text: Griffiths Intro to QM (2008)
--class1--
problems: 1.3, 1.4
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
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. (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 must 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.
--class2--
problems: 2.4,2.5
(these are fairly long problems, but they are also quite valuable)
--class3--
problems: 2.7
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
HW:
on page: http://www.physics.csbsju.edu/QM/shm.16.html
do problem: 1 a&b
Eq. 2.89 (p.59 in Griffiths) is a "generating function" for Hermite
polynomials. We can test this formula easily with mathematica:
In[1]:= Exp[-z^2 +2 z x]
In[2]:= Series[%,{z,0,11}]
In[3]:= Simplify[%]
Show that the H_9, H_10, and H_11 obtained from this
series match those found above.
Note: SeriesCoefficient[%,10] 10!
--class4--
Griffiths: 2.12, 2.13
The aim of these problems is practice the skill of "integration by operator
methods". These problems should not be super-hard, but you can test your
algebra by explicit mathematica integration. E.g.:
psi[n_,x_]=HermiteH[n, x] Exp[-x^2/2] /Sqrt[ Sqrt[Pi] 2^n n!]
Integrate[psi[3,x] x^2 psi[3,x],{x,-Infinity,Infinity}]
7
Out[2]= -
2
Integrate[psi[3,x] psi[3,x],{x,-Infinity,Infinity}]
Out[3]= 1
Integrate[psi[3,x] psi[4,x],{x,-Infinity,Infinity}]
Out[4]= 0
(where of course we are doing things in dimensionless coordinates).
--class5--
read:
http://www.physics.csbsju.edu/QM/fall.11.html
http://www.physics.csbsju.edu/QM/fall.12.html
HW:
Griffiths: 2.19 & 2.21
RE: "Discuss"-- as a grows, how does the uncertainty in x change? uncertainty in k?
problem #19 on http://www.physics.csbsju.edu/QM/fall.15.html
note that in these dimensionless coordinates, time-dependent
Schrödinger's equation reads:
i D_t psi = -D_y^2 psi
and that, in this solution, sigma and k_0 are constants.
Again, I don't think you want to try this by hand: use a
mathematics program....a start:
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]
The following problem is essentially problem#20 on http://www.physics.csbsju.edu/QM/fall.15.html:
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.
Note solutions to real-vs-complex-constants problem:
phi[x_,t_]=A Exp[I (k x -h k^2/(2 m) t)]
Conjugate[phi[x,t]] D[phi[x,t],x]-Conjugate[D[phi[x,t],x]] phi[x,t]
Simplify[%,Element[{x,t,h,m,k},Reals]]
Out[3]= (2 I) A k Conjugate[A]
or:
$Assumptions=$Assumptions && Element[{x,t,h,m,k},Reals]
phi[x_,t_]=A Exp[I (k x -h k^2/(2 m) t)]
Conjugate[phi[x,t]] D[phi[x,t],x]-Conjugate[D[phi[x,t],x]] phi[x,t]
Simplify[%]
Out[4]= (2 I) A k Conjugate[A]
or:
phi[x_,t_]=A Exp[I (k x -h k^2/(2 m) t)]
phic[x_,t_]=A Exp[-I (k x -h k^2/(2 m) t)]
phic[x,t] D[phi[x,t],x]-D[phic[x,t],x] phi[x,t]
Out[3]= (2 I) A^2 k
--class6--
Griffiths: 2.24a, 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.
read:
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
--class7--
Read: Appendix (review of vector spaces), start Ch 3
HW:
3.39a&b
Appendix A: A.8, A.9, A.26abc
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}}
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.
--class8--
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)
3.31, 37
--class9--
3.38
old exam #2,#6
http://www.physics.csbsju.edu/346/2005/346t105.pdf
Help session Wed?
--class10--
We covered systems for which the 3d problem reduced to three 1d problems, with
\psi=X(x)Y(y)Z(z)
H=H_x+H_y+H_z
E=E_x+E_y+E_z
For the infinite 3d square well: p^2=p_x^2+p_y^2+p_z^2, i.e., the Laplacian
could be written as a three term sum of second derivatives w.r.t. x, y, and z
The potential was zero: nothing to separate out.
For the 3d SHO: r^2=x^2+y^2+z^2, so the Hamiltonian can split up:
p^2/2m+kr^2/2=(p_x^2/2m+kx^2/2)+(p_y^2/2m+ky^2/2)+(p_z^2/2m+kz^2/2)
E=E_x+E_y+E_z=\hbar \omega(n_x+.5 + n_y+.5 + n_z+.5)
--class11--
first exam Ch. 1-3
--class12--
start H atom: ch 4
http://www.physics.csbsju.edu/QM/H.01.html
http://www.physics.csbsju.edu/QM/H.02.html
http://www.physics.csbsju.edu/QM/H.03.html
http://www.physics.csbsju.edu/QM/H.04.html
http://www.physics.csbsju.edu/QM/H.05.html
HW:
http://www.physics.csbsju.edu/QM/H.13.html
Problems #3 & #4 gives general results for the expectation value of powers of r
for H-atom wavefunctions specified by n,l,m -- these can be proved using
the recursion relation for Laguerre polynomials, but that's not the aim
of your homework. Rather I want you to use Mathematica to test these
general formulae for 3 specific values of n & l. For this it is handy to
have the normalized version of the u(r) wavefunction:
u[n_,l_,r_]=1/(n Sqrt[Pochhammer[(n-l), 2 l +1]]) (2 r/n)^(l+1) LaguerreL[n-l-1,2 l +1,2 r/n] Exp[-r/n]
a) Check that this wavefunction is normalized for 3 different wavefunctions (i.e., (n,l) values):
e.g.,: Integrate[u[5,3,r]^2,{r,0,Infinity}]
Note that if this were a R wavefunction rather than a u wavefunction, normalization would read:
Integrate[R[5,3,r]^2 r^2, {r,0,Infinity}]
Note: these are unitless problems where r is being measured in units of the Bohr radius.
b) Now for each formula in web problems #3 & #4 provide 3 checks of that formula
using different wavefunctions (i.e., (n,l) values)
Web problem #11 a and b (Note that this is very similar previous orthonormal problems
(like on your recent exam) and that no actual integration is required.)
Modify web problem #1...Griffiths' Laguerre polynomial differs a bit from Mathematica's
(you may have noticed that the u 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
--class13--
read about the 3d SHO:
http://www.physics.csbsju.edu/QM/shm.12.html
http://www.physics.csbsju.edu/QM/shm.13.html
important result: E=\hbar\omega(2n_r+l+3/2)
read about 3d spherical infinite square well:
http://www.physics.csbsju.edu/QM/square.14.html
http://www.physics.csbsju.edu/QM/square.15.html
important result: E depends on the locations of zeros of spherical Bessel functions: "roots"
our energy unit e=hbar^2/(2ma^2) and
E=(root of spherical Bessel function)^2 e
FYI: I see the current version of Mathematica now defines a standalone spherical Bessel function:
SphericalBesselJ[n, z] gives the spherical Bessel function of the first kind j_n(z).
HW: According to square.14.html:
sBesselj[l_,x_]=Sqrt[Pi/(2 x)] BesselJ[l+(1/2),x]
so the roots of j_l are the roots of BesselJ[l+(1/2),x], which Mathematica knows under the name:
In[2]:= ?BesselJZero
BesselJZero[n, k] represents the k^{th} zero of the Bessel function J_n (x)
so the first (n_r=0) zero of j_l for l=0 is
BesselJZero[1/2,1]
to have Mathematica report the numerical value of this zero:
N[BesselJZero[1/2,1]]
(A) Use Mathematica to find at least 4 sigfigs of the first 6 roots listed in the table
near the bottom of square.14.html. Plot sBesselj[l,x] for l=0 and 1, domain: {x,0,10}
and confirm that those zeros look about correct.
(B) Figure 2.11 in the handout gives energy levels for a spherical well, but uses a
different energy unit. Use your zeros to confirm that figure. (I actually find
2 that are mis-rounded...can you find them?)
HW: According to shm.12.html
R[n_,l_,r_]=Sqrt[2 Factorial[n]/Factorial[l+1/2+n]] r^l LaguerreL[n, l +1/2, r^2] Exp[-r^2/2]
is the normalized R wavefunction for the 3d SHO.
Check that this formula produces normalized wavefunctions that are orthogonal if different n (but the same l)
I want you to discover a possible formula for like those you tested for the H-atom last time.
Use the following integrations to guess the formula for
Table[Integrate[R[n,0,r] r^2 R[n,0,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,1,r] r^2 R[n,1,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,2,r] r^2 R[n,2,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,3,r] r^2 R[n,3,r] r^2, {r,0,Infinity}],{n,0,10}]
Extra Credit (hard): try to find the formula for using:
Table[Integrate[R[n,0,r] r R[n,0,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,1,r] r R[n,1,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,2,r] r R[n,2,r] r^2, {r,0,Infinity}],{n,0,10}]
Table[Integrate[R[n,3,r] r R[n,3,r] r^2, {r,0,Infinity}],{n,0,10}]
HW: old exam #6
http://www.physics.csbsju.edu/346/2005/346t205.pdf
(Do not bother to make the report requested in the last sentence.)
HW: Figure 2.10 in the handout gives the energy levels for an infinite cube
square well. Confirm the lowest six listed energies.
HW: The radial part of the (unnormalized) 3d H-atom wavefunction is given by:
u[r_]=r^3 Exp[-r/(3 a)]
where a is Bohr's radius. Plug this into the radial equation 4.53 (p. 145) and show that it is
an eigenfunction with the proper eigenenergy. (The formula for Bohr radius is Eq. 4.72.)
Note that "3d" should tell you the suggested value for n and l.
--class14--
angular momentum
HW: rotation.pdf
HW: 4.22, 4.23, 4.44 (see below)
4.44b asks you to find , but we've done lots of those, so instead
find . Note that since z=r cos(theta), you are going to end up with
an integral:
Integrate[R[4,3,r] r^2 R[4,3,r] r^2, {r,0,Infinity}]
Integrate[Integrate[Conjugate[SphericalHarmonicY[3,3,theta,phi]] Cos[theta]^2 SphericalHarmonicY[3,3,theta,phi] Sin[theta],{phi,0,2 Pi}],{theta,0,Pi}]
Since I've given you the answer, what do I want you to turn in?
(A) Provide the Mathematica code that shows that this beast is normalized, i.e., <1>=1
(B) Write out (by hand) the above Mathematica integral (use R and Y directly: no need to expand out things)
in normal mathematical notation
(C) Report what is *including units*
(D) Calculate again, but this time for n=4,l=1,m=0
--class 15--
spin, addition of angular momentum, Clebsch-Gordan coefficients
HW: 4.27
HW: http://www.physics.csbsju.edu/346/2005/346t205.pdf #1, #4
HW: Find here 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}]
--class16--
FYI: next exam is class20 (on chapters 4,5)
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: 4.49, 4.55, 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. 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". Explain why these calculations support that statement.
Please do not turn in any hardcopy produced by a Mathematica notebook.)
By comparing Mathematica's (n=1,l=2) answers to Eq. 5.22,
determine the numerical value of the last term in Eq. 5.22:
2|_ab|^2
Note: Mathematica will require some time (3 minutes) to do these integrals!
(As you will see, Mathematica is worried about n,l not being
integers.)
u[n_,x_]=Sqrt[2/a] Sin[Pi n x/a]
psid=u[n,x1] u[l,x2]
psif=(u[n,x1] u[l,x2]-u[n,x2] u[l,x1])/Sqrt[2]
psib=(u[n,x1] u[l,x2]+u[n,x2] u[l,x1])/Sqrt[2]
Integrate[Integrate[(x1-x2)^2 psid^2,{x1,0,a}],{x2,0,a}]
Simplify[%,Element[{l,n},Integers]]
N[%] /. {n->1,l->2}
Integrate[Integrate[(x1-x2)^2 psif^2,{x1,0,a}],{x2,0,a}]
Simplify[%,Element[{l,n},Integers]]
N[%] /. {n->1,l->2}
Integrate[Integrate[(x1-x2)^2 psib^2,{x1,0,a}],{x2,0,a}]
Simplify[%,Element[{l,n},Integers]]
N[%] /. {n->1,l->2}
2
Out[7]= 0.103341 a
2
Out[10]= 0.168232 a
2
Out[13]= 0.0384498 a
--class17--
HW: cm.pdf
HW: 5.16
HW: Consider the two dimensional infinite square well (i.e., a rectangle):
http://www.physics.csbsju.edu/QM/square.07.html
Follow the argument to the final diagram on that page. Find the formula
for the number of states with energy less than some value E. If we have
N non-interacting electrons in the rectangle, packed in such a way as to minimize the total energy,
find the formula for the "Fermi Energy" (i.e., the highest energy filled state).
Express your answer in terms of the number of electrons per area. (This problem
is essentially 5.34, just with more words).
HW: Figure 58 on "nebular lines" in the atomic physics handout shows the energy levels for OII, OIII, NII
Write down the electron configuration for each and report the expected terms, e.g.:
The electron configuration for OII should report X equivalent (all n=2) p electrons; Table 11 on the flip side of
the handout reports the LS terms for given equivalent electrons. For each LS term in that table you can
find the possible total j. Given the term symbols, order the states according to Hund's rule and see
if the order matches the given energy level diagram. How many degenerate states are contained
in each term?
FYI: http://physics.nist.gov/PhysRefData/Handbook/periodictable.htm
is a good source of atomic energy levels, but the presentation is tabular not graphical.
Read about bands:
http://www.physics.csbsju.edu/QM/delta.03.html
http://www.physics.csbsju.edu/QM/delta.04.html
http://www.physics.csbsju.edu/QM/delta.05.html
http://www.physics.csbsju.edu/QM/delta.06.html
http://www.physics.csbsju.edu/QM/delta.07.html
http://www.physics.csbsju.edu/QM/delta.08.html
http://www.physics.csbsju.edu/QM/delta.09.html
--class18--
Help: Tuesday 7:30pm
no homework due to arm-twisting by Ryan
what I was going to assign is:
http://www.physics.csbsju.edu/QM/delta.10.html #5
and
The eigenenergy for these delta function problems has been expressed in terms of kappa (k):
E=-(hbar k)^2/(2m)
Rewrite this equation in terms of s=ka/2.
Data: Free H-atom energy=-13.6 eV
H2+ molecule energy -(13.6+2.8)eV with a separation of 1.06 Angstrom
(A) To get a solitary delta function to match the isolated H-atom energy,
what value of k is required? What value of alpha does that require?
Working in the units of eV and nm (as then hbar c = 197.33 eV·nm, a "nice" number)
(hbar c k)^2/(2mc^2) = 13.6
k= sqrt(13.6*2*.511e6)/197.33 = 18.893 unit:nm^-1
k=mc^2 alpha/(hbar c)^2; alpha = k*197.33^2/.511e6 = 1.4397 unit: eV·nm
(B) If we have a pair of such delta functions separated by the known H2+ separation,
what is the energy of the ground state? Is there an excited state? If so what is its energy?
a' =mc^2 a alpha/(hbar c)^2 = .511e6*.106*1.4397/197.33^2 = 2.0027
FindRoot[s(1+Tanh[s])==2.0027,{s,1}]
Out[1]= {s -> 1.11009}
E= -(hbar k)^2/(2m) = - (hbar c 2 s)^2/(2mc^2a^2) = - (197.33*2*1.11009)^2/(2*.511e6*.106^2)=-16.715 eV
compare actual: -16.4 eV
FindRoot[s(1+Coth[s])==2.0027,{s,1}]
{s -> 0.798624}
E=- (197.33*2*0.798624)^2/(2*.511e6*.106^2)= -8.6510
This is above the energy of H-atom + separated proton, so it is not a stable state.
--class19--
magnetic fields (not on examII)
unassigned problems: 4.59,4.60,4.61
--class20--
examII
--class 21--
perturbation theory (Ch 6)
6.2,6.3,6.5a
read:
http://www.physics.csbsju.edu/QM/shm.15.html
--class 22--
degenerate perturbation theory (Ch 6)
6.8,
old exam#9: http://www.physics.csbsju.edu/346/2005/346t304.pdf
read:
http://www.physics.csbsju.edu/QM/H.10.html
--class23--
variational (Rayleigh-Ritz) method
7.1, cosh2.pdf #1&2
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!
--class24--
HW: old exam: 2005/346t304.pdf #5
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=....
plote=Table[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}]
ListAnimate[plote]
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
21 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:
plote2=Table[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}]
ListAnimate[plote2]
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 25--
http://www.physics.csbsju.edu/QM/shm.05.html
http://www.physics.csbsju.edu/QM/fall.08.html
HW: cosh2.pdf #5, old exam #1 & #2 (2005/346t304.pdf)
(Print out page one of the old exam and draw your sketch directly on that sheet.)
Let's assume we use Griffiths version of WKB with (n-.5), so the ground state
is n=1, 1st excited is n=2, etc. This is a continuation of 2005/346t304.pdf #1:
Find 5 wavefunction plots math/346t304#1A.pdf thru math/346t304#1E.pdf.
What value of n goes with each plot?
(If you do not understand this type of wavefunction plot consult QM/fall.04.html)
--class 26--
Consider again the potential of #1 2005/346t304.pdf; for the record this potential is
V=x^2(x^2-1)
Mathematica has problems doing the WKB integral (but Gradshteyn & Ryzhik, report results)
The turning points are of course found from PE=E (so KE=0):
Solve[x^2(x^2-1)==e,x]
bb = x /. Last[%]
aa=Sqrt[e]/bb
alpha=Sqrt[2628.8256429]
the WKB integral is directly: Integrate[Sqrt[e-x^2(x^2-1)],{x,-b,b}] but Mathematica thinks
for a while but comes up with no answer.
For some reason Mathematica will do the equivalent integral
Integrate[Sqrt[(a^2+x^2)(b^2-x^2)],{x,-b,b}]
Simplify[%,{a>0,b>0}]
% /. {a->aa,b->bb}
Simplify[%]
wkb=alpha * %
Now when the wkb integral equals Pi*(n-.5) we have a solution. You can graphically
find those spots from the graph of the integral as a function of e:
Plot[wkb/Pi,{e,0,.3}]
You can look at softcopy of that graph at math/346t304#1WKB.pdf
Graphically estimate the energy e for n between 12 and 22.
Mathematica can improve an estimate with code like:
FindRoot[wkb/Pi==21.5,{e,your initial estimate here}]
Improve your estimate for the odd n (13,15,17,19,21)
and compare to the exact results: (.028015, .069755, .118303, .17171, .22925)
For e<0 the problem becomes more interesting as then we have two allowed regions separated
by a disallowed region...the wavefunction will tunnel through that disallowed region.
Since the problem is reflection symmetric the exact solutions will be either even or odd.
Take a look at: math/346t304#1g.pdf math/346t304#1h.pdf
for a pair of nearly degenerate even&odd wavefunctions.
SO if we apply WKB to e<0 it will only know about one well, and it will end
up reporting the energy of what in fact turns out to be a pair of even/odd solutions.
An additional problem is Mathematica cannot integrate even the simple version of the WKB integral:
Integrate[Sqrt[(x^2-a^2)(b^2-x^2)],{x,a,b}]
Gradshteyn & Ryzhik come to our rescue and report that this integral is:
(b*((a^2+b^2)*EllipticE[(b^2-a^2)/b^2]-2*a^2*EllipticK[(b^2-a^2)/b^2]))/3
% /. {a->Sqrt[-e]/bb,b->bb}
Simplify[%]
wkb2=alpha * %
Plot[wkb2/Pi,{e,-.25,0}]
You can look at softcopy of that graph at math/346t304#1WKB2.pdf
Use code like:
FindRoot[wkb2/Pi==.5,{e,-.2}]
to find the five low paired states
The exact answers are: (-.2228, -.1701, -.1202, -.073, -.032)
FYI: Griffiths 8.15 aims to improve these guesses, by using WKB to
estimate the energy difference between the pair of nearly degenerate
even&odd wavefunctions.
HW: tdpt.pdf #1 a-c
--class 27--
FYI: exam 3 has long been scheduled for Monday 24-Nov (before Thanksgiving).
morse.pdf: #1,3,5
9.11 (just find the lifetime for the transition: nlm=210->100)
see class 12 for a H-atom u wavefunction in Mathematica form. Recall that
in this form, r is being measured in units of Bohr radius and that the
full wavefunction is: u(r)/r Y(theta,phi), where Y is a spherical harmonic:
In[1]:= ?SphericalHarmonicY
SphericalHarmonicY[l, m, theta, phi] gives the spherical harmonic function of
theta and phi for parameters l and m.
128 Sqrt[2]
Out[4]= -----------
243
Reduce your answer to a numerical value in seconds.
--class 28--
HW: see file: 2005/scatter.problem.txt
READ:
http://www.physics.csbsju.edu/QM/square.14.html
http://www.physics.csbsju.edu/QM/square.15.html
http://www.physics.csbsju.edu/QM/square.16.html
http://www.physics.csbsju.edu/QM/square.17.html
http://www.physics.csbsju.edu/QM/square.18.html
--class 29--
HW: see file square_U=30Q.pdf which refers to plots in square_U=30.pdf
do 11 a,d,e
Help 8:20am Friday 21-Nov-08
--class30--
Exam: approximation methods
HW: find the file: square_U=33,36,39,42.pdf which follows the same conventions as square_U=30.pdf
See the four columns of plots; they correspond to increasing square well depth: 33,36,39,42 in dimensionless units
The first plot shows phase shifts (l=0:red, l=1:orange, l=2:green, l=3:cyan, l=4:blue, l=5:violet)
The following plot shows total cross section
The bottom plot (if present) shows an expanded view of a resonance. Note particularly the resonance in
the first column at k=.3934: it is so narrow it does not show up in the cross section plot above it.
The x-axis for all plots is dimensionless momentum k
How many zeros would you expect to find in the differential cross section for U=33 at k=.3934?
How many zeros would you expect to find in the differential cross section for U=42 at k=2.23?
What is the cause of the k=.4 resonance for U=39?
The step in delta4 occurs for smaller k as U increases...please explain
QM/square.16.html (near bottom)
shows the bound state energy levels for U=50; Using this source:
report the number of l=0 bound states, the number of l=1 bound states, the number of l=2 bound states,
the number of l=3 bound states, the number of l=4 bound states, all for U=50.
If U is reduced from 50, some bound states may be lost. How many bound states (for each l) do you
suppose exist at U=42?
For small k, each phase shift approaches a multiple of pi. For U=42, for each value of l, report that multiple of pi.
--class31--
While the following pages deal with 2d scattering, they aim to help you understand 3d scattering
QM/square.10.html
QM/square.11.html
QM/square.12.html
HW: partial_waves.pdf
HW: find the file: square_U=45,48,51,30.pdf which follows the same conventions as previous files.
The above file combined with square_U=33,36,39,42.pdf display six expanded resonances.
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 data table of (X,Y)=(k location, maximum/(2l+1))
Fit this 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
The bottom plot on QM/square.16.html shows how the bound states for U=50 compare
to those of the infinite square well. Compare the actual number of bound states
for U=50 to the low-k multiple-of-pi for U=51. If U were increased beyond 51, guess
which state will be next added to the list of bound states. Sketch how the partial wave
phase shift would change when that occurs.
--class32--
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.
HW: Review the data in square_U=54,57,60,63.pdf. What states in fact were added to the list
of bound states as the potential deepened from U=51 to U=63? Recall what we said about the scattering length
(or look at wiki on scattering length...essentially the scattering length is defined as
minus the slope of delta0(k) at k=0 and it determines the small-k value of the total cross-section).
Report the sign of the scattering length for U from 30 to 63 and report whether the magnitude
of the scattering length seems to have increased, decreased, or stayed about the same compared
to its value at the prior (smaller) value of U.
--class33--
HW: 11.10, 11.13
HW: square_U=30Q.pdf (c) refers to plots in square_U=30_PvB.pdf
HW: In the previous homework you discovered that the scattering length was doing interesting things
between U=60 and U=63. Find in square_U=59,61,61.5,62.pdf data on the intermediate region and report how
the sign and magnitude of the scattering length change and the corresponding effect on the low-energy
total cross-section. Among all of the plots, which shows the smallest low-energy
total cross-section; what scattering length causes small low-energy total cross-section?
Among all of the plots, which shows the largest low-energy
total cross-section; what scattering length causes large low-energy total cross-section?
Report the number of hits Google gives for the following search
"scattering length"
try the search:
"scattering length" Nobel
--class34--
HW: breit-wigner.txt
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?
HW: square_U=30Q.pdf (f)
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.
--class35--
class evaluations
discussion of relativistic QM: Klein-Gordon, Dirac
"many-body" theory, quantum field theory, Lagrangian density
Note: at some point I must have dropped a day as we had 36 class periods this semester!
Help: Monday 2pm