This channel is a collection of physics video lectures recorded for or delivered at Carthage College. As for me, I am a professor of physics at Carthage specializing in lightning and other forms of atmospheric electricity and high-energy atmospheric phenomena.
How to find the normalization constant B: psi is normalizable implies that the integral over all space (-infty to infty) of psi* psi = 1 Here psi = B*e^(kx) if x < 0 and B*e^(-kx) if x > 0. Note that psi* = psi in both cases. The integral over all space of psi* psi can be split as the sum integral1 + integral2, where integral1 goes from -infty to 0 of psi* psi and integral2 from 0 to infty of psi* psi. integral1 psi* psi = B^2 e^(2kx) = B^2 e^(2kx)/2k evaluated at lower bound -infty and upper bound 0: B^2/2k - 0 = B^2/2k integral2 psi* psi = B^2 e^(-2kx) = B^2 e^(-2kx)/-2k evaluated at lower bound 0 and upper bound infty = 0 - (B^2/-2k) = B^2/2k integral1 + integral2 = B^2(1/2k + 1/2k) = B^2/k which must equal one. Hence B^2 = k and we get B = sqrt(k) = sqrt(ma^2/h_bar) Alternatively one can use psi = B*e^(-k|x|) and that psi is an even function. The integral over all space then becomes 2 times the integral from 0 to infty, which is 2*integral2 in the above calculation. This leads to the same result 2*(B^2/2k) = 1 -> B = sqrt(k).
The wave function collapses and becomes narrow, right? Could that possibly have any links with the idea that particles are different when unobserved and when observed, or that you can only know the velocity or the position but not both, or quantum superposition? Or maybe it's just a coincidence.
I'm really not sure if I'm right, but I think the issue is that we aren't taking the integral of the function. When we don't take the integral, I don't think X1*X2 necessarily equals 0. If I'm right, it means that the time dependencies don't cancel each other out when we multiply out. I'd love if you could give me a real answer though, this is just what I got.
@@phillipdavis2290 You are correct. Most people here are confusing the expectation value with the probability density. The question asks for the probability density of the superposition of the two states. This then leads to the integrand at 18:05 (without the energies E1 and E2 as coefficients, since we don't use the hamilton operator for the probability density). So there remains a time-dependent oscillation term in the cross terms.
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
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.
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.
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?
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.
Momentum is a vector quantity. It has a magnitude certainly but if you say 'momentum' then you are talking about magnitude and direction which on average is definitely zero.
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?
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
