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Quantum Mechanics I

Solutions 3.

HS 2012Prof. Ch. Anastasiou

Exercise 1. Bra-ket formalism in a two-state system

Let us denote by | the (normalised) eigenstates of the spin operator in the z-direction witheigenvalues /2.

(a) How does Sz act on the states |?(b) Find a way to express Sz out of | and |.

Hint. The states | are orthonormal.

We want now to measure the spin in another direction, say along the x-axis. Let Sx be theoperator corresponding to this measurement with eigenvectors |Sx, and eigenvalues /2.

(c) Why must|Sx, +

be a superposition of

|only?

(d) Recalling the Stern-Gerlach experiment, what is the value ||Sx, + |?(e) Compute |Sx, + using this information. Is it uniquely determined? In the same spirit

deduce |Sx, and obtain the operator Sx.(f) Do the same for Sy and obtain |Sy, . What should be the values of the four different

transition probabilities | Sy, |Sx, |2 (think of a Stern-Gerlach apparatus again)? Doesit put further constraints on Sx,y?

We introduce the raising and lowering operators, S+ and S respectively, which have the re-

spective properties S+ | =

|+ and S+ |+ = 0, S | = 0 and S |+ =

|.(g) Construct these operators from | and |. Are they Hermitian?

Hint. Recall that an operator A is called Hermitian if A = A. It is called anti-Hermitian if

A = A.

(h) Show that S = Sx iSy.(i) Let us represent our kets by vectors in C2, namely

|+

10

, |

01

.

Compute the matrix representation of |, Sx,y,z and S in this basis.

Finally, we want to study the commutation properties of the spin operators.

(j) Show that [Si, Sj ] = iijkSk, where ijk is the totally antisymmetric tensor with 123 = 1.

(k) Defining S2 = S2x + S2y + S

2z , show that it commutes with each Si.

Solution. The solution to this exercise is attached on page 5.

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Exercise 2. Observables and Measurements

Solution. In the lectures and in previous exercises we understood that in quantum mechanics the state of aphysical system is described by a vector in a Hilbert space.

From the Stern-Gerlach experiment, we understood that probing the state of a physical system may result in

discrete outcomes. We also understood these outcomes to b e generally unpredictable: we are not generally ableto tell in advance whether a single Ag atom of the beam will bend up or down in the original 1922 experiment.On the other hand, we saw that we can select, i.e., filter, one of the two outcomes by blocking the other path,producing then a polarized beam.

A Stern-Gerlach filter realizes what in quantum mechanics is called a measurement of anobservable. An observable is a physical quantity, i.e., something that exists in nature and canthen be measured. An observable is described in quantum mechanics by a linear Hermitianoperator A with a complete set of eigenstates. We will play a bit with these kind ofoperators in this and in next week exercises.

Let A be a hermitian operator, i.e. A = A where the conjugate is defined via the inner product

on our vectorspace.

(a) For a normalized eigenstate |a, i.e. that A |a = a |a and a = 1, show thata| A |a = a R and a| A = a a| (1)

Hint: calculate a|A|b for arbitrary |b.

Solution.

a|A |a = a a|a = a||a|| = aa|A |a = a a|a = a = a|A |a = a|A |a = a

(S.1)

and a|A |b = b|A |a = a b|a = a a|b = a a| |b a|A = a a| (S.2)

(b) For two eigenstates A |a1 = a1 |a1 and A |a2 = a2 |a2 show thatif a1 = a2 then a1|a2 = 0 . (2)

Solution.

a1| (A A) |a2 = 0 = (a1 a2) a1|a2 since a1 = a2 then a1|a2 = 0 (S.3)

In series 2 we understood what a basis of a Hilbert space H is. In finite dimensions, if A ishermitian a basis of H can be given in terms of eigenvectors of A. In quantum mechanics, themeasurement of the observable A over the state | = i ci |ai yields one of the eigenvalues aiwith probability |ci|2. Immediately after the measurement is performed, the state collapses onthe eigenvector |ai, whose corresponding eigenvalue is the measurement outcome.

(c) CalculateA = |A| (3)

in terms of ci and ai. If

|

=

|aj

for a certain j, calculate (3) and

|cj

|2.

How do we interpret the Stern-Gerlach experiment now?

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Solution.

|A | =ij

cicj aj |A |ai =

ij

cicjai aj|ai =

ij

cicjaiij =

i

|ci|2ai (S.4)

when | = |aj for a certain j|A | = aj and cj = 1 (S.5)

In the spin 1/2 case, when the beam is unpolarized both outcomes are equally likely (up, down) so thatprobabilities are P = 1/2 = P. When the beam is polarized in one of the two directions, e.g., up, P = 1

and no beam splitting is further observed.

(d) Show that transformations from one orthonormal basis to the other are given by unitarytransformations, i.e. for |bi = U|ai and ai|aj = ij , prove

(i) UU = 1 bi|bj = ij and (ii) bi|bj = ij UU = 1 . (4)Verify this for the transformation matrices calculated in Exercise 1.

Solution.

bi|bj = ai|UU|aj = ai|aj = ij (S.6)proves (i). For (ii), we write (summation over repeated indices)

|bk = |aj aj |U|ak =: Ujk |aj bk| = aj|U|ak aj | = ak|U|aj aj | = Ukj aj| , (S.7)and thus,

mk!

= bm|bk = am|U|al al|aj aj|U|ak = am|U|al al|U|ak = UmlUlk , (S.8)

so U!

= U1, i.e. U must be unitary. In the case of the transformations from | to |Sx, in exercise 1,the transformation matrix is given by

U =1

2

1 11 1

UU = 1

2

1 11 1

1 11 1

=

1 00 1

. (S.9)

Now, we will consider an example of a two-state quantum mechanical system where a smartchoice of basis can give us more information about the systems sate. Consider the quantumstate,

| = 12

|0 + ei

2

|1 (5)

(e) Performing a measurement in the {|0 , |1} basis, what is the probability of measuring|0? What is the probability of measuring |1?

Solution. measure |0 with probability | 12|2=1/2 and measure |1 with probability | ei

2|2=1/2

(f) Performing a measurement in the {|+ , |} basis, what is the probability of measuring|+? What is the probability of measuring |?

Solution. Recall |+ = 12

(|0 + |1) and | = 12

(|0 |1). We rewrite | in this new basis byexpressing |0 and |1 in the new basis |0 = 1

2(|+ + |) and |1 = 1

2(|+|) and plugging into |

get | = 1+ei2

|+ + 1ei2

| We then see that the probability of measuring |+ is | 1+ei2

|2 = cos2(/2)and that the probability of measuring | is | 1ei

2|2 = sin2(/2). Thus measurement in this basis reveals

information about the phase.

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Exercise 3. Commutator [, ] Anticommutator {, } and the Uncertainty RelationGiven two operators A and B, we define:

[A, B] = AB BA , {A, B} = AB + BA . (6)

(a) Show that AB = 12 ([A, B] + {A, B})(b) Show that

i) [A, A] = 0 ii) [A, B] = [B, A] iii) [A, c] = 0where c is a number.

(c) Show thati) [A + B, C] = [A, C] + [B, C] ii) [A, B + C] = [A, B] + [A, C]

(d) Show thati) [A,BC] = B[A, C] + [A, B]C ii) [AB,C] = A[B, C] + [A, C]B

where C is an operator, too.

(e) Show that the Jacobi identity [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 is satisfied by thecommutator.

(f) Do you know something else that behaves in the same way?

(g) Show that if [[A, B], A] = 0, then [Am, B] = mAm1[A, B], m > 0.

(h) Show by explicit construction that any operator could be decomposed into a sum of pureHermitian and anti-Hermitian operators.

(i) What corresponds to the adjoint for the complex numbers?

(j) Given A and B are Hermitian. What can you say about the adjoint ofi) cA ii) AB iii) [A, B] iv) {A, B} v) i[A, B]

where c is a complex number.

Given an operator A we define A A A, where A is the expectation value of A.

(k) Prove that the expectation value of a (i) Hermitian operator is purely real, while for an(ii) anti-Hermitian operator is purely imaginary.

(l) Show that [A, B] = [A, B](m) Prove Schwarz inequality: | | | | |2.(n) Prove the uncertainty relation: (A)2 (B)2 1

4| [A, B] |2, where A,B are Hermi-

tian.

Solution. The solution to this exercise is attached on page 7.

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Solution to Exercise 1.

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