P113A Homework 5

Peter D Alison

May 18, 2009

1 Problem 3.5

The hermitian conjugate (or adjoint) of an operator Q is the operatorQ† such that

< f |Qg >=< Q†f |g > (for all f and g).

(A hermitian operator, then is equal to its hermitian conjugate: Q = Q†.)

(a) Find the hermitian conjugates of x, i, and d/dx.

We can rearrange the integrand of the inner product as such

< f |xg >=∫ ∞−∞

f ∗xg dx =∫ ∞−∞

(xf)∗g dx =< xf |g >

The adjoint of the operator x† = x. Besides x is an observable so it is her-mitian so this was obvious.

< f |ig >=∫ ∞−∞

f ∗ig dx =∫ ∞−∞

(−if)∗g dx =< −if |g >

i† = −i

The adjoint of i is its complex conjugate, seems logical, makes sense.

For the operator ddx

we can use integration by parts to find the adjoint.

< f |∂g∂x

>=∫ ∞−∞

f ∗∂g

∂xdx = f ∗g|∞−∞ −

∫ ∞−∞

(∂f

∂x)∗g dx

1

The adjoint of ddx

is − ddx

.(b) Construct the hermitian conjugate of the harmonic oscillator raising op-erator, a+ (Equation 2.47).

a+ =1√

2hmω(−ip+mωx) =

1√2hmω

(−i(−ih ∂∂x

) +mωx)

a+ =1√

2hmω(−h ∂

∂x+mωx)

< f |a+g >=1√

2hmω(−h

∫ ∞−∞

f ∗(∂g

∂x) dx+mω

∫ ∞−∞

f ∗(xg) dx)

We can use the method of integration by parts to deconstruct the first inte-gral.

=1√

2hmω(−h(f ∗g|∞−∞ −

∫ ∞−∞

(∂f

∂x)∗g dx) +mω

∫ ∞−∞

(xf)∗g dx)

a†+ =1√

2hmω(h

∂

∂x+mωx) = a−

The adjoint of a+ is a−.

(c) Show that (QR)† = R†Q†.

< f |QRg >=< Q†f |Rg >=< R†Q†f |g >=< (QR†f |g >

(QR)† = R†Q†

2 Problem 3.7

(a) Suppose that f(x) and g(x) are two eigenfunctions of an operator Q, withthe same eigenvalue q. Show that any linear combination of f and g is itselfan eigenfuction of Q, with eigenvalue q.We assume that Q is a linear operator by definition, we are in the realm oflinear algebra, by the way. Q acting on f and g yield the eigenvalue q.

Qf = qf, and Qg = qg

2

Let us now take a linear combination of f and g and have Q act on it.

af(x) + bg(x)

Q(af(x) + bg(x)) = aQf(x) + bQg(x)

= aqf(x) + bqg(x) = q(af(x) + bg(x))

As we can see q is still an eigenvalue of the linear combination. QED.

(b) Check that f(x) = exp(x) and g(x) = exp(−x) are eigenfunctions ofthe operator d2/dx2, with the same eigenvalue. Construct two linear combi-nations of f and g that are orthogonal eigenfunctions on the interval (−1, 1).

d2

dx2[ex] = ex

andd2

dx2[e−x] = e−x

These functions both have the eigenvalue of 1. We can make the linearcombinations

1

2(ex + e−x) and

1

2(ex − e−x)

These functions are orthonormal because∫ ∞−∞

1

2(ex + e−x)

1

2(ex − e−x) dx = 0

3 Problem 3.11

Find the momentum-space wave function, Φ(p, t), for a particle in the groundstate of the harmonic oscillator. What is the probability (to 2 significantdigits) that a measurement of p on a particle in this state would yield a valueoutside the classical range (for the same energy)? Hint: Look in a math tableunder ”Normal Distribution” or ”Error Function” for the numerical part -or use Mathematica.The ground state of the harmonic oscillator is

ψ0(x) = (mω

πh)

14 e−

mω2h

x2

3

The time-dependent ground state wave function is then

Ψ0(x, t) = (mω

πh)

14 e−

mω2h

x2

e−12ıωt

The momentum-space wave function is given the Fourier transform

Φ(p, t) =1√2πh

∫ ∞−∞

e−ıpx/hΨ(x, t) dx

Φ(p, t) =1√2πh

∫ ∞−∞

e−ıpx/h(mω

πh)

14 e−

mω2h

x2

e−12ıωt dx

Using Mathematica this integral turns to be

Φ(p, t) = (1

mωπh)

14 e−

p2

2mωh e−12ıωt

The momentum p that is outside the classical range is√mωh. Because

the momentum-space wave function is even around zero and the probabilitymust sum up to 1, the probability that a measurement of p would outsidethe classical range can written as

1− 2∫ √mωh

0|Φ(p, t)|2 dp

Although the true way to write this would be∫ −√mωh

−∞|Φ(p, t)|2 dp+

∫ ∞√

mωh|Φ(p, t)|2 dp

Mathematica yields the Error Function value of

Erfc[1]

which numerically isP = 0.157.

4 Problem 3.14

Prove the famous ”Alison Uncertainty Principle”, relating the uncertainty inposition (A = x) to the uncertainty in energy (B = p2/2m+ V ):

σxσH ≥h

2m| < p > |.

4

For stationary states this doesn’t tell you much - why not?

The general Heisenberg uncertainty relation is

σ2Aσ

2B ≥ (

1

2ı< [A, B] >)2

We now must find the commutator of x and H.

x = x and H = − h2

2m

∂2

∂x2+ V

I will stick a dummy function on the commutator to make things easier.

[x, H]g(x) = x(− h2

2m)∂2g

∂x2− (− h2

2m)∂2

∂x2(xg)

= − h2

2mx∂2g

∂x2+h2

2m[∂

∂x(g + x

∂g

∂x)]

= − h2

2mx∂2g

∂x2+h2

2m(∂g

∂x+∂g

∂x+ x

∂2g

∂x2)

=h2

m

∂g

∂xRemoving the dummy function we obtain

[x, H] =h2

m

∂

∂x

which can be rewritten as

ıh

m(−ıh ∂

∂x) =

ıh

mp

Plugging into the uncertainty relation

σ2xσ

2H ≥ (

1

2ı| < ıh

mp > |)2

σ2xσ

2H ≥ (

1

2ı

ıh

m| < p > |)2 ⇒ σ2

xσ2H ≥ (

h

2m| < p > |)2

Finally,

σxσH ≥h

2m| < p > |

The Alison Uncertainty Relation, now if only we could find the elementAlisonium, atomic number 1200.

5

5 Problem 3.17

Apply Equation 3.71 to the following special cases: (a) Q = 1; (b) Q = H;(c) Q = p. In each case, comment on the result, with particular reference toEquations 1.27, 1.33, 1.38, and conservation of energy (comments followingEquation 2.39).Equation 3.71 states that

d

dt< Q >=

i

h< [H, Q] > + <

∂Q

∂t>

(a)Q = 1d

dt< 1 >=

ı

h< [H, 1] > + <

∂(1)

∂t>

=ı

h< H − H > +0 = 0

The conservation of normalization as in chapter 1.(b)Q = H

d

dt< H >=

ı

h< [H, H] > + <

∂

∂t(− h2

2m

∂2

∂x2+ V ) >

=ı

h< HH −HH > +0 = 0

The conservation of energy.(c)Q = x

d

dt< x >=

ı

h< [H, x] > + <

∂x

∂t>

=ı

h< (− h2

2m

∂2

∂x2+ V )x− x(− h2

2m

∂2

∂x2+ V ) >

From a previous problem

[x, H] =ıh

mp so [H, x] = − ıh

mp

d

dt< x >=

ı

h< − ıhp

m>

d

dt< x >=

< p >

m

6

This is the classical equation for momentum otherwise known as p = mv.(d)Q = p

d

dt< p >=

ı

h< [H, p] > + <

∂

∂t(−ıh ∂

∂x) >

What is [H, p]? Let’s use a dummy function to help a bit.

[H, p]g(x) = [(− h2

2m

∂2

∂x+ V (x))(−ıh ∂

∂x)− (−ıh ∂

∂x)(− h2

2m

∂2

∂x2+ V (x))]g(x)

= (− h2

2m

∂2

∂x+ V (x))(−ıh ∂

∂x)g(x)− (−ıh ∂

∂x)(− h2

2m

∂2

∂x2+ V (x))g(x)

= (− h2

2m

∂2

∂x+ V (x))(−ıh ∂g

∂x)− (−ıh ∂

∂x)(− h2

2m

∂2g

∂x2+ V (x)g)

ıh3

2m

∂3g

∂x3− ıhV (x)

∂g

∂x− ıh3

2m

∂3g

∂x3+ ıhV (x)

∂g

∂x+ ıhg

∂V

∂x

So

[H, p] = ıh∂V

∂x

d < p >

dt=ı

h< ıh

∂V

∂x> +0

d < p >

dt= − <

∂V

∂x>

which is Newton’s Law concerning conservative forces.

6 Problem 3.23

The Hamiltonian for a certain two-level system is

H = E(|1 >< 1| − |2 >< 2|+ |1 >< 2|+ |2 >< 1|),

where |1 >, |2 > is an orthonormal basis and E is a number with the dimen-sion of energy. Find its eigenvalues and (normalized) eigenvectors (as linearcombinations of |1 > and |2 >). What is the matrix H representing H withrespect to this basis?Let us take the wave function that is a linear combination of the vectors

ψ = a|1 > +b|2 >

7

H|ψ >=

E(a|1 >< 1|1 > +b|1 >< 1|2 > −a|2 >< 2|1 > −b|2 >< 2|2 >

+a|1 >< 2|1 > +b|1 >< 2|2 > +a|2 >< 1|1 > +b|2 >< 1|1 >)

= E(a|1 > −b|2 > +b|1 > a|2 >) = E((a+ b)|1 > +(a− b)|2 >)

The term on |1 > went from a to a + b, and the |2 > went from b to a − b.The matrix H is then

E

(1 11 −1

)

7 Problem 3.27

Sequential Measurements. An operator A, representing observable A, hastwo normalized eigenstates ψ1 and ψ2, with eigenvalues a1 and a2, respec-tively. Operator B, representing observable B, has two normalized eigen-states φ1 and φ2, with eigenvalues b1 and b2. The eigenstates are relatedby

ψ1 = (3φ1 + 4φ2)/5, ψ2 = (4φ1 − 3φ2)/5.

(a) Observable A is measured, and the value is a1 is obtained. What is stateof the system (immediately) after this measurement?

The measurement value of a1 corresponds to ψ1, so the system is in thestate ψ1.

(b) If B is now measured, what are the possible results, and what are theirprobabilities?

The possible results for B are b1 and b2 and because we are in the stateψ1 their probabilities are their corresponding φ1 and φ2 constants squared.

Prob.b1 =9

25

Prob.b2 =16

25

(c) Right after the measurement of B, A is measured again. What is theprobability of getting a1? (Note that the answer would be quite different if

8

I had told you the outcome of the B measurement.)

First we solve for φ1 and φ1 in terms of ψ1 and ψ2.

φ1 =1

5(3ψ1 + ψ2)

φ2 =1

5(4ψ1 − 3ψ2)

Now, we need to find the probabilities of measuring a1 in each of possiblestates for B.

Prob.a1inφ1 =9

25

Prob.a1inφ2 =16

25

Now we multiply the probability of a1 with the probabilities of the possibleB states and then add. So the probability of finding a1 is

(9

25)2 + (

16

25)2 =

337

625= 0.5392

8 Problem 3.38

The Hamiltonian for a certain three-level system is represented by the matrix

H = hω

1 0 00 2 00 0 2

Two other observables, A and B, are represented by the matrices

A = λ

0 1 01 0 00 0 2

,B = µ

2 0 00 0 10 1 0

where ω, λ, and µ are positive real numbers.

(a) Find the eigenvalues and (normalized) eigenvectors of H, A, and B.

9

H:

(H− cI3) =

hω − c 0 00 2hω − c 00 0 2hω − c

det(H− cI3) = 0 = (2hω − c)(2hω − c)(hω − c)

c = 2hω, 2hω, hω

These are the values down the diagonal so the eigenvectors are

|h1 >=

100

, |h2 >=

010

, |h3 >=

001

A:

(A− cI3) =

−c λ 0λ −c 00 0 2λ− c

det(A− cI3) = 0 = (2λ− c)(c2 − λ2) = (2λ− c)(c− λ)(c+ λ)

The eigenvalues arec = 2λ, λ,−λ

Plugging in ±λ we obtain the systems −λ λ 0λ −λ 00 0 0

and

λ λ 0λ λ 00 0 0

which gives the eigenvectors

|a1 >=1√2

110

, |a2 >=1√2

1−10

and by default

|a3 >=

001

10

B:

(B− cI3) =

2µ− c 0 00 −c µ0 µ −c

det(B− cI3) = 0 = (2µ− c)(c2 − µ2) = (2µ− c)(c− µ)(c+ µ)

The eigenvalues arec = 2µ, µ,−µ

Plugging in ±µ we obtain the system 0 0 00 −µ µ0 µ −µ

and

0 0 00 µ µ0 µ µ

which yield the eigenvectors

|b2 >=1√2

011

, |b3 >=1√2

01−1

by default the last eigenvector is

|b1 >=

100

(b) Suppose the system starts out in the generic state

|S(0) >=

c1

c2

c3

with |c1|2 + |c2|2 + |c3|2 = 1. Find the expectation values (at t = 0) of H, A,and B.

< H >=< S(0)|H|S(0) >

11

= hω(c∗1 c∗2 c∗3

) 1 0 00 2 00 0 2

c1

c2

c3

= hω(c∗1 c∗2 c∗3

) c1

2c2

2c3

< H >= hω(|c1|2 + 2|c2|2 + 2|c3|2)

< A >=< S(0)|A|S(0) >

= λ(c∗1 c∗2 c∗3

) 0 1 01 0 00 0 2

c1

c2

c3

= λ(c∗1 c∗2 c∗3

) c2

c1

2c3

< A >= λ(c∗1c2 + c∗2c1 + 2|c3|2)

< B >=< S(0)|B|S(0) >

= µ(c∗1 c∗2 c∗3

) 2 0 00 0 10 1 0

c1

c2

c3

= µ(c∗1 c∗2 c∗3

) 2c1

c3

c2

< B >= µ(2|c1|2 + c∗2c3 + c∗3c2)

(c) What is |S(t) >? If you measured the energy of this state (at timet), what values might you get, and what is the probability of each? Answerthe same questions for A and for B.The time-dependent state is

|S(t) >= c1e−ıE1/ht|h1 > +c2e

−ıE2/ht|h2 > +c3e−ıE3/ht|h3 >

= c1e−ıωt|h1 > +c2e

−ı2ωt|h2 > +c3e−ı2ωt|h3 >

12

The possible values are the eigenvalues of the operator H.

hω, 2hω

So the probabilities are the corresponding constants absolute value squared.

E = hω ⇒ Prob. = |c1|2

E = 2hω ⇒ Prob. = |c2|2 + |c3|2

To calculate the probabilites for the operators A and B, we perform a pre-form a projection on |S(t) > with corresponding eigenvectors.

On the operator A the possible values are the eigenvalues

2λ, λ,−λ

For c = λ

< a1|S(t) >=1√2

(1 1 0

) c1e−ıωt

c2e−2ıωt

c3e−2ıωt

=

1√2

(c1e−ıωt + c2e

−2ıωt)

Prob. = | 1√2

(c1e−ıωt + c2e

−2ıωt)|2

=1

2(|c1|2 + |c2|2c∗1c2e

−ıωt + c∗2c1eıωt)

For c = −λ

< a2|S(t) >=1√2

(1 −1 0

) c1e−ıωt

c2e−2ıωt

c3e−2ıωt

=1√2

(c1e−ıωt − c2e

−2ıωt)

Prob. = | 1√2

(c1e−ıωt − c2e

−2ıωt)|2

=1

2(|c1|2 + |c2|2 − c∗1e−ıωt − c∗2c1e

ıωt)

13

For c = 2λ

< a3|S(t) >=(

0 0 1) c1e

−ıωt

c2e−2ıωt

c3e−2ıωt

= c3e

−2ıωt

Prob. = |c3e−2ıωt|2

= |c3|2

The possible values for the operator B are the eigenvalues

2µ, µ,−µ

For c = µ

< b2|S(t) >=1√2

(0 1 1

) c1e−ıωt

c2e−2ıωt

c3e−2ıωt

=

1√2

(c2e−2ıωt + c3e

−2ıωt)

Prob. = | 1√2

(c2e−2ıωt + c3e

−2ıωt)|2

=1

2(|c2|2 + |c3|2 + c∗1c2 + c∗2c1)

For c = −µ

< b3|S(t) >=1√2

(0 1 −1

) c1e−ıωt

c2e−2ıωt

c3e−2ıωt

=

1√2

(c2e−2ıωt − c3e

−2ıωt)

Prob. = | 1√2

(c2e−2ıωt − c3e

−2ıωt)|2

=1

2(|c2|2 + |c3|2 − c∗2c3 − c∗3c2)

For c = 2µ

< b1|S(t) >=(

1 0 0) c1e

−ıωt

c2e−2ıωt

c3e−2ıωt

14

= c1e(−ıωt

Prob. = |c1e−ıωt|2

= |c1|2

A lot of probabilities finally done. QED.

15