Quantum Mechanics I / Quantum theory I. 02 October, 2015
Assignment 2: Solution
1. Consider the following two kets: −3i 2 |ψ = 2 + i , |φ = −i . 4 2 − 3i
(a) Find the bra φ|. (b) Evaluate the scalar product φ|ψ . (c) Why the products |ψ |φ and φ| ψ| do not make sense? Answer (a)
|ψ = 2 + i 2 + 3i .
(b) −3i φ|ψ = 2 + i 2 + 3i 2 + i . 4 = −6i + 2i − 1 + 8 + 12i, = 7 + 8i.
(c) The |ψ and |φ are column vectors and φ| and ψ| are row vectors. And we know that matrix multiplication is performed only when we have the number of columns of first matrix equal to the number of rows to the second matrix. Therefore the multiplication of the form |ψ |φ and φ| ψ| does not make sense. Neither do these objects make nay physical sense.
2. Consider the states
|ψ = 3i|φ1 −7i|φ2
|χ = −|φ1 +2i|φ2 ,
where |φ1 and |φ2 are orthonormal. Calculate the scalar products ψ|χ and χ|ψ . Are they equal?
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Answer
ψ|χ = −3i φ1| + 7i φ2| −|φ1 +2i|φ2 , ψ|χ = −14 + 3i.
Similarly,
χ|ψ = − φ1| − 2i φ2| 3i|φ1 −7i|φ2 , = −14 − 3i.
Hence ψ|χ = χ|ψ ∗ as expected.
3. (a) Discuss the hermiticity of operators (Aˆ + Aˆ†), i(Aˆ + Aˆ†) and i(Aˆ − Aˆ†). (b) Find the Hermitian adjoint of
(1 + iAˆ + 3Aˆ2)(1 − 2iAˆ − 9Aˆ2) f(Aˆ) = · 5 + 7Aˆ (c) Show that the expectation value of a Hermitian operator is real and that of an anti-Hermitian operator is imaginary. Answer (a)(Aˆ + Aˆ†):
Aˆ + Aˆ††= Aˆ† + A.ˆ
So this is hermitian. i Aˆ + Aˆ†:
i Aˆ + Aˆ††= −i Aˆ + Aˆ†.
So this is not hermitian, rather anti-Hermitian. i Aˆ − Aˆ†:
i Aˆ − Aˆ†† = −i Aˆ† − Aˆ,
= i Aˆ − Aˆ†
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So, this is hermitian. (b)
(1 + iAˆ + 3Aˆ2)(1 − 2iAˆ − 9Aˆ2) f(Aˆ) = · 5 + 7Aˆ (1 + iAˆ + 3Aˆ2)(1 − 2iAˆ − 9Aˆ2)† f(Aˆ)† = · 5 + 7Aˆ (1 − 2iAˆ − 9Aˆ2)†(1 + iAˆ + 3Aˆ2)† = . (5 + 7Aˆ)† (1 + 2iAˆ† − 9Aˆ2†)(1 − iAˆ† + 3Aˆ2†) = . 5 + 7Aˆ† † This can be simplified further if Aˆ is hermitian, then Aˆ = Aˆ† and Aˆ2 = Aˆ2 . (c) ˆ Suppose A is a hermitian operator with real eigenvalues λi and eigenvector |λi , i.e
ˆ A|λi = λi|ψ .
The expectation value is,
ˆ ˆ A = λi|O|λi , X ˆ = ψ|A|λi λi|ψ , i X = λi ψ|λi λi|ψ , n X = λi ψ|ψ , n X = λi. n The expectation value is the sum of eigenvalues which must be real. Hence Aˆ = ˆ P Tr A = n λi. Now consider an anti-Hermitian operator Bˆ, i.e.,Bˆ† = Bˆ. Suppose Bˆ has eigenvalues
bi with eigenvectors |bi .
ˆ B|bi = bi|bi (1)
ˆ† ∗ ˆ bi|B = bi bi| = − bi|B (2)
From 1, we obtain
ˆ bi|B|bi = bi bi|bi = bi (3)
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ˆ ∗ But from 2, bi|B = −bi bi|. Hence the LHS of can be written as,
ˆ ∗ bi|B|bi = −bi bi|bi = −bi (4)
∗ Comparing 3 and 4, bi = −bi which is only possible if bi is imaginary. Now
Bˆ = ψ|Bˆ|ψ X ˆ = ψ|B|bi bi|ψ i X = bi ψ|bi bi|ψ i X = bi Which must be imaginary i
Since all the bi’s are imaginary.
4. What conditions must the parameter ε and the operator Gˆ satisfy so that the operator Uˆ = eiεGˆ is unitary? Answer
UˆUˆ † = eiεGˆ eiεGˆ†, = eiεGˆ eiεGˆ† ,
ˆ ˆ† = eiε G−G = 1ˆ.
This means that Gˆ = Gˆ† or Gˆ must be hermitian to ensure that eiεGˆ is unitary. Remember: iˆ2 iˆ3 eiεGˆ = 1ˆ + iˆ+ + + ...... 2! 3!
ˆ ˆ and eiεAˆeiεBˆ = eiε A+B only if [A,ˆ Bˆ] = 0.
5. Show that if Aˆ−1 exists, the eigenvalues of Aˆ−1 are just the inverse of those of Aˆ. Answer If Aˆ−1 exists then we have
AˆAˆ−1 = 1ˆ = Aˆ−1A,ˆ
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So,
1ˆ|ψ = |ψ ,
= Aˆ−1Aˆ|ψ , = Aˆ−1a|ψ , = aAˆ−1|ψ . 1 Aˆ−1a|ψ = |ψ , a
Hence the eigenvalues of Aˆ−1 are the inverse of the eigenvalues of Aˆ.
6. Consider the following two kets: 5i 3 |ψ = 2 , |φ = 8i . −i −9i
(a) Find |ψ ∗ and ψ|. (b) Is |ψ normalized? If not, normalize it. (c) Are |ψ and |φ orthogonal? Answer (a) −5i ∗ |ψ = 2 , i ψ| = −5i 2 i .
5i ψ|ψ = −5i 2 i 2 , −i = 10 + 4 + 1, = 15.
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|ψ So, normalized ket will be , r ψ|ψ
5i 1 = √ 2 . 15 −i
(c) 5i φ|ψ = 3 − 8i 9i 2 , −i = 15i − 16i + 9, = 9 − i.
As φ|ψ is not zero, so these are not orthonormal.
7. (a) Show that Tr(AˆBˆ) = Tr(BˆAˆ). (b) Show that the trace of a commutator is always zero. Answer (a)
X Tr AˆBˆ = i|aˆBˆ|i , i X X = i|Aˆ|j j|Bˆ|i , i j X X ˆ ˆ = AijBji, i j
Similarly,
ˆ ˆ X X ˆ ˆ ˆ ˆ Tr BA = BjiAij = Tr AB i j
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(b)
Tr [A,ˆ Bˆ] = Tr AˆBˆ − BˆAˆ, = Tr AˆBˆ−Tr BˆAˆ,
= Tr AˆBˆ−Tr AˆBˆ, = 0.
8. Show that the trace of an operator does not depend on the basis in which it is expressed. Answer Let we take two basis |i and |˜j , with |˜j = Sˆ|i where Sˆ is a similarity transform ˆ with elements Sji = ˜j|i .
X Tr Aˆ= ˜j|Aˆ|˜j , j
Since |˜j = Sˆ|i ,
X Tr Aˆ = i|Sˆ†AˆSˆ|i , i = Tr Sˆ†AˆSˆ, ˜ = Tr Aˆ,
˜ Where Aˆ is the operator in the transformed bases.
9. Consider the two states,
|ψ = i|φ1 +3i|φ2 −|φ3 ,
and
|χ = |φ1 −i|φ2 +5i|φ3 ,
where |φ1 , |φ2 and |φ3 are orthogonal and normalized. (a) Calculate |ψ χ| and |χ ψ|. Are they equal? Calculate their traces and compare them.
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(b) Find the Hermitian conjugates of |ψ χ|. Answer
(a) These states in the |φ1 , |φ2 , |φ3 basis are given by,
i |ψ = 3i , −1
1 |χ = −i , −5i
Now i |ψ χ| = 3i 1 i 5i , −1 i − 1 − 5 = 3i − 3 − 15 , −1 − i − 5i
and similarly, 1 |χ ψ| = −i −i − 3i − 1 , −5i −i − 3i − 1 = −1 − 3 i , −5 − 15 5i
(b) Hermitian Conjugate, also called the adjoint is,
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† i − 1 − 5 † |ψ χ| = 3i − 3 − 15 , −1 − i − 5i −i − 3i − 1 = −1 − 3 i . −5 − 15 5i
10. Consider a state which is given in terms of three orthonormal vectors |φ1 , |φ2 , and
|φ3 as follows:
1 1 1 |ψ = √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5
ˆ ˆ 2 where |φn are eigenstates to an operator B such that B|φn = (3n − 1)|φn with n = 1, 2, 3. (a) Find the norm of the state |ψ . (b) Find the expectation value of Bˆ for the state |ψ . (c) Find the expectation value of Bˆ2 for the state |ψ . Answer (a) The norm is,
2 1 1 1 1 1 1 |ψ = ψ|ψ = √ φ1| + √ φ2| + √ φ3| √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 15 3 5 1 1 1 = φ |φ + φ |φ + φ |φ , 15 1 1 3 2 2 5 3 3 1 1 1 = + + , 15 3 5 1 + 5 + 3 = , 15 3 = . 5
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(b)
ˆ ˆ 1 1 1 B|ψ = B √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 2 11 26 = √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 ˆ 1 1 1 2 11 26 ψ|B|ψ = √ φ1| + √ φ2| + √ φ3| . √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 15 3 5 2 11 26 = √ φ1|φ1 +√ φ2|φ2 + φ3|φ3 , 15 3 5 2 11 26 = √ + √ + √ , 15 3 5 2 + 55 + 78 = , 15 135 = = 9. 15
(c).
ˆ2 ˆ ˆ 1 1 1 B |ψ = BB √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 ˆ 2 11 26 = B √ |φ1 +√ |φ2 +√ |φ3 , 15 3 5 4 121 676 = √ |φ1 + √ |φ2 + √ |φ3 , 15 3 5 1 1 1 4 121 676 = √ φ1| + √ φ2| + √ φ3| . √ |φ1 + √ |φ2 + √ |φ3 , 15 3 5 15 3 5 4 121 676 = φ |φ + φ |φ + φ |φ , 15 1 1 3 2 2 5 3 3 4 121 676 = + + , 15 3 5 2637 = = 175.8. 15
11. In the following expressions,where Aˆ is an operator, specify the nature of each expres- sion (i.e., specify whether it is an operator, a bra, or a ket); then find its Hermitian conjugate.
(a) φ|Aˆ|ψ ψ| (b) Aˆ|ψ φ| (c) |φ φ|Aˆ−i Aˆ|ψ ψ|.
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Answer
(a) φ|Aˆ|ψ ψ| Bra.
(b) Aˆ|ψ φ| Operator. (c) |φ φ|Aˆ−i Aˆ|ψ ψ| Operator.
12. Consider a two-dimensional space where a Hermitian operator Aˆ is defined by
ˆ A|φ1 = |φ1 and ˆ A|φ2 = −|φ2 ,
where |φ1 and |φ2 are orthonormal.
(a) Do these states |φ1 and |φ2 form a basis? ˆ ˆ ˆ2 (b) Consider the operator B = |φ1 φ2|. Is B Hermitian? Show that B = 0. (c) Show that the products BˆBˆ† and Bˆ†Bˆ are projection operators. (d) Show that the operator BˆBˆ† − Bˆ†Bˆ is unitary. ˆ ˆ ˆ† ˆ† ˆ ˆ ˆ (e) Consider C = BB + B B. Show that C|φ1 = |φ1 and C|φ2 = |φ2 . Answer (a) Yes, They form a basis, because they are orthogonal states (and hence linearly independent). (b)
ˆ B = |φ1 φ2|, ˆ† ˆ ˆ B = |φ2 φ1|= 6 B. So, B is not Hermitian.
Now,
ˆ2 B = |φ1 φ2|φ1 φ2|, = 0.
(c)
ˆ ˆ† BB = |φ1 φ2|φ2 φ1|,
= |φ1 φ1| Projection operator.
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and
ˆ† ˆ B B = |φ2 φ1|φ1 φ2|,
= |φ2 φ2| Projection operator.
BˆBˆ††= and Bˆ†Bˆ= Bˆ†Bˆ. (d)Let
ˆ ˆ† ˆ† ˆ ˆ |φ1 φ1| − |φ2 φ2| = BB − B B = C.
ˆ† C = |φ1 φ1| − |φ2 φ2| ,
Now let’s find out Cˆ†Cˆ.
ˆ† ˆ C C = |φ1 φ1| − |φ2 φ2| |φ1 φ1| − |φ2 φ2| ,
= |φ1 φ1| + |φ2 φ2|, = 1ˆ by completeness relation.
(e)
ˆ ˆ ˆ† ˆ† ˆ C = BB + B B = |φ1 φ1| + |φ2 φ2|.
(i)
ˆ C|φ1 = |φ1 φ1| + |φ2 φ2| |φ1 ,
= |φ1 ,
(ii)
ˆ C|φ2 = |φ1 φ1| + |φ2 φ2| |φ2 ,
= |φ2 ,
13. Consider an operator
ˆ A = |φ1 φ1| + |φ2 φ2| + |φ3 φ3| − i|φ1 φ2| − |φ1 φ3| + i|φ2 φ1| − |φ3 φ1|,
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where |φ1 , |φ2 , and |φ3 form a complete and orthonormal basis. (a) Is Aˆ Hermitian? Calculate Aˆ2. Is it a projection operator? ˆ (b) Find the 3×3 matrix representing A in the |φ1 , |φ2 , |φ3 basis. Answer (a)(i)
ˆ† ˆ A = |φ1 φ1| + |φ2 φ2| + |φ3 φ3| + i|φ2 φ1| − |φ3 φ1| − i|φ1 φ2| − |φ1 φ3| = A
(ii) It’s much easier to solve this question in a matrix representation. 1 0 0 |φ1 = 0 , |φ1 = 1 , |φ1 = 0 . 0 0 1
1 − i − 1 ˆ A = i 1 0 . −1 0 1
1 − i − 1 1 − i − 1 ˆ2 A = i 1 0 i 1 0 , −1 0 1 −1 0 1 1 + 1 + 1 − i − i − 1 − 1 = i + i 1 + 1 − i . −1 − 1 + i 1 + 1 3 − 2i − 2 = 2i 2 − i . −2 i 2
Hence Aˆ is not a projection operator. (b) 1 − i − 1 i 1 0 . −1 0 1
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