1.
a) Evaluate $L^2 \Psi_{n,l,m,s_z}$ where L is the angular momentum operator.
b) Evaluate $S^2 \Psi_{n,l,m,s_z}$ where S is the spin operator.
c) Evaluate $L_z \Psi_{n,l,m,s_z}$ where L_z is the z component of L.
2. In the first excited states of H, what 4 $\Psi_{n,l,j,j_z}$ eigenstates are made from only one $\Psi_{n,l,m,s_z}$ eigenstate?
3. Before including the spin-orbit and p^4 terms, there are 8 1st excited states, right? How about after? If we group states with the same energy together, are there 2 groups or 3 groups? How many states are in each group?
4. Consider the new 1st excited states*. (*This means only the lowest energy group from the original 8.)
a) How many are there?
b) What are the wave-functions for each (expressed in terms of $\Psi_{n,l,m,s_z}$)?
c) What is the energy of each state in this group in eV?
5. extra credit: How many groups of different energies do the 2nd excited states of H break up into? How many states are in each one and what are their energies?(How many states are there to start with?)
6. In a time varying potential, U(x,t), one can get mixed states whose coefficients depend on time (for example, a 1D harmonic oscillator where, in addition to the $k x^2/2$ static confining potential, you have also a time varying electric field $ E_x cos(2 \pi f t)$).
a) Describe the nature and behavior of the state:
$\Psi(x,t) = (1-\gamma^2 t^2)^{1/2} \Psi_o(x,t) + \gamma t \Psi_1(x,t)$
as a function of time. Assume that $\gamma$ is relatively small. (What are the units of $\gamma$, by the way.)
b) Use sketched graphs and words to describe this evolving state. What is it like at low t? How does it evolve? What does it become?
c) Is there a value of t that should perhaps not be exceeded with this state? Why?
extra credit: d) sketch a rough plot of what the expectation value of x might look like, as a function of t, for an electron in this state.
7. ( Matrix algebra) The vector (1,1) is an eigenvector of the matrix M =
1., c
c , 1.
a) Show that this is true. Determine the eigenvalue associated with (1,1) for c=0.01.
[hint: multiply]
b) What is the other eigenvector of this matrix and what is its eigenvalue (for c=0.01)?
c) is this matrix Hermitian? Are its eigen-vectors orthogonal?
d) is the trace of the matrix equal to the sum of its eigenvalues?
e) Show that the if c=0 then (1,0) and (0,1) are eigenvectors of M. What are their eigenvalues?
Discuss the following: Any non-zero value of c, no matter how small, can really have a big effect on the eigen-vectors of this matrix.
8. On the other hand consider the matrix M2 =
1., c
c , 2.
a) Show that (1,0) and (0,1) are each eigen-vectors of this matrix for c=0 and that they are barely changed at all if c=0.01. Why is that? What is the difference between this matriz and the one in problem 7?
In problem 1, our answers are supposed to be in a general form, correct? Also, where does the term L(L+1) or s(s+1) come from?
ReplyDeleteYes, general form. exactly.
ReplyDeletethe l(l+1) and s(s+1) come from books*... we didn't derive it. (usually people use lower case for the numbers and upper case for the operators.
Well, here is where the L^2 one ultimately comes from: L=rxp (definition of L), p_x = i hbar d/dx...
So you work through the 3 components of L in terms of the components of r (which are x, y and z) and the components of p. You square it and multiply it (operate) times the state, and 3 days later, you get hbar^2 l(l+1) as the eigenvalue for that eigenstate.
Also, the L^2 operator arises when you solve the Schrodinger equation by separation of variables in spherical coordinates.
ReplyDeleteFor number 3: doesn't including the spin-orbit term bump up our number of states from 4 to 8? If so then we would only have four states before including that term and the p^4 term.
ReplyDeleteI think I answered my own question. Did you mean there are 8 original states $\Psi_{n,l,m,s}$ that all have the same energy because we recognized that there is spin but didn't account for it in the Schrödinger equation?
DeleteI think that there are still eight states, they are just different now, or adjusted for spin and some slight relativistic corrections. It's true that $L$ (orbital angular momentum) provides us with eight possible quantum states for the first excited state, but in accounting for spin (intrinsic angular momentum) we need the total angular momentum $J = L + S$. This adjusts our states relative to the z-axis and sometimes increases (like in the ground state of H-atom where $l = 0$, but $m_s = \pm{1/2}$) or decreases the expected number of states. Do I have this right?
DeleteI think that just having the spin degree of freedom on our radar increases the number from 4 to 8 (of (orthogonal)1st excited states). The four spatial states x 2 = 8, and that math works the same whether the 1st excited states are in the {x, y, z, r} form or the $\psi_{2,l,m}$ form.
DeleteIncluding the spin-orbit interaction goes a stop farther than just including spin and that rearranges the states into two groups of 4 with two different energies 9slightly different).
"I think I answered my own question. Did you mean there are 8 original states Ψn,l,m,s that all have the same energy because we recognized that there is spin but didn't account for it in the Schrödinger equation?"
Deleteyes, exactly.
So, in a way, looking at question 4, there are really only four states in this new 1st excited state* when considering the spin-orbit interaction, since the anti-aligned $\vec{L}$ and $\vec{S}$ decrease the energy in this new first excited state*? So do we still consider there to be 8 different states in the first excited state, or do we want to only consider four?
DeleteFor number 6, are we to assume that we are looking at a mixed state for a 1D harmonic oscillator?
ReplyDeleteI see that a 1D HO is mentioned as part of the example at the top of the problem but am unclear about whether that is what we are looking at for parts a-d.
Thanks.
6 might as well be all 1DHO.
DeleteFor 7, is (-1,1) the other eigenvector? It seems to work. And what about (-1,-1), is this too similar to the first vector (1,1)? (-1,1) is certainly orthogonal to (1,1) as the angle is $\frac{\pi}{2}$. If we are not considering (-1,-1), then we need not consider (1,-1). Then the number of eigenvectors is 2, which is equal to the trace of the matrix, correct? I am a bit fuzzy on my linear algebra, so I'm not sure if I am interpreting everything correctly.
ReplyDeleteWas this the assignment that was not collected?
ReplyDeleteyes
Delete