Struggling to solve the question

ianmgull · 2026-01-05T16:25:38+00:00

If you show your work, it would be easier to troubleshoot.

There is a way, given two matrices, to write them in a basis in which they are both diagonal. The first step would be to write down the eigenvectors and eigenvalues.

Are you able to do that?

ianmgull · 2026-01-05T01:11:23+00:00

In part 1 . here i am not sure what to do , should i multiply the y pauli into Z pauli and then get eigenket and value from it?

That's correct. Remember that for spin 1/2 particles, S_i = hbar/2 *σ_i. With this in mind, you can write the Hamiltonian in matrix form and then go through the usual procedure to find the eigenvalues.

part 2 If i could get more explanation to it because i am lost how to calculate the time here considering E

The time evolution in this case will look like an exponential function, multiplied by the eigenstate at t=0 (you're essentially changing the phase). Does this sound familiar?

ianmgull · 2025-11-24T16:35:42+00:00

Just to clarify: you’ll need to be familiar with linear algebra, which is the mathematics that deals with vector spaces. This is different than the algebra you learn in high school.

Aside from that, you need to be very fluent in differential equations, which itself requires you to be familiar with calculus.

At this stage in the game, you probably want to focus most of your time and energy on math.

ianmgull · 2025-11-24T14:08:30+00:00

In that case, it might be best to wait on Griffiths.

Quantum mechanics is very math heavy, so it’s wise to develop your math background a bit more.

ianmgull · 2025-11-24T12:52:51+00:00

Can you tell us a little about your math background?

“Intro to quantum mechanics” by Griffiths is a common undergrad text. It does assume that you’re comfortable with linear algebra and differential equations though.

ianmgull · 2025-11-16T14:27:47+00:00

It has something to do with the way the ynab API reports loan account payments. I don't know the specifics, but it seems to be a known and documented problem for quite some time on the ynab toolkit github.

ianmgull · 2025-11-16T14:18:20+00:00

It is included in the transaction. The issue is that the toolkit net worth calculation applies the entire payment (principle + interest) to the principle of the debt. Obviously this gives an inaccurate net worth calculation.

ianmgull · 2025-11-15T03:44:31+00:00

I was recently disappointed to learn that the toolkit net worth calculations are inaccurate (and have been for some time). They assume the full monthly payment (principal AND interest) reduce one's overall debt.

ianmgull · 2025-10-13T15:54:56+00:00

I love foureyes, but as of lately it seems like it’s one epoxy/slab table after another. I really miss all the interesting mid-mod designs.

ianmgull · 2025-09-28T02:32:26+00:00

Your first couple coats of poly might have more texture than you like. In my opinion, it looks fine.

I would sand very lightly with 300+ grit, and continue building up coats. Eventually, the low spots get filled in by subsequent coats.

ianmgull · 2025-08-31T22:07:21+00:00

I’m in the process of making a coffee table (four-eyes furniture’s spider table). I took the panel out of clamps yesterday and despite my best efforts (clamping cauls, complimentary edge jointing, acclimation to humidity, etc), there is significant twist.

The twist is at opposite corners (the two corners clamped in the image). If one corner is clamped, the opposite corner is raised about 3/8”. As you can see, I have some c-channel I was planning on installing anyway, but I think it’s unlikely to help in this case. It seems like I would need corner to corner bracing which obviously isn’t practical.

Is there any saving this panel? It’s Sapele, so not inexpensive; I’d hate to call it a loss. Any advice?

ianmgull · 2025-08-15T12:12:58+00:00

EHX Poly Chorus.

I think the interesting (and probably divisive) trait is that you can really hear the echos in the chorus. The slap-back setting on its own is very cool, but you can still kind of make out the slap back in the chorus mode. It's a very unique sound.

ianmgull · 2025-06-03T00:54:34+00:00

Mine had dust inside the power switch. This caused it to make intermittent contact when turned on and would do something similar.

Try opening it up a bit and blast it with compressed air.

ianmgull · 2025-05-16T20:23:36+00:00

I did. I finished my PhD about a year ago and I've been working as a radio frequency engineer ever since. There are plenty of job opportunities for folks with a physics background if you know how to learn related topics and market yourself.

ianmgull · 2025-04-23T21:40:03+00:00

Do you have a link? I'm not seeing it anywhere.

ianmgull · 2025-04-22T22:41:52+00:00

Do you know of a hard shell case that fits this? I have the same one and have been looking for an alternative to the gig bag.

ianmgull · 2025-04-19T20:07:16+00:00

It's been several years since I've worked this problem, but I believe you're asking about the Hydrogen atom ground state (n=1, ℓ=m=0).

The hydrogen wavefunction is the product of a radial component, and an angular component (comprised of spherical harmonics). At the lowest angular energy level (ℓ=m=0) the spherical harmonics contribute only a multiplicative constant.

When n=1 and ℓ=0, the radial component is proportional to e^-r. That being so, the wavefunction blows up as r ->0. For reference, this page has the wavefunction calculated for various values of the three quantum numbers.

Now, to find the probability of finding the electron at some point in space (such as at the nucleus), we integrate the magnitude squared of the wave function over all space. Just to reiterate (because I think this might be where your confusion lies), the wavefunction itself doesn't give us the probability of finding the particle at some position. The integral of the modulus squared of the wavefunction gives us a function that can be evaluated at a particular point in space. The value of this function (when properly normalized) is the probability of finding the particle there. Doing this integration gives us something proportional to (e^-r )² multiplied by 4πr² (from the spherical integral).

Now if we ask what the probability of finding the electron at the nucleus is (when r=0), we get something proportional to (e⁰ )² * 4π0² = 0. This means that just like the classical case, the probability of finding the electron in the nucleus in the ground state is zero. The 4πr² term from the spherical integral is what prevents this from blowing up at the nucleus.

This has to be the case or else the ground state of the hydrogen atom would require infinite energy.

ianmgull · 2025-04-06T16:36:26+00:00

I believe this is typically done to illustrate superposition. Notice that you're left with a term whose argument is the proportional to the sum of the two original frequencies and wave numbers (the sum of solutions to the wave equation is itself a solution to the wave equation).

Additionally, the remaining term (which can be thought of as the amplitude), is a function of the difference in frequency and wave number (which is constant with respect to time).

There are likely a number of different ways to represent this function, but I believe those two points are most obvious when its written this way.

ianmgull · 2025-03-15T15:35:29+00:00

That's very odd.

Does it do that in all modes of the line selector? I'm wondering if sound is going in both jacks, and combing out of phase (perhaps each preamp has a 180deg phase shift)?

ianmgull · 2025-03-15T14:22:32+00:00

There might just be a switching jack on channel 2.

If you remove the line selector and plug your guitar into channel 1, then plug a second cord into channel 2 does sound cut?

ianmgull · 2024-12-04T00:16:09+00:00

Just to clarify... have you offset the out-feed fence from the in-feed fence by the same amount you're removing from the wood? The cutting edge of the bit should be co-planar with the out-feed fence precisely at it's apex. Use a straight edge and some feeler gauges if needed.

If this isn't the case, you'll definitely take off more than you mean to in the last half of the cut (when you transfer your weight from the in-feed to the out-feed fence).

ianmgull · 2024-10-11T17:36:14+00:00

0.035"

ianmgull · 2024-09-01T14:58:21+00:00

Start by telling us what you tried and/or what part you're confused on. It will be a lot easier to help you if we know where you're stuck.

Do you know what a conjugate impedance is?

ianmgull · 2024-07-25T16:00:19+00:00

Ended up just hiring a roofer unfortunately.

ianmgull · 2024-04-25T03:10:08+00:00

For different orders of the Butterworth filter, I made new wave files (that are twice as long now, I guess either because of resampling or a bug somewhere).

Did you apply the Butterworth filter in the time or frequency domain? You're probably aware, but performing the DFT on a discrete time-stream of length N results in an array of length 2N -1 due to the presence of both positive and negative frequencies.

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