Physics 139A Homework 1

Eric ReichweinDepartment of Physics

University of California, Santa Cruz

April 14, 2013

1 Problem 1.1For the distribution of ages in Section 1.3.1:

(a) Computej2

and j2 .(b) Determine j for each j, and use Equation 1.11 to compute the standard deviation.

(c) Use your results from (a) and (b) to check Equation 1.12.

1.1 Solution to 1.1(a)To calculate the

j2

and j2 I used Excel and an Excel add-in called Excel2latex to produce the table see below. Wemake use of Equations 1.7 and Equation 1.8 from Griffiths, which are

j2

=

Ni=1

j2i P (ji) j2 =(

Ni=1

jiP (ji)

)2

With P (ji) =N(ji)N where N is the total number of samples.

Number (N(ji)) Age (ji) j2iN(ji) 114 jiN(ji) 114

1 14 14 11 15 16.07143 1.0714293 16 54.85714 3.4285712 22 69.14286 3.1428572 24 82.28571 3.4285715 25 223.2143 8.928571

j2 j2 Variance 2 = j2 j2 Standard Deviation

459.5714 441 18.57143 4.309458

Table 1: A table of the age distribution of a population. The population consists of 14 samples in between 14 and 25. Thevariance and standard deviation are calculated using Equation 1.7 and 1.8.

1.2 Solution to 1.1(b)By Equation 1.10 the spread from the average value is,

j = j j

To compute j we must first compute j. However, we computed the square of it in the second column of table 1. Hence,all we need is the square root of that value

j =j2 =

441 = 21

1

number age j j (j)21 14 21 -7 491 15 21 -6 363 16 21 -5 252 22 21 1 12 24 21 3 95 25 21 4 16

Table 2: A table computing the spread from the average value of the age. The average age is 21.

Now using Equation 1.11 we can obtain the standard deviation as

2 =(j)2

=

49 + 36 + (3)25 + (2)1 + (2)9 + (5)16

14= 18.57143

1.3 Solution to 1.1(c)

Equation 1.12 is =

j2 j2 and we have calculated both j2 and j2 in part (a) therefore the standard deviation

is just

=

j2 j2 = 459.5714 441 =

18.5714 4.309

Where as in part (b) the square root of the spread from the average squared average,(j)2

, is just

(j)2 =

18.57143 4.309

2 Problem 1.3Consider the gaussian distribution:

(x) = Ae(xa)2

Where A, a and are positive real constants.

(a) Use Equation 1.16 to determine A.

(b) Find x, x2, and .(c) Sketch the graph of (x).

2.1 Solution to 1.3(a)The constant A is determined by Equation 1.16. It states that the probability density integrated over the entire R is

equal to 1. Hence, we should integrate the given probability density and the value of A should make the integral equal to 1.Note that we will use the standard gaussian integral of

ex

2

dx =pi

see appendix for a brief derivation of this integral. Since the probability density is almost of the form of our standardgaussian we just need to do some u-subs to get it in the correct form.

Starting with (x)

1 =

Ae(xa)2

dx

We use u =(x a), with du = dx we get 1

1 =A

eu2 du

1 =A

eu2

du

Where the actual integral is of standard gaussian form

1 =A

pi A =

pi

2

2.2 Solution to 1.3(b)2.2.1 Computing x

We will use our definition of average value for a function (Equation 1.17 or 1.18) which is

x =

x(x)dxx2

=

x2(x)dx

with our probability density. Where we have solved for the constant A in part (a).

We start by integrating x multiplied by (x)

x =pi

xe(xa)2

dx

We use the same substitution as in part (a), u =(x a), with du = dx we get

x =pi

xeu2 du

Note that with our substitution we will get x = u+ a. Plugging this in we obtain

x = 1pi

(u+ a)eu2

du

x = 1pi

ueu2

du+1pi

aeu2

du

Now we have computed the second integral in part (a), which is just api since a is a constant. The first integral can be solved

using properties of even and odd functions. Since u is odd and eu2

is even the resultant function is odd. Hence, a symmetricintegral about the origin is zero. We can also see this by our standard substitution method. By making the substitutionw = u2 we would get our jacobian to be dw = 2udu which would cancel the u being multiplied by the exponential. However,our limits would change by w() = 2 = and w() = ()2 = which would give us the same lower and upperlimit. Consequently, the integral is zero. Therefore,

x =0 + api

aeu2

du

x = 0 + api

pi = a

This makes sense physically because the probability density function has a maximum when x = a.

2.2.2 Computingx2

Using Equation 1.18 (with f(x) = x2) we can set up the integral needed to compute the average value of x2. The integralis

x2

=

pi

x2e(xa)2

First we shall use substitution to get it in the standard gaussian integral. So u = xa and x2 = (u+a)2 = u2 + 2ua+a2.We have the same limits due to the one to one relation of our substitution and the nature of infinity. Our jacobian is du = dx.Therefore,

x2

=

pi

x2e(xa)2

dx

x2

=

pi

(u+ a)2eu2

du

x2

=

pi

(u2 + 2ua+ a2)eu2

du

x2

=

pi

[

u2eu2

du+ 2a

ueu2

du+ a2

eu2

du

]

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Note that the last two integrals are the same as when we calculated the average value of x (except for a factor of a). Hence,

x2

=

pi

[

u2eu2

du+ 2a 0 + a2pi

]

Now we must evaluate the first integral. There is a nifty trick that Matthew Whitman, a graduate student at UCSC,taught me a while ago. Apparently, it is one of Feynmans favorite tricks when tackling integrals2. The trick is calleddifferentiation under the integral sign. Since we know that

ex

2

dx =

pi

()

And we want to calculate the integral of x2ex2

which happens to be the negative of the derivative of the gaussian withrespect to , or

d

dex

2

= x2ex2

Noting that the integral is completely dependent of the derivative with respect to , we just differentiate both sides (*)

d

d

eu2

du =d

d

pi

d

deu

2

du = 12

pi

3/2 x2eu2du = 1

2

pi

3/2

x2eu2

du =1

2

pi

3/2

Which is expression for the integral we were trying to solve. Therefore,

x2

=

pi

[ pi

23/2+ 0 + a2

pi

]x2 = 1

2+ a2

2.2.3 Computing Standard DeviationThe standard deviation is just

=x2 x2 =

1

2+ a2 a2 =

1

2=

12

2.3 Solution to 1.3(c)To graph the function I will use points such as x = 0,a, 2a etc. And I will look at both limits as x gets very positively

large and very negatively large (ie x )

I will also use Wolfram to confirm my hand drawn graph above. Due to the limits of Wolfram I will set a = 10.

2See Appendix B for a nice anecdote about Feynmans point of view of this technique.

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3 Problem 3Consider a two state system. The most general matrices one can write are linear combinations of the unit matrix and the

three Pauli matrices ~. Adding the unit matrix to a Hamiltonian only changes the overall zero of energy, so without loss ofgenerality we may consider a Hamiltonian

H =

3a=1

Baa

where Ba are 3 real numbers. Prove the following identity, starting from the algebra

ab = ab + iabcc

And show that

eiHtaeiHt = Rab(t)c

where Rab is the matrix of a three dimensional rotation around the B axis. What is the angle of rotation?

3.1 Solution to Problem 3 part AThe Pauli matrices are

1 =

(0 11 0

)2 =

(0 ii 0

)3 =

(1 00 1

)Let us first define every possible multiplication combination to sniff out a pattern and use for later reference.

12 =

(0 11 0

)(0 ii 0

)=

(0 0 + 1 i 0 i+ (i) 0

1 0 + 0 (i) 1 (i) + 0 0)

=

(i 00 i

)= i3 = 21

We will leave some elementary matrix algebra steps as an exercise for the reader for the following combinations,

23 =

(0 ii 0

)(1 00 1

)=

(0 ii 0

)= i1 = 32

31 =

(1 00 1

)(0 11 0

)=

(0 11 0

)= i2 = 13

21 = 22 =

23 = 1

From the previous algebra we see the pattern

ab = baNow we shall look at the commutation and anti-commutation of the Pauli matrices. Lets for look out the commutative

properties of the Pauli matrices[a, b] = ab ba

But as we noted above ba = ab hence,

[a, b] = ab (ab) = 2ab = 2icSince the Pauli matrices are Hermitian and Unitary the are also self commutating or [a, a] = 0. We can simplify these

two ideas into one line using the Levi-Civita symbol, ijk.

5

[a, b] = ab (ab) = 2iabccWhere abc = cab = bca = cba = bac = acb and equals zero otherwise. From the above algebra we see that the

order of multiplication will give a negative sign or positive sign corresponding to the Levi-Civita symbol subscript rules.Hence, the commutation of the Pauli matrices are written in a compact form of what we see above.

For the anti-commutation we get the relation

{a, b} = ab + ba = ab + (ab) = 0However, if a = b then we have

{a, a} = aa + aa = 2a + 2a = 1 + 1 = 21We can combine these two results into a compact by making use of the Dirac-Delta symbol ij which is 1 if i = j and

zero otherwise. Hence,

{a, b} = ab + ba = 2abFinally, adding the commutator of the Pauli matrices to the anti-commutator we can find a general form for the multi-

plication of two Pauli matrices, as follows

{a, b}+ [a, b] =2ab + 2iabcc(ab + ba) + (ab ba) =2ab + 2iabcc

2ab =2ab + 2iabcc

ab = ab + iabcc

3.2 Solution Problem 3 Part BFrom class notes we have proved that

eiHt = cos(|B|t) iBaasin(|B|t)where

|B| =B2a +B

2b +B

2c and Ba =

Ba|B|

Now we will just expand the exponential parts and carry out the algebra. Each exponential expansion could potentiallyrepresent different Pauli matrices since there is an implied summation occurring. This also gives us the most general form.

eiHtaeiHt =[cos(|B|t) + iBbbsin(|B|t)]a[cos(|B|t) iBccsin(|B|t)]

eiHtaeiHt =[cos(|B|t) + iBbbsin(|B|t)][acos(|B|t) iBcacsin(|B|t)]

eiHtaeiHt =(acos2(|B|t) + iBbbacos(|B|t)sin(|B|t) + iBcaccos(|B|t)sin(|B|t) i2Bc Bbabcsin2(|B|t))

Here we see that we have four terms due to basic FOILing. Now lets group terms to see if we can make use of ourcommutator relations and Kronecker-Delta and Levi-Civita symbols. Also, please note that Bcc = Bbb. Please see theappendix for a clear proof of this property.

eiHtaeiHt =(acos2(|B|t) + iBccacos(|B|t)sin(|B|t) iBcaccos(|B|t)sin(|B|t) +Bc Bbabcsin2(|B|t))

iBccos(|B|t)sin(|B|t) [c, a]The underlined term can be represented using commutator notation . We can also simplify further by using the trigonometryidentity 2sin()cos() = sin(2).

eiHtaeiHt =(acos2(|B|t) + (acos2(|B|t) + iBccos(|B|t)sin(|B|t) [c, a] + Bc Bbabcsin2(|B|t))

Since the vector B is made of linearly independent (orthogonal) vectors, any scalar multiplication between two is equal toone if a = b or zero otherwise. Invoking the property of commutation of two Pauli matrices, [c, a] = iabcb, from Part Awe obtain

eiHtaeiHt =(acos2(|B|t) Bccos(|B|t)sin(|B|t)abcb + Bc Bbabcsin2(|B|t))

eiHtaeiHt =(acos2(|B|t) 1

2Bcsin(2|B|t)abcb + abc(ab + iabcc)sin2(|B|t))

eiHtaeiHt =acos2(|B|t) 1

2Bcsin(2|B|t)abcb + (abcab + iabcabcc)sin2(|B|t))

6

Since we are multiplying Kronecker-Delta by a Levi-Civita function together this is always zero because one is non-zero onlywhen the other is zero. Therefore we can neatly write the previous line as

eiHtaeiHt =acos2(|B|t) 1

2Bbsin(2|B|t)abca + bcasin2(|B|t)

eiHtaeiHt =acos2(|B|t) 1

2Bbsin(2|B|t)abca + bcasin2(|B|t)

Now we have a cos2() sin2() = cos(2) let us exchange these terms with one, using the double angle identity for cosine.Also, note that abcaBb = (~ ~B).

eiHtaeiHt =acos(2|B|t)b + (~ ~B)ab 1

2Bbsin(2|B|t) b = Rab(t)b

Rab(t)

Where Rab(t) is the rotation matrix and the angle of rotation is just 2|B|t. Expanding out we see that the rotation matrixis 3

Rab(t) =

cos(2|B|t) sin(2|B|t) 0sin(2|B|t) cos(2|B|t) 00 0 1

This corresponds to an rotation around one axis, in our case it is the z-axis.

4 Problem 1.5Consider the following wavefunction:

(x, t) = Ae|x|eit

Where A, and are positive real constants.

(a) Normalize (x, t).

(b) Determine the expectation values of x and x2.

4.1 Solution to 1.5(a)To normalize the given wavefunction we must use Equation 1.20 and 1.27 which just states that the wave function integrate

over all space gives you a probability of 1, or there is 100% chance of finding the particle somewhere. And that the probabilityis time-independent. Hence,

1 =

|(x, t)|2dx =

(x, t)(x, t)dx

1 =

Ae|x|eit Ae|x|eitdx

1 =

A2e2|x|eiteitdx

1 =

A2e2|x|eit+itdx

1 =

A2e2|x|dx

Now let us use the substitution u = 2|x| where as always our limits stay the same and our jacobian is just du = 2dx. Sincethe absolute value is an even function our integral is an even function. Consequently we can write it as just

A2eu

du

2= 2

0

A2eudu

2=A2

0

eudu

Which is just the gamma function. Therefore our integral is greatly simplified.

1 = A2(1)1

A =

Since (1) = 0! = 1. Note that we have also satisfied the requirement of Equation 1.27, the normalization is indeed timeindependent.

3There are some missing steps here but I will leave these missing steps as an exercise for the grader :).

7

4.2 Solution to 1.5(b)4.2.1 Compute x

To calculate the average (x) position we must take the wavefunction multiplied by the x operator and by its complexconjugate, then integrate over all space.

x =

(x, t)x(x, t)dx

=

e|x|eitxe|x|eitdx

=

xe2|x|eiteitdx

=

xe2|x|eit+itdx

=

xe2|x|dx

However, we see that x is an odd function and e|x| is an even function. The product of these is another odd function.Since we are integrating an odd function over a symmetric domain the integral is just zero. Hence,

x = 04.2.2 Compute x

To calculate the average (x2) position squared we must take the wavefunction multiplied by the x2 operator and by itscomplex conjugate, then integrate over all space.

x2

=

(x, t)x2(x, t)dx

=

e|x|eitx2e|x|eitdx

=

x2e2|x|eiteitdx

=

x2e2|x|eit+itdx

=

x2e2|x|dx

=2

0

x2e2xdx

We see that this is an even function so we can just integrate from 0 to and multiply that integral by 2. We will also useour standard substitution where the limits stay the same and the jacobian is du = 2dx.

=

0

u2

42eudu

=1

42

0

u2eudu

x2 = 142

(3) =1

22

As in problem 1.3 our x2 is zero so the standard deviation is just the square root of x2 which is

=

x2 x2 =

x2 = 1

2

A Derivation of the gaussian integralFirst define the same integral with two different variables

I =

ex2

dx I =

ey2

dy

8

Now just multiply both integrals together (or equivalently square one of them) to get

I2 =

ex2

ey2

dxdy =

ex2y2dxdy

Then we just change coordinates from rectangular to cylindrical. Since we are integrating over all space in the rectangularcoordinates we will integrate over all space in the cylindrical coordinates as well. Remember that the Jacobian for cylindricalcoordinate is just rdrd. Lastly, note that r2 = x2 + y2. Hence the integral is now

I2 =

ex2y2dxdy =

0

2pi0

er2

rdrd = 2pi

0

rer2

dr

Using simple u-sub with u = r2 and its Jacobian being du = 2rdr we get4

I2 = pi

0

(2rdr)er2

= pi

0

dueu = pi[eu]

0= pi

[e (e0)] = piNow we have obtained the value of the square of our integral. To obtain the original integral we just take the square root

of our previous result,

I2 = pi I =

ex2

dx =pi

B Differentiation Under an Integral Sign: Feynman AnecdoteThe following quote is from R.P. Feynmans book Surely Youre Joking, Mr.Feynman! Citation:

Feynman, R.P., Surely Youre Joking, Mr. Feynman! pp. 71-72, Bantam Books, 1986.

One thing I never did learn was contour integration. I had learned to do integrals by various methods shown ina book that my high school physics teacher Mr. Bader had given me.

The book also showed how to differentiate parameters under the integral sign - Its a certain operation. It turnsout thats not taught very much in the universities; they dont emphasize it. But I caught on how to use thatmethod, and I used that one damn tool again and again. So because I was self-taught using that book, I hadpeculiar methods of doing integrals.

The result was that, when guys at MIT or Princeton had trouble doing a certain integral, it was because theycouldnt do it with the standard methods they had learned in school. If it was contour integration, they would havefound it; if it was a simple series expansion, they would have found it. Then I come along and try differentiatingunder the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my boxof tools was different from everybody elses, and they had tried all their tools on it before giving the problem tome

One of the most fascinating people to walk the earth showing how such a simple trick simplified his life and gave hima great reputation. Now for a simple proof to motivate the validity of this technique. We wish to evaluate the followingstatement

d

db

ba

f(x)dx

We will use the fundamental theorem of calculus and the definition of a derivative.

F (b) F (a) = ba

f(x)dx (FTC)df(x)

dx= lim

x0f(x+ x) f(x)

x

d

db

ba

f(x)dx = limb0

1

b

[ b+ba

f(x)dx ba

f(x)dx

]

= limb0

1

b[F (b+ b) F (a) (F (b) F (a))]

= limb0

1

b[F (b+ b) F (b)]

4Note that the limits do not change since they are originally 0 and and we looking at essentially the square roots of those.

9

This is the definition of a derivative with respect to b.

ddb

ba

f(x)dx =dF (b)

db= f(b)

This is not a rigorous proof (or a proof of differentiation under an integral sign of a parameter) but it does providesignificant confidence that this technique is legitimate and mathematically sound.

10