For the 18:30 Check your understanding problem I got: ρ(x, t) = c₁²•|X₁(x)|² + c₂²•|X₂(x)|², so, I don't find the time-dependence.

Thank you

Thank you

Thank you

I guess at 6:55 the inequality for a and b has to be reversed

Adieu

And that's a really ugly Psi Must fix.

Ill-defined?

did anyone understand the spin =wtf thing

I be watching this on my pz

my guess a) no istnt smooth, curves wrongon Left side b) smooth curves wrong on left side. looks good on right sidde c) All correct d) Has to curve When E<V so also bad

PSi looks very communist

I dont understand the check questions answer/ What is correct?

Anybody got the answer to the Understaning Check?

14:48 There is a star missing on the last blue psi for those who are confused :)

So is the continous distribution is not a wave funktion. Youd would have to square the wave funktion, and then integrate to get the expctation values for that right?

Check your understanding: . . . 1. Eigenvalue of Sy : hbar/2 and -hbar/2 ( just like Sx and Sz ) 2. Eigenvectors of Sy: y_plus = 1/sqrt(2) * ( 1, i ) # with eigenvalues hbar/2 y_minux = 1/sqrt(2) * ( 1, -i ) # with eigenvalues -hbar/2 3. x_plus state can be expressed in Sy eigenstates, as x_plus = ( 1/2 - i/2) * y_plus + ( 1/2 + i/2 ) * y_minus. So the possibilities is both 0.5 to attain one of the states of corresponding eigenvalue. 4. Same as before, 0.5 possibility to get both states and corresponding eigenvalues.

check your understanding: . . . 1. L2|f> = 2hbar^2|f>, L = 1, Lz could be -1, 0 , 1 ( all * hbar ) . A total of 3 states. 2. L2|f> = 15/4 hbar^2|f>, L = 3/2, Lz could be -3/2, -1/2, 1/2, 3/2 ( all * hbar ). A total of 4 states. 3. Lz|f> = 0, L2|f> could be any (hbar)^2 * l * ( l+1 ), l = 0, 1/2, 1, 3/2, .... 4. Lz|f> = 3/2 * hbar |f> could be any (hbar)^2 * l * ( l+1 ), l = 3/2, 2, 5/2, 3, ... Shouldn't there be a |f> at the end for the 4th question?

Check your understanding: . . . . 1. [x, Py] = 0 2. So we could measure x and Py as precise as we want. 3. [Px, Py] = 0 also. So we could measure Px and Py as precise as we want.

I kind of liked that proof of the roots of unity. that x^3=1 bit. You make y=0, the unity thing kicks in anyway.

Check your understanding: . . . . 1: H applies on the right state |ψm>, and return Em|ψm>. Em is just and number and get outside of the braket. So Hnm = Em<ψn|ψm> 2: Hnm is a diagonal matrix with values E1,E2...En on its diagonal line. 3: a+|ψm> and returns 1*|ψm+1>. a+nm = <ψn|ψm+1>. 4: a+nm is a off-diagonal matrix with values 1,1,1,1,1.... on the line just below the diagonal line.

Check Your Understanding at 19:30: . . . Part 1: a. Got (QR+RQ-2(µQ)(µR))/2, So it's not a commutator I guess. b. -(µQ)(µR) is an extra term, what does it mean? Part 2: Because µR is just a number, it could be moved outside of the braket to the front, and the rest part is just expectation of Q.

It is now 2024... I already forgot how much difference you can feel between 480p and 720p. This video is much sharper than the previous one.

Check Your understanding: Part 1: C2=0, C3=1, C4=0 Part 2: -a/pi

How do we know that psi-0 is equal to B? Isn't it that we can not use psi-1 or psi-2 equations to solve for psi-0 because they do not include x=0? Then why are we using them to find psi-0?

kzbin.info/www/bejne/gp6chI2cZ715ecUsi=5h5AElm81LIXkFdT took your videos

is this is the part of jee syllabus

Ah yes, my go to data set when I'm trying to understand discrete probability distributions. ~sexual partners of the general public

is this guy a dentist on some naruto shit

Had to set the playback speed to 0.75 that was way too fast man

Thank you for these lectures, these make so much more sense!!

Amazing, just check the comments for certain errors or keep the Griffiths book next to you to verify. Thank you for the video

At time 5:17 is said that the average momentum of an electron is zero because the Hydrogen atom is not moving: this is true only as for as the vecton momentum is conserned not as its value though which is constant.

Thank you Dr. Carlson! Just blazed through your PHY4200 playlist to study for my final. You're an amazing teacher and your explanations really helped build intuition.

When you write p-hat squared psi, does that mean p-hat(p-hat(psi)) or does it mean (p-hat(psi))(p-hat(psi)), i.e. squared in the traditional sense. For that matter, it looks like you're using (x-hat)(p-hat)psi = x-hat(p-hat(psi)), otherwise they would commute. But earlier, you definitely treated (p-hat)(p-hat) as p-hat squared. Can someone clarify please?

why isn't at 22:30 the phi function an arbitrary sum of opposite exponentials as opposed to just one exponential nonetheless m ends up a whole number

so from the wave animation the wavelet flattens out and there is no more wavelet and the probability animation suggest that the particle can be anywhere. Is that right?

at around 6:00 you assume the function only contains sines and cosines instead of letting it be complex exponentials, it ends up the same because once you assume those complex exponentials form an even function, then you arrive at sine anyways

Maaaaaaaaaany thanks

YAYY the static wavey has no momentum YUPPIIEEE

What the fuckkkkkkkkkkkkk 😢

that's what I neeeeeeed!

why can you write a+ in the left side? isn't that implying that the ladder and the hamiltonian commute? (which they seem not to)

you mixed up minus and plus signs in the opperators, but ill let it slide ;)

Thank you about the last part in particular. It does seem quite fishy, but it is very necessary in QFT. That one part caught me by surprize.

5:20 Maybe the correct approach is to say: It's NOT a point particle. If you do a measurement that collapses the wave function, then its position collapses to a point. But the rest of the time, it's a spread out thing. How does quantum field theory describe the electron? NOT as a point particle, right? But as a spread-out disturbance in a field, right? So maybe SPIN is some kind of phenomenon in what the electron field is doing.

Maaaaaaaany thanks

Great lecture. Thank you!

19:00 "It SORT OF goes to zero". I'm thinking it might DEFINITELY GO TO ZERO if we use the Lebesgue Integral.

You saved me man, thank you so much