# Electron in a Coulomb Field

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1. Dec 18, 2016

### MxwllsPersuasns

1. The problem statement, all variables and given/known data

An electron in the Coulomb Field of the proton is in the state:

|ψ> = (4/5)|1, 0, 0> + (3i/5)|2, 1, 1> with |n, l, m> as the quantum numbers defining the state

a) What is <E> for this state? What are <L2> and <Lz>?

b) What is |ψ(t)>? Which expectation values from a) vary with time?

c) Show that the expectation value <x> will be time dependent provided the matrix element <ψ100|xˆ|ψ211 ≠ 0.

d)Show that this matrix element is indeed nonzero. You don’t need to calculate a value for the matrix element, although you can if you want.

2. Relevant equations
Basically laid out in the problem statement

3. The attempt at a solution

So I completed part a) and part b) except I haven't shown whether or not the Energy commutes with the hamiltonian. The reason I haven't is I haven't been able to find the representation of the energy (in order to commute it with the hamiltonian) anywhere. Part of me thinks that the energy operator has the Hamiltonian as the generator of time translations (right?) and as such since the energy operator involves the Hamiltonian when it is commutated with the actual Hamiltonian operator it would be, in effect, like commuting the Hamiltonian with itself, which obviously commutes. Can anyone comment on this line of thinking and let me know where I went astray or if it is indeed logical thinking? Also if anyone could provide the representation of the Energy operator I would be very thankful.

Moving on I need help with parts c) and d). Part c) asks to show <x> is time dependent (which means it doesn't commute with H) given that a particular matrix element is nonzero.. I imagine that when you commute H with x these matrix elements (<100|x|100>, <211|x|100>, <100|x|211> and <211|x|211>) play a role in the calculation and I'm assuming since we're tasked to show one particular element is nonzero than that means the other three are zero and this one term is what keeps x from commuting with H (and thus making it time dependent) can anyone comment on this as well and maybe give me some advice moving forward? I would truly appreciate it. Thanks!

2. Dec 18, 2016

### Gan_HOPE326

About your first question, the energy is the eigenvalue of the Hamiltonian. So by definition the energy is going to be given by the Hamiltonian, which obviously commutes with itself. That makes all too much sense: the energy is a conserved quantity. If it was time dependent in a closed system we'd have a huge problem on our hands.

About part c. The expression for the time dependence of an expectation value for a time independent operator is

$$\frac{d}{dt}\langle \psi |x| \psi \rangle = \frac{i}{\hbar}\langle \psi |[H,x]| \psi \rangle$$

which comes about quite naturally when you apply the composite derivative formula to the bra/ket wavefunctions and keep in mind the time dependent Schroedinger equation. Since in your case $\psi$ is expressed as a sum of eigenstates you can always bring the Hamiltonian out of the brakets and replace it with the energy (remember that $\langle211|H = \langle211|E_{211}$ for example), so you're going indeed to have those four terms you mentioned. The two homogeneous terms (with the same state on both sides) cancel each other because they appear twice with opposite signs (due to the commutator). The mixed terms instead do something like

$$E_{211}\langle211|x|100\rangle-E_{100}\langle211|x|100\rangle$$

hence don't commute and cause the time derivative to be non-zero. Which makes sense: in the end pure states are time-independent (except for their phase factor, which has no bearing on expectation values anyway), and whenever you have a mix of two states you observe oscillations with frequency $\frac{\Delta E}{\hbar}$.

Last edited: Dec 18, 2016
3. Dec 18, 2016

### MxwllsPersuasns

Thanks so much for your incredibly lucid and clear explanation of the problem Gan HOPE326, I really appreciate the insight! If you might help me with the next part I'm a little stuck on that as well. Basically it's Asking to show that <100|x|211> is nonzero (though we aren't required to calculate its value) however no information (other than the probability amplitudes associated with each wavefunction) about the two wavefunctions are given so I'm not sure what to do exactly.

The only inclination I would have is perhaps taking the x(op) out of the inner product then having something like x<100|211> and then perhaps showing that the wavefunctions |100> and |211> are NOT orthogonal.

Any ideas? Like I said I'm a bit unsure about how to approach this as there aren't value associated to these wavefunctions and so I'm not quite sure how I can prove that the inner product of <100| and |211> is not equal to zero with the information given.

4. Dec 18, 2016

### Gan_HOPE326

Well, we're talking hydrogenoid wavefunctions, so their form is well known. Basically it's an exponentially decaying radially multiplied by a spherical harmonic defined by the L,M quantum numbers. So you can certainly try and compute the integral by hand. A simpler way to obtain the result would just be to write down the integral and demonstrate the parity of the argument. Since x is an odd function, integrating over infinity the product $\langle211|100\rangle$ needs to be odd as well for the integral to be non-zero. I think that should be easily the case because |100> is the s orbital (even) and |211> is the p-orbital with m=1, which for what I remember should lie in the XY-plane, and therefore is likely to be odd in the X axis. I don't remember the exact form right now but you can check them anywhere and work out the result by yourself.

5. Dec 18, 2016

### MxwllsPersuasns

Wow, again, a clear explanation of the problem which addresses my faults perfectly! I shall follow your advice and I'm certain I'll be able to get everything to work out. Thanks again Gan HOPE326, I can't tell you how much I appreciate the help. You should be a physics instructor if you aren't already .

6. Dec 18, 2016

### Gan_HOPE326

Ahahaha, thanks . I've had an experience giving tutorial lessons in a course of structure of matter but that was it.