x dx −∞
− x )dx =
∞
g (x)dθ( dθ(x
−∞
= θ(x
∞
∞
g (x)δ (x)dx
−∞
Finally, we found that:
∞
θ(x
−∞
∞ − g(x)dx g(∞) − g(∞) + g(0) = g(0) 0
=
−
− x)
∞ − − x )g(x)|−∞
= g( ) =
−
d θ(x dx
− x) = δ (x)
6
−
− x )g (x)dx