Class 1 M Aug 26
read chapter 1
problems: 1.3, 1.4, and below Web problem
Note: in order to complete problem 1.3 you'll need the "gaussian integral".
The third integral from the bottom on the back cover of Griffith (for the case n=0)
is one version of this famous integral. Another version can be found as the first
entry in qmcheat.pdf. For this latter version, don't be tricked into expanding
the problem's (x-a)^2 to give a linear term in x (like the beta term on the cheat sheet).
Rather deal with (x-a) with a change-of-variables. In any case this gaussian
integral (in whatever form you like) should be tattooed on your left arm
just below where you tattooed Maxwell's equations last year; it will occur frequently.
(The gaussian integral is featured in many problems this semester because it
provides an easy way to write an equation for a particle blob; you see the
alternative in problem 1.4: a discontinuous function.)
Note2: Mathematica does a reasonable job dealing with this integral:
In[1]:= Integrate[Exp[-lambda (x-a)^2],{x,-Infinity,+Infinity}]
Sqrt[Pi]
Out[1]= ConditionalExpression[-------------, Re[lambda] >= 0]
Sqrt[lambda]
but it cannot afford to make the reasonable assumption that lambda is
real and positive (lambda negative is a real disaster; for lambda=I
the integrand does not go to zero for large |x|). I encourage you
to use Mathematica on homework; do include printouts or carefully
copied expressions that display exactly what you entered so I can grade them
properly. To use Mathematica you may have to renew your Mathematica license
on your personal computer or use the campus computers.
Web problem:
On this web site, in the data folder, you'll find 10 distinct data files
each with 1000 data points. Assume each file shows the measured location (in cm)
of a particle in an ensemble of 1000 identical systems. (The data in these
files have been sorted from low to high to ease the analysis; of course,
when measured the values were not so ordered, instead seemingly random.)
Plot the probability density for this wave function; include units of probability
density. (For extra credit, show the error bars for your estimate of the
probability density function.)
Select the data file 1000_X.dat where "X" matches the last digit of
your ID number. Note that the files 1000_Xw.dat are identical
to 1000_X.dat, except that the data have been wrapped for easy
printing. (There is no reason you should print the data, in fact,
for many purposes the long list of numbers is easier to work with.)
Recall: This problem is very similar to one you did in the Bubble Chamber Lab,
except here we seek probability density not just a histogram.
In a spreadsheet it would be very easy to calculate the AVERAGE() and STDEV()
of these locations. Write down the formal expression that would give these same
quantities from the probability density.
Class 2 W Aug 28
read chapter 2 thru p.40
http://www.physics.csbsju.edu/QM/square.01.html
1.17, 2.4
Class 3 F Aug 30
http://www.physics.csbsju.edu/QM/square.02.html
http://www.physics.csbsju.edu/QM/square.03.html
2.5, 2.7
Class 4 M Sep 2
read 2.3; my suggestion is to initially skip 2.3.1 jumping to 2.3.2 and then come back to 2.3.1
read web site:
start: http://www.physics.csbsju.edu/QM/shm.01.html
final: http://www.physics.csbsju.edu/QM/shm.05.html
also: http://www.physics.csbsju.edu/QM/shm.08.html
In the notation of Mathematica the normalized SHO wavefunctions (in dimensionless units)
can be expressed as:
psi[n_,x_]=HermiteH[n, x] Exp[-x^2/2] /Sqrt[ Sqrt[Pi] 2^n n!]
partA:
for three different values of n (your choice) use Mathematica to demonstrate that:
=1
partB:
for three different distinct pairs {n,m} (your choice) use Mathematica to demonstrate that:
=0
i.e., you've now presented examples of the orthonormality of the wavefunctions
partC:
the TISE is: H psi_n = E_n psi_n, which we can rearrange as:
(H psi_n)/psi_n = E_n
where the main point is E_n (eigenenergy) is a number (constant) not a function
The dimensionless version of (H psi_n) in the notation of Mathematica is:
(-D[D[psi[n,x],x],x] + x^2 psi[n,x])
For n=0,1,2,3 use Mathematica to calculate
(H psi_n)/psi_n
from your result confirm the expression the web site has for the dimensionless E_n
Note: Simplify[ ]
partD:
For n=0,1,2,3
Use Mathematica to calculate each of the following integrals. Use your results to
suggest what the result would be for any value of n
Note: with these dimensionless coordinates in the notation of Mathematica
p psi = (1/I) D[psi,x]
p^2 psi = -D[D[psi,x],x]
partE:
For n=2 and m=0,1,2,3,4 calculate the following integrals. Use your results to
suggest what the result would be for any value of m
Class 5 W Sep 4
Griffiths: 2.12, 2.13
The aim of these problems is practice the skill of "integration by operator
methods", so you must use the ladder operators a+ a- (or class notation a^dagger a)
to solve these problems. Feel free to use the dimensionless forms we used in class
or the dimensioned forms in Eq. 2.69: its really exactly the same.
Class 6 F Sep 6
read 2.4
Griffiths: 2.19 & 2.21
RE: "Discuss"-- as a grows, how does the uncertainty in x change? uncertainty in k?
The following problem is essentially problem#20 on http://www.physics.csbsju.edu/QM/fall.15.html:
The following solves the TDSE for a gaussian wave function with typical wavenumber k
(k and sigma are constants). [proving this was basically #19 in fall.15.html]
psi[y_,t_]=Exp[-(2 k t -y)^2/(4 (sigma^2 + I t)) + I k y - I k^2 t]/Sqrt[sigma^2 + I t]
Using your own words define group velocity and phase velocity.
For sigma=1, k=2 Pi, plot the real part (Re[]) of the above function at
t=0 and t=1. Estimate the group velocity from these plots.
Plot again at t=.05 and t=.1 to estimate the phase velocity.
Try:
Animate[Plot[Re[psi[x, t]], {x, -5, 20}, PlotRange -> {-1, 1}], {t, 0, 1}]
note that the probability density |psi|^2 lacks the waves:
Animate[Plot[Abs[psi[x, t]]^2, {x, -5, 20}, PlotRange -> {-1, 1}], {t, 0, 1}]
Can you see that the wave spreads out as it travels?
Class 7 M Sep 9
read 2.5
http://www.physics.csbsju.edu/QM/delta.01.html
Griffiths: 2.23 & 2.24a
(I'm not sure why 23 gets a * ; if it seems at all difficult, come talk to me)
Class 8 W Sep 11
finish reading ch 2
http://www.physics.csbsju.edu/QM/square.03.html
http://www.physics.csbsju.edu/QM/square.04.html
http://www.physics.csbsju.edu/QM/square.05.html
http://www.physics.csbsju.edu/QM/square.06.html
Griffiths: 2.34
Find the bound state energy levels (i.e., stationary states
with E<0) of a proton in a finite square well with depth=50MeV,
full width=2a=10fm. Begin by finding the dimensionless quantity z_0,
using Mathematica to solve Eqn 2.156 and the equivalent equation
for the odd states, and then converting the resulting z
to energy (in MeV). Having Mathematica plot both sides
of the equation to be solved (with FindRoot[]), allows you to
to see the solutions and thus give Mathematica a better initial
guess at a solution.
Class 9 F Sep 13
Read: Appendix (review of vector spaces), start Ch 3
HW:
Appendix A: A.8, A.9, A.25abce, A.26abc
Rotation matrices are used to calculate the components of a vector in a new rotated coordinate system.
Memorably rotation in 2d results in the matrix:
cos(theta) sin(theta)
-sin(theta) cos(theta)
Note that the determinant of this matrix is 1, the trace is 2*cos(theta), and the transpose of the
matrix is its inverse.
Rotation matrices in 3d operate the same way but are more complicated.
But consider: for any rotation there should be a direction (the rotation axis) which in
left unchanged by the rotation, and if we put that axis on the z axis the matrix must be like:
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1
and thus still have determinate equal 1 and now trace equal 1+2*cos(theta)
Consider (using Mathematica) the matrix:
m={{0.75, -0.612372, 0.25}, {0.612372, 0.5, -0.612372}, {0.25, 0.612372, 0.75}}
Find the rotation axis, i.e., the vector v such that m.v=v
Hint: what eigenvalue is being displayed?
Find the rotation angle involved.
Find:
MatrixPower[m,6]
and explain why the result could have been predicted given the previous answers.
Find:
Transpose[m].m
Note that Mathematica is good at matrices. The matrix:
a b c
M = d e f
g h i
in Mathematica is spelled:
m={{a,b,c},{d,e,f},{g,h,i}}
Technically a column vector should be written:
u={{w},{x},{y}}
but usually
v={r,s,t}
will work both with m.v and v.m
Note:
Eigensystem[], Eigenvectors[], Eigenvalues[]
Det[], Inverse[], Tr[]
Transpose[], ConjugateTranspose[], Conjugate[]
Note2: Mathematica is up for some complex tricks like:
sig={{0,1},{1,0}} I
Sum[MatrixPower[sig,k]/k!,{k,0,Infinity}]
ConjugateTranspose[%].%
Simplify[%]
which is behind problem A.28
You really should be able to most of these problems "by hand"
(think: exam). I admit that A.26bc do require quite a bit of algebra.
While I hope you will do all of these by hand (and perhaps
check results with Mathematica), I will not object to Mathematica
only solutions.
Class 10 M Sep 16
chapter 3
old exam: http://www.physics.csbsju.edu/tk/346/346t108.pdf #6
3.13
prove the following formulae for commutators:
[p,x^2]f(x)= (hbar/i)2x f(x)
[p,V(x)]f(x)= (hbar/i)V' f(x)
[p^2,x]f(x)= 2 (hbar/i)p f(x)
[p^n,x]f(x)= n(hbar/i)p^(n-1) f(x) (hint: use induction)
Class 11 W Sep 18
more chapter 3
old exam: http://www.physics.csbsju.edu/tk/346/346t108.pdf #1
3.37, 3.38
Class 12 F Sep 20
3.31
Class 13 M Sep 23
separation of variables in 3d: infinite square wells & SHO
http://www.physics.csbsju.edu/QM/square.07.html
http://www.physics.csbsju.edu/QM/square.13.html
start of pages which deal with xyz separation of SHO
http://www.physics.csbsju.edu/QM/shm.09.html
http://www.physics.csbsju.edu/QM/shm.12.html
the diagram at the top of
http://www.physics.csbsju.edu/QM/shm.10.html
shows that there 5 degenerate states with E'=10 for the 2-d SHO
Record the nx & ny values of these 5 degenerate states
there are 15 degnerate states with E'=11 for the 3-d SHO
Record the nx, ny, & nz values of these 15 degenerate states
http://www.physics.csbsju.edu/QM/square.19.html #9
Class 14 W Sep 25
no homework!
Class 15 F Sep 27
In the below I'm using the dimensionless versions of the L operators;
multiply by hbar if you really want to get the dimensioned version of these operators
We showed in class that:
Sin[theta]^L Exp[I L phi] is an eigenfunction of L^2 with eigenvalue L(L+1)
and an eigenfunction of L_z with eigenvalue L
Note possible confusion: I'm using above "L" both for the integer eigenvalue and
for the operator. The integer is elsewhere always denoted with a lowercase L=l
but that looks so much like a one=1 that for this paragraph I've used uppercase=L.
We will soon get to the case L=l=5 so this confusion exists just in the above sentence.
and that LegendreP[L,Cos[theta]] is an eigenfunction of L^2 with eigenvalue L(L+1)
and an eigenfunction of L_z with eigenvalue 0
We showed in class the lowering operator is
lower[f_]:=Exp[-I phi](-D[f,theta]+I Cot[theta] D[f,phi])
and L^2 operator is
L2[f_]:=-(Csc[theta]^2*D[f,{phi,2}] + Cot[theta]*D[f,theta] + D[f,{theta, 2}])
Consider the following example:
l5m5=Sin[theta]^5 Exp[I 5 phi]
Check that this is an eigenfunction of L^2 (and report the eigenvalue) by:
Simplify[L2[l5m5]/l5m5]
QUESTIONS:
why does the above show that l5m5 is an eigenfunction of L^2?
How is the eigenvalue being displayed?
Describe how you can immediately see the eigenvalue of L_z for l5m5
Lower l5m5 with code:
Simplify[lower[l5m5]]
Check that the result still is an eigenfunction of L^2 (report the eigenvalue)
Simplify[L2[%]/%]
Describe how you can immediately see the eigenvalue of L_z for this function
You should be able to lower & check again with code:
Simplify[lower[%%]]
Simplify[L2[%]/%]
Careful! the word %% refers to the twice previous result, if you do additional
operations between the first lowering and this lowering you must add additional %
Continue to lower and check L^2 a total of 5 lowerings from l5m5 with repeated code like:
Simplify[lower[%%]]
Simplify[L2[%]/%]
If we start with L_z=5 and lower five times we should be at L_z=0 which is
supposed to be a multiple of
LegendreP[5,Cos[theta]]
Check this out with code like:
Simplify[%%/LegendreP[5,Cos[theta]]]
Proceed to lower five more times (reaching L_z=-5); one additional lowering
should annihilate the function; show this!
Sequential lowering can proceed with code like the below, but you must be careful
how you start this process following the LegendreP[5,Cos[theta]] check
Simplify[lower[%]]
l5m5 was not normalized; calculate:
Integrate[Integrate[Evaluate[l5m5 /. phi->-phi] l5m5 Sin[theta],{phi,0,2 Pi}],{theta,0,Pi}]
QUESTION: why is this Sin[theta] here................^^^^^^^^^
Note: Evaluate[l5m5 /. phi->-phi] is the stupid way I've used to generate the
Conjugate since Mathematica will assume phi is itself complex
and use the result to define l5m5N which IS normalized
IF we lower l5m5N we are supposed to get Sqrt[5*6-5*4] l5m4N, where l5m4N is the normalized
eigenfunction with L=5 and m=4
Thus
lower[l5m5N]
Integrate[Integrate[Evaluate[% /. phi->-phi] % Sin[theta],{phi,0,2 Pi}],{theta,0,Pi}]
should produce 10. Check this and explain why 10 is the expected result.
If we
Simplify[lower[l5m5N]]/Sqrt[5*6-5*4]
we should have l5m4N, and if we
lower that result we should get Sqrt[5*6-4*3] l5m3N
Simplify[lower[%]]
Integrate[Integrate[Evaluate[% /. phi->-phi] % Sin[theta],{phi,0,2 Pi}],{theta,0,Pi}]
What does this produce? Why is that value expected?
The functions you've been generating have differenct eigenvalues for L_z and
hence must be orthogonal (recall general proof: nondegenerate eigenvectors of hermitian operators must be orthogonal).
Provide an example showing this.
Remark: These angular momentum eigenfunctions you've been generating are available in
Mathematica under the name
SphericalHarmonicY[l, m, theta, phi]
Help: 4pm Saturday
Class 16 M Sep 30
Exam 1
Class 17 W Oct 2
section 4.1.1 separation of variables
sections 4.1.2 & 4.3 on angular momentum
example 4.1: spherical infinite square well
3d spherical infinte square well:
http://www.physics.csbsju.edu/QM/square.14.html
http://www.physics.csbsju.edu/QM/square.15.html
The wavefunction for the 3d spherical infinite square well problem are products:
SphericalBesselJ[l,k r] times SphericalHarmonicY[l, m, theta, phi]
Since its hard to visualize things in 3d, lets review each of these terms separately
Claim: the SphericalHarmonicY are like standing waves on the surface of a sphere.
The easiest way to look at the surface of a sphere is in a Mercator-like map...Consider
ContourPlot[Re[SphericalHarmonicY[5, 2, theta,phi]],{phi,-Pi,Pi},{theta,0,Pi},AspectRatio->Automatic]
Note: theta=0 (north pole) is at the bottom of this chart...Australian upside-down view
plot several different values of l,m and answer:
QUESTION: how does the number of azimuthal nodes depend on m? what formula gives the number of zonal nodes?
REMARK: theta has "fixed ends" and so we can squeeze and add a half-wavelength (which comes with one node)
phi=0 is exactly the same spot as phi=2 Pi so we can only add full-wavelengths (which comes with two nodes)
ContourPlot[Abs[SphericalHarmonicY[5, 2, theta,phi]],{phi,-Pi,Pi},{theta,0,Pi},AspectRatio->Automatic]
QUESTION: why are there no azimuthal nodes in the above?
R[n_,l_,r_]=Sqrt[2] SphericalBesselJ[l,BesselJZero[l+1/2, n] r]/Abs[SphericalBesselJ[l+1,BesselJZero[l+1/2, n]]]
Consider the below and additional values of n,l
Plot[R[3,2,r],{r,0,1}]
QUESTION: How does changing n change the wavefunction? How does changing l change the wavefunction?
QUESTION: which wavefunctions suggest a non-zero chance of being at the origin (r=0)?
orthonormality: if we consider full wavefunctions specified by quantum numbers n,l,m, wavefunctions
with different values of l,m will be orthogonal because of the orthogonality of SphericalHarmonicY.
The r integral must supply orthogonality for wavefunctions with the same value of l,m but different values of n
Integrate[R[3,2,r] R[4,2,r] r^2,{r,0,1}]
QUESTION: why this r^2 here ^^^^?
Provide three examples of orthogonality and 3 examples of normalization
NOTE: RE: normalized: Mathematica has problems simplifying the result. Request the numerical value with:
N[Integrate[R[3,2,r]^2 r^2,{r,0,1}]]
A common textbook problem in QM is: given an initial psi that is not an energy eigenfunction
find the probabilities of measuring various eigenenergies, (e.g., Griffiths 2.37, 2.38).
Consider an initial wavefunction that is uniformly distributed inside r=.3:
psi = Sqrt[1000/9] times SphericalHarmonicY[0, 0, theta,phi] and zero for r>.3
Note: Sqrt[1000/9] is the radial function..uniform for r<.3
Check that this initial radial function is normalized and find which eigenenergy (nl) is most likely to be measured
Which is the highest energy state which has more than a 1% chance of being measured.
FYI: the total probability of measureing an energy above n=100 is about 1%
Print out or duplicate the final energy level diagram on page:
http://www.physics.csbsju.edu/QM/square.14.html
label each state with the usual (but stupid) notation: nl where instead of l you use the corresponding letter.
Compare this diagram to the 3d SHO (second to bottom on http://www.physics.csbsju.edu/QM/shm.12.html).
REMARK: "compare"... because the energy units used in these two plots may be wildly different it makes no sense
to say, for example, that the energies for infinite square well are more than for SHO. Rather
you must make comparisons that would be unchanged by a linear transformation E=m E' + b.
Energy differences ("splittings") are insensitive to b; ratios of energy splittings
will be insensitive to m. (See for example problem 27 http://www.physics.csbsju.edu/QM/square.19.html)
REMARK2: Energy level splittings are what can be easily measured and what is related to classical
orbit periods and hence for both of these reasons plays a big role.
Class 18 F Oct 4
H-atom: read Griffiths 4.2
2d SHO: http://www.physics.csbsju.edu/QM/shm.09.html
http://www.physics.csbsju.edu/QM/shm.10.html
3d SHO: http://www.physics.csbsju.edu/QM/shm.12.html
H atom: http://www.physics.csbsju.edu/QM/H.01.html (and thru page 5)
This problem is similar to http://www.physics.csbsju.edu/QM/shm.16.html #7 except done via Mathematica.
The issue is to see how two distinct basis of the 2d SHO (XY & r phi) span the same space.
The eigenfunction [for E'=2(2 n + Abs[m] +1)] of the 2d SHO in polar coordinates:
shoR[n_,m_,r_,phi_]=r^Abs[m] LaguerreL[n, Abs[m], r^2] Exp[-r^2/2] Exp[I m phi]/Sqrt[ Pi Pochhammer[n+1, Abs[m]]]
Begin by checking orthonormality, i.e., the integrals .
Note that for m' not equal m, the phi integral will provide orthogonality and for m'=m but n' not equal n
the r integral will provide orthogonality. Also note we can provide Conjugate via phi->-phi
Calculate 3 examples of orthogonality of the form:
Integrate[Integrate[shoR[n',M,r,-phi] shoR[n,M,r,phi] r ,{phi,-Pi,Pi}],{r,0,Infinity}]
QUESTION: why is there an r here.....................^^^
Calculate 3 examples of normality of the form:
Integrate[Integrate[shoR[n,M,r,-phi] shoR[n,M,r,phi] r ,{phi,-Pi,Pi}],{r,0,Infinity}]
The orthonormal solution to the 1d SHO are:
shoX[n_,x_]=HermiteH[n,x] Exp[-x^2/2] /Sqrt[Sqrt[Pi] 2^n n!]
Calculate 3 examples of orthogonality of the form:
Integrate[shoX[n',x] shoX[n,x] ,{x,-Infinity,+Infinity}]
Calculate 3 examples of normality of the form:
Integrate[shoX[n,x] shoX[n,x] ,{x,-Infinity,+Infinity}]
In the XY separation of the 2d SHO we have product eigenfunctions [for E'=2(nx+ny +1)]
shoX[nx,x]shoX[ny,y]
Note: shoX even for y as same function form
Verify that these product wavefunctions are "square" by a ContourPlot:
ContourPlot[sho[5,x] shoX[0,y],{x,-5,5},{y,-5,5},AspectRatio->Automatic]
whereas the r phi wavefunctions are "round"
ContourPlot[Re[shoR[1,3,Sqrt[x^2+y^2], ArcTan[x, y]]],{x,-5,5},{y,-5,5},AspectRatio->Automatic]
Note: we must take the Re[] part of the wavefunctions as via Exp[I m phi] they are complex
ContourPlot[Abs[shoR[1,3,Sqrt[x^2+y^2], ArcTan[x, y]]],{x,-5,5},{y,-5,5},AspectRatio->Automatic]
QUESTION: why does the above plot show no phi oscillation whereas the previous plot showed phi oscillation?
The (nx,ny)=(5,0) and the (nr,m)=(1,3) actually have the same energy: E'=12
Whether we count in XY or r phi there are six wavefunctions with this energy:
(nx,ny)=(0,5); (1,4); (2,3); (3,2); (4,1); (5,0)
(nr,m) = (2,1); (2,-1); (1,3); (1,-3); (0,5); (0,-5)
CLAIM: these two sets of six functions span the same space, and in particular any XY wavefunction
can be expressed as a sum of r phi wavefunctions. Fouriers Trick will provide the expansion coefficients
SO if:
(nx,ny)=(5,0) can be expressed as a linear combination or r phi wavefunctions:
a1*(2,1)+a2*(2,-1)+a3*(1,3)+a4*(1,-3)+a5*(0,5)+a6*(0,-5)
NOTE: it will turn out a1=a2, a3=a4, and a5=a6 so we can rewrite this as:
= a1*[(2,1)+(2,-1)] + a3*[(1,3)+(1,-3)] + a5*[(0,5)+(0,-5)]
= a1*2*Re[(2,1)] + a3*2*Re[(1,3)] + a5*2*Re[(0,5)]
By Fourier's Trick the integral gives the coefficient
a1=<(nr,m)=(2,1)|(nx,ny)=(5,0)>
a1=Integrate[Integrate[shoR[2,1,r,-phi] shoX[5,r Cos[phi]] shoX[0,r Sin[phi]] r ,{phi,-Pi,Pi}],{r,0,Infinity}]
QUESTION: explain in a sentence why this is the correct integral
Note: we can automate the collection of the required integrals via:
a=Table[Integrate[Integrate[shoR[k,(5-2 k),r,-phi] shoX[5,r Cos[phi]] shoX[0,r Sin[phi]] r ,{phi,-Pi,Pi}],{r,0,Infinity}],{k,0,2}]
REPORT: the values of these coefficients
and then put together the sum:
f[r_,phi_]=a[[1]] 2 Re[shoR[0,5,r,phi]]+a[[2]] 2 Re[shoR[1,3,r,phi]]+a[[3]] 2 Re[shoR[2,1,r,phi]]
Make a contour plot to confirm that this linear combination looks just like shoX[5,x] shoX[0,y] which you did near the start
ContourPlot[f[Sqrt[x^2+y^2], ArcTan[x, y]],{x,-5,5},{y,-5,5},AspectRatio->Automatic]
Class 19 W Oct 9
Note: principal quantum number n= nr+l+1; nr is the degree of the poly: LaguerreL
Note: l must be <= n-1
R[n_,l_,r_]=2/(n^2 Sqrt[Pochhammer[n-l,2l+1]]) (2r/n)^l LaguerreL[n-l-1,2l+1,2r/n] Exp[-r/n]
Provide three example of orthogonality
Provide three exampels of normalization
The following code checks that R[5,2,r] is an eigenfunction of the Hamiltonian:
n5l2=R[5,2,r]
Simplify[(-1/2 (D[D[n5l2,r],r]+2/r D[n5l2,r] -2*3/r^2 n5l2)-1/r n5l2)/n5l2]
Provide three examples like that above checking that R[n,l,r] is an eigenfunction
QUESTION: what/why does "2*3" become in your examples?
QUESTION: what/why do you get the value you display from the Simplify[]?
web: http://www.physics.csbsju.edu/QM/H.13.html #3
Provide one example for each of the three provided expectation values
web: http://www.physics.csbsju.edu/QM/H.13.html #11
Similar to web #1:
Modify web problem #1...Griffiths' Laguerre polynomial differs a bit from Mathematica's
(you may have noticed that the R defined above does not match Eq. 4.89 in the book),
so I need to change the H(x) defined in this problem to match Griffiths v(x)
Consider:
v(x)= -945 + 1575*x - 900*x^2 + 225*x^3 - 25*x^4 + x^5
Show that the coefficients satisfy the recursion relation 4.63 (for what value of rho_0 and l?)
Show that this satisfies differential equation 4.61
Class 20 F Oct 11
old exam: 346t208.pdf #1, #2
Class 21 M Oct 14
Using the table clebsch-gordan_pdg.lbl.gov.pdf:
Consider the problem of adding orbital angular momentum l=2 to spin angular momentum s=3/2
Write the state that has total angular momentum j=1/2 and mj=1/2 in terms of a linear combination of product wavefunctions.
Write the product wavefunction |lm>|sm>=|20>|3/2 1/2> as a linear combination of total angular momentum eigenstates.
FYI Mathematica: ClebschGordan[{j1 , m1 }, {j2 , m2 }, {j, m}]
HW: 4.27,4.49,4.55
Class 22 W Oct 16
starting chapter 5
read about spin-statistics and the periodic table:
http://www.physics.csbsju.edu/QM/H.06.html
http://www.physics.csbsju.edu/QM/H.07.html
http://www.physics.csbsju.edu/QM/H.08.html
http://www.physics.csbsju.edu/QM/H.09.html
HW: 5.6, 5.7
5.6 is a real mess without Mathematica. I imagine Griffiths
would have you apply his Eq 5.19 (p 207) and Eq 5.21 (p.208) to come up with answers.
Instead make a direct attack on the integrals using Mathematica.
Explain why the below code is solving this problem. Select and print out
the below code; write out--in human readable format directly adjacent to
the equivalent Mathematica code--what each line is calculating, and comment
why what is being calculated answers the question. For example, what/why these particular functions u?
The problem involves (a) distinguishable (b) bosons and (c) fermions: which code
relates to which parts? Write out--in your own hand--the formulas that
are the answers to parts (a), (b) and (c). On the top of p. 209
Griffiths writes: "Identical bosons...tend to be somewhat closer together,
and identical fermions...somewhat further apart, than distinguishable particles
in the same two states". Write down why these calculations support that statement.
By comparing Mathematica's (n=1,l=2) answers to Eq. 5.21 or 5.22,
determine the numerical value of the last term in Eq. 5.21 or 5.22:
2|_ab|^2
Note: Mathematica will require some time (3 minutes) to do these integrals!
If you fail to include the Assumptions line, compute time will be measured in hours
$Assumptions = Element[{l,n},Integers]
u[n_,x_]=Sqrt[2/a] Sin[Pi n x/a]
psi1=u[n,x1] u[l,x2]
psi2=(u[n,x1] u[l,x2]+u[n,x2] u[l,x1])/Sqrt[2]
psi3=(u[n,x1] u[l,x2]-u[n,x2] u[l,x1])/Sqrt[2]
Integrate[Integrate[(x1-x2)^2 psi1^2,{x1,0,a}],{x2,0,a}]
N[%] /. {n->1,l->2}
Integrate[Integrate[(x1-x2)^2 psi2^2,{x1,0,a}],{x2,0,a}]
N[%] /. {n->1,l->2}
Integrate[Integrate[(x1-x2)^2 psi3^2,{x1,0,a}],{x2,0,a}]
N[%] /. {n->1,l->2}
2
Out[7]= 0.103341 a
2
Out[9]= 0.0384498 a
2
Out[11]= 0.168232 a
$
Class 23 F Oct 18
old exam: 346t208.pdf #5
Class 24 M Oct 21
finish read ch 5
delta.04.html thru delta.09.html
HW: 5.16
Help: Sunday 9pm
Class 25 W Oct 23
Magnetic fields into Hamiltonian
Class 26 F Oct 25
Zeeman Effect calculation
Class 27 M Oct 28
Exam 2: chapters 4 thru but not including section 5.4
Class 28 W Oct 30
perturbation theory (Ch 6)
6.2,6.3,6.5a
read:
http://www.physics.csbsju.edu/QM/shm.15.html
Class 29 F Nov 1
degenerate perturbation theory (Ch 6)
6.8,6.37
we'll do 6.37 with Mathematica.
Recall the H-atom wavefunction in Mathematica form:
psi[n_,l_,m_,r_]=2/(n^2 Sqrt[Pochhammer[n-l,2l+1]]) (2r/n)^l LaguerreL[n-l-1,2l+1,2r/n] Exp[-r/n] SphericalHarmonicY[l, m, theta,phi]
it will help to have the complex conjugate of this wavefunction:
psic[n_,l_,m_,r_]=2/(n^2 Sqrt[Pochhammer[n-l,2l+1]]) (2r/n)^l LaguerreL[n-l-1,2l+1,2r/n] Exp[-r/n] SphericalHarmonicY[l, m, theta,-phi]
In general there are n^2 degenerate states for principal quantum number n
For n=3->9. here is a list of the lm values:
lm={{0,0},{1,1},{1,0},{1,-1},{2,2},{2,1},{2,0},{2,-1},{2,-2}}
QUESTION: using the spd encoding, circle and label the 3s, three 3p, five 3d states.
Check the orthonormality of these states by calculating the 81 integrals:
Do this in one chunk:
Table[Table[
Integrate[Integrate[Integrate[psic[3,First[lm[[i]]],Last[lm[[i]]],r] psi[3,First[lm[[j]]],Last[lm[[j]]],r] r^2 Sin[theta],{phi,-Pi,Pi}],{theta,0,Pi}],{r,0,Infinity}],
{i,1,9}],{j,1,9}]
MatrixForm[%]
QUESTION: explain (write down) why each term in the integral is the correct term.
For the Stark Effect we seek the z matrix elements:
Calculate these 81 integrals in one chunk:
Table[Table[
Integrate[Integrate[Integrate[psic[3,First[lm[[i]]],Last[lm[[i]]],r] r Cos[theta] psi[3,First[lm[[j]]],Last[lm[[j]]],r] r^2 Sin[theta],{phi,-Pi,Pi}],{theta,0,Pi}],{r,0,Infinity}],
{j,1,9}],{i,1,9}]
MatrixForm[%]
QUESTION: explain why each term in the integral is the correct term.
0 0 -3 Sqrt[6] 0 0 0 0 0 0
0 0 0 0 0 -9/2 0 0 0
-3 Sqrt[6] 0 0 0 0 0 -3 Sqrt[3] 0 0
0 0 0 0 0 0 0 -9/2 0
0 0 0 0 0 0 0 0 0
0 -9/2 0 0 0 0 0 0 0
0 0 -3 Sqrt[3] 0 0 0 0 0 0
0 0 0 -9/2 0 0 0 0 0
0 0 0 0 0 0 0 0 0
QUESTION: print out the above answer for the matrix. Note that in the problem Griffiths gives
three check integrals. Find and circle each one of these example integrals making sure
you get the order right...i.e., is at a different location from ** but they both have the same value.
Eigensystem[%%]
QUESTION: for each eigenvector: Does every non-zero term in the vector have the same value of m? If so write that m value
adjacent to the vector. If not write NO adjacent to the eigenvector.
Display these eigenvalues as a function of the corresponding eigenvector's m value in a chart similar to the one at the bottom of:
http://www.physics.csbsju.edu/QM/H.10.html
QUESTION: The degeneracy is not completely "broken", e.g., there remain 3 states sharing the eigenvalue zero.
I claim that nevertheless if we calculate the matrix element where |A> and |B> are still degenerate
the result is zero. (So 2nd order is safe!) Report a commutator that guarantees this outcome.
read:
http://www.physics.csbsju.edu/QM/H.10.html
Class 30 M Nov 4
no assignment; finished ch 6
Class 31 W Nov 6
variational (Rayleigh-Ritz) method
ch 7
cosh2.pdf #1 & #2
7.1,
11c on page:http://www.physics.csbsju.edu/QM/fall.15.html
This problem relates to the linear potential problem described in:
http://www.physics.csbsju.edu/QM/fall.08.html
...I'd use Mathematica..some BIG hints:
psi[z_]=z Exp[-a z^(3/2)]
Integrate[psi[z]^2,{z,0,Infinity}]
Integrate[psi[z]^2 z,{z,0,Infinity}]
Integrate[psi'[z]^2,{z,0,Infinity}]
Simplify[(%+%%)/%%%,Re[a] > 0]
Solve[D[%,a]==0,a]
%% /. Last[%]
N[%]
2.34723
Mostly what I've left you to do is figure out the units for your answer!
Class 32 F Nov 8
HW: #19 on http://www.physics.csbsju.edu/QM/H.13.html
Consider the problem of finding the minimum with three variables.
While Mathematica has a FindMinimum function, it requires a good starting
guess. (This is exactly the same problem with minimizing chi-square in
non-linear fit problems: a guess is required, and the program will fall
into the first minimum it finds, even if there is a much lower minimum
nearby.) Thus it helps to have a graphical display of the function
in order to by-eye find a minimum. With one parameter minimizations--
say finding the minimum of f(a)--a plot of f (y-axis) vs a (x-axis) allows you
to see (approximately) the minimum. With two parameter minimizations, the best bet
is a contour plot of f(a,b) with a (x-axis) and b (y-axis) and contours of
constant f. This is a topographical map where we're seeking valleys.
Mathematica can remind you of the function values at each point in the plane
by coloring the plot: red for small f(a,b) through blue for big.
(The alternative of a 3d mountain range plot [a perspective plot with
f(a,b) on the z-axis], is visually stunning, but its not up to seeing details.)
For three parameter minimizations f(a,b,c), there are no perfect choices. What I recommend
is making a stack of contour plots where each contour plot is for a slightly different value
of c. Using a slider to go through the stack, you can find the minimum.
Here we go...start by copy&paste the formula in #19 for e into Mathematica.
e=....
Animate[ContourPlot[Evaluate[e /. c-> c1], {a, .15, 1.25}, {b,.15, 1.25},
Contours->{-.52,-.5,-.48,-.46,-.44,-.42,-.4,-.38,-.36,-.34,-.32},
ColorFunction->Hue,DisplayFunction->Identity],{c1,0,1,.05}]
This makes a stack of contour plots. Each frame displays the contours for
a and b in the range [.15,1.25] for a different value of c. The stack includes
c values from 0 to 1 in steps of .05. Red is small and blue is large.
You should find two red eyes in an orange mask which are the desired minimums.
You can use the slider to go through the stack by hand.
The fact that the red eyes begin at Frame#3 (c=.15), grow and disappear at Frame#21,
(c=1) tells you that this low-valued region is something like a sausage. You should
guess that the sausage center is near the true minimum, and that the sausage center is
on-screen when the apparent radius (red eye) is largest...at about Frame#8,
where c=.05*(8-1)=.35. That center is at about (a,b)=(1.1,.5) or (.5,1.1) [the function
should be symmetric in a and b].
Once you have your starting guess, have Mathematica determine the best (lowest) minimum:
FindMinimum[e,{a,1.1},{b,.5},{c,.35}]
If you cut the parameter space volume along the plane a=1.07
you can see the sausage in cross-section:
ContourPlot[Evaluate[e /. a-> 1.07], {c, .05, 1.05}, {b,.15, 1.25},
Contours->{-.52,-.5,-.48,-.46,-.44,-.42,-.4,-.38,-.36,-.34,-.32},
ColorFunction->Hue,DisplayFunction->Identity]
You can make series of cuts for various values of a:
Animate[ContourPlot[Evaluate[e /. a-> a1], {c, .05, 1.05}, {b,.15, 1.25},
Contours->{-.52,-.5,-.48,-.46,-.44,-.42,-.4,-.38,-.36,-.34,-.32},
ColorFunction->Hue,DisplayFunction->Identity],{a1,.15,1.25,.05}]
The two red eyes now become two sausages (or perhaps carrots), one for small a (slider left) and
an equal sausage for large a (slider right)
HW: #18...Note that if c=0 in #19, we are doing problem #18 (which is the same as Griffiths 7.18):
e2=e /. c->0
Find the minimum of e2.
Convert the dimensionless energies from #19 and #18 to eV and report the ionization
energy of the H- ion.
Do note that our trial wavefunctions have been symmetric wrt particle exchange, so they
would need to be combined with an antisymmetric (singlet) spin state to make an overall
wavefunction appropriate for these two fermions.
Class 33 M Nov 11
WKB: ch 8
8.7,8.11
morse.pdf #5
old exam 346t308.pdf #4
Class 34 W Nov 13
2wells.pdf
alpha=Sqrt[2628.8256429]
wkbB[e_]=alpha*(a*((b^2+a^2)*EllipticE[(a^2-b^2)/a^2]-2*b^2*EllipticK[(a^2-b^2)/a^2]))/3
Class 35 F Nov 15
HW: tdpt.pdf #1 a-c
Class 36 M Nov 18
old exam: 346t308.pdf #5
Class 37 W Nov 20
9.11
see class 29 for a H-atom wavefunction in Mathematica form. Recall that
in this form, r is being measured in units of Bohr radius.
Reduce your answer to a numerical value in seconds
(this is the hard part!)
Class 38 F Nov 22
http://www.youtube.com/watch?v=ta09WXiUqcQ
Class 39 M Nov 25
Exam 3
Class 40 M Dec 2
HW: in folder scatter see file: scatter.problem.txt
READ:
http://www.physics.csbsju.edu/QM/square.10.html
http://www.physics.csbsju.edu/QM/square.11.html
http://www.physics.csbsju.edu/QM/square.12.html
http://www.physics.csbsju.edu/QM/square.17.html
Class 41 W Dec 4
http://www.physics.csbsju.edu/QM/square.18.html
HW: in folder scatter see file: partial_waves.pdf
HW: In the partial wave handout find 11 expanded resonances (for U=18,27,30,33,39,42,45,48,57,60,59).
For each resonance find (by reading the graphs) the maximum cross section, the k location
of that maximum, and the l value of the resonance.
Make a spreadsheet table of (U, k location, maximum, l, maximum/(2l+1))
Fit (k,maximum/(2l+1)) data to a power law and make a log-log plot of the results.
What is the expected value of B? What is the fit value of B?
What you are finding is called the unitary limit
HW: Looking through the partial wave handout, report the U that show a negative scattering length
HW: A not uncommon sight is a pale halo around the Sun or Moon. I've saved an image of this:
Solar-halo-inca-trail-20080928.jpg
Describe how this scattering phenomenon works. Include a sketch of the graph of
differential cross-section vs. scattering angle
that would be required for this phenomenon. Draw a picture showing (labeling) the relative
location of the material causing the effect, the observer, and the Sun.
Note: what is going on here is abit different from what was described above (square.12.html:
db/d theta -> infinity): instead of random impact parameters puting lots of light into a particular angle theta
we have lots of differently rotated ice crystals puting lots of light into a
particular angle. sun_dog.pdf is a start at the ray optics involved.
Class 42 F Dec 6
no homework
Class 43 M Dec 9
in scatter folder: scatter_problem2.pdf
Class 44 W Dec 11
HW: The form factor of the proton obtained from electron scattering approximately fits the simple
formula:
F(q)=1/(1+(q/q0)^2)^2
where constant q0=4.3 fm^-1
What is the corresponding charge density?
What is the resulting rms charge radius of the proton?
The main diffraction peak for k=2.1 fm^-1 electrons-on-oxygen ends at about theta=44 degree.
Estimate the radius of an oxygen nucleus.
Class 45 F Dec 13
Final T Dec 17 10:30 am
**