Infinite square well states, orthogonality and completeness (Fourier series)
ฝัง
- เผยแพร่เมื่อ 13 พ.ค. 2013
- (This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at th-cam.com/users/playlist?list=...)
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Beautiful and clearly expressed lectures. Thank you so much, Doctor Brant @BrantCarlson, for making your lectures public to everyone.
22:39 The blue curve should be going up to n = 2 since there are two local extrema. The function should only be able to reach a local extrema once if n = 1.
I was about to comment precisely this :)
For this function (saw tooth or linear ramp), all odd terms (n=1, n=3, etc.) have coefficient c[n]=0
I have one small remark. If we made the analogy of inner product of functions with the dot product of vectors, we could have also explained the meaning of formula for Cn trough the idea of projection, which is also related to dot product. Cn is a number we get by doing orthogonal projection of f(x) to suitable „basis vector". The function space bears close resemblance to the ordinary Euclidean space. Later the jump at 0 and a is the so called "Gibbs phenomenon" for those that would like to google it. The overshoot remains whatever is the number of sines we add, the amount of overshoot remains the same it just gets narrower.
For my PhD studies, I had to write an essay on the Gibbs phenomenon; Then I was not experienced enough on Fourier Analysis to realise that it is closely related to convolution ... (nevertheless, I got a good grade on that work).
Very accessible explained, especially the Fourier sine transform. But first of all I admire your ability to write with the mouse, it really isn't easy. Good old school!...
It's probably a drawing tablet. Otherwise, that would be really impressive!
Fourier was a badass.
nice! really liked this lecture thanks a lot
Thank you so much! I love you!
What software do you use to record these, if you don't mind me asking?
really very good explanation .... Thanks Sir
Please can you add the answers to the questions?
amazing work
Thank you so much sr you are awesome thanks a lot it helped me lot
In part 2, thats -a/(2+x) or (-a/2)+x?
shouldnt the second term (Ψm) at 12:57 be without the complex conjugate? plz reply
+patrinos13 Of course, there is a typo there. He didn't mean it.
Great. Thanks
Thank you so much 😘💖
I love you
Correct me if I'm wrong but isn't 2/a supposed to have a square root at 17:51 ?
I think so too. that's what Griffiths has as well
Part 1:
C2=0, C3=1, C4=0 cus duh.
Part 2:
C2 = -a/π
but the general solution is: f(x) = sigma (n=0, to infinity) [-2asin(2nπx/a)/(2nπ)]
i.e. Cn = -2a/(nπ) where n is even (that's why I multiply it by 2 in the sum)
hiii
17:16
the integration part became a\2
why?
i think it has to be 1 if it is normalized
any help!!!
This is explained in the previous video. Essentially sin^2 is rewritten by using trigonometric identities. Solving the integral that way gives you a/2. Video with timestamp here: th-cam.com/video/nFHhLJGDNHA/w-d-xo.html
Check Your understanding:
Part 1: C2=0, C3=1, C4=0
Part 2: -a/pi
13.33 why two conjugate?
prabhakar oli he made a mistake. It should be Psi* Psi
why do we need psi*.. Even (psi)(psi) would give the same result
raghunath karri (psi)*(psi) would give the same result as (psi)(psi) if psi is a real function. We know that it can be complex so we use the conjugate rather than square
answer to part 01 : c2=0, c3=1 and c4=0
answer to part 02 : c2=-a^2/2*pi
ps. correct me if i am wrong
I get the same mate
I think for part 2, you forgot to multiply by 2/a that was outside the integral. I got -a/pi
shouldn't n+1 be the number of nodes
If we consider the end points that is, otherwise it's n-1.
@@1ashad1 ooh ok makes sense
08:48, that m≠m laugh stinks of unwelcome foreshadowing
22:22 Take a look at the boundaries of the box: can you explain _how on Earth_ did you get a _positive_ value of the combined wavefunction (black) at that boundary if ALL those sines have a value of 0 there? How is it possible to add a bunch of zeros (even infinitely many) and get a non-zero value? :q
Let alone that previously you used the same argumentation to rule out the cosines as valid solutions (since they cannot be 0 at boundaries), and yet now you're saying that the black wavefunction have a non-zero value there :q Don't you think that this logic isn't sound? You said that the wavefunction _cannot_ be nonzero at boundaries, because it is zero outside of the box, and by the continuity condition, it must also be zero at the boundary. By the same argument, it must tend to 0 in the close vicinity of that boundary inside the box. The example wavefunction (black) doesn't do that - it would have to be discontinuous at the boundary (i.e. abruptly change its value from positive to 0), which violates the continuity requirement. Can you explain?
You're mixing up examples. The example at the end is only to show how you can approximate any smooth function as a summation of many sine function. The black function is just given. That's what we're trying to approximate
@@OmnipotentO, I'd say that Brant was a bit quick jumping form QM waves to waves in general; more so after having discussed the infinite potential square well. I don't intend to critizise; I'm having a great time with this course (some lapsus may haphazardly creep in, though) :)
Try this video: th-cam.com/video/yBvsNueXIw4/w-d-xo.html
blehbleh