The bound state solution to the delta function potential TISE
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- เผยแพร่เมื่อ 20 ก.ค. 2024
- Solutions away from the delta function are connected with boundary condition matching at the delta function to give an expression for the bound state solution to the delta function potential TISE. (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=...)
Have watched all 51 videos in the play list. Brilliant !
I have learned a lot from them, thank you.
Do you have plans (Time) to make videos associated with the applications chapters (6 - 11) of Griffiths ?
Thanks for your videos. I'm taking a course in quantum physics right now and they are super helpful for studying
Thanks man!!! It really helped. Is there a video for the double delta function potential done by you? I just love your way of explaining because you talk about the mathematical reasoning as well.
How would you go about finding the probability that the measurement of the particle would give that energy you found E=-ma^2/2hbar^2?
Great explanation. What software do you use to write?
1) Because you don't have any variable you can change to get other states? Or is there some deeper answer.
2) No because you only have one bound state, so it's not complete.
Awesome man 🧠
For check your understanding part 1 , you are talking more mathematical. I think we have only one energy state to make psi normalizable (In other words to make it kiss the axis) as for others it may not?
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?
thank you for you awesome video
THANK YOU!
Good explanation 😍😍
Sorry for asking the previous question. I forgot that the coefficients can be complex, since u were only solving for real coefficients.
Why do we not solve for the imaginary part of psi(x). Is is some kind of initial condition for our stationary state to be real at t=0. Because a complex stationary state solution for the wavefunction in this case could be at any angle in the complex plane at t=0, however the probability density is not affected as magnitude is same for all angles.
Btw these videos are great. Really appreciate the good work
Sir I have the same doubt, did you ever get answer to your ques?
@@AkshitSharma0 If a self-adjoint operator has multiplicity 1 for all eigenvalues (a non-degenerate spectrum) then the eigenvectors / eigenfunctions can all be taken to be real. You can prove this by assuming phi(x) = u(x) + i v(x) is an eigenfunction, then use the non-degeneracy to show that u(x) and v(x) are linearly dependent. Note that although the energy eigenfunctions can always be taken to be real, we will still need complex combinations of them to represent the time-evolution of an arbitrary state.
At 6:20 - 6:38.. Shoudn't it be psi_1 = Be^kx and psi_2 = Ae^-kx ??
You are right.
Yes, you are Right. But it doesnt matter. Because A and B are just constants and you can replace them. ;)
yes but it makes it confusing it looks like solutions are zero on graph
was gonna comment the same thing
The function must be continuous, so at zero, the e^kx and e^-kx are both 1 thus their normalization factors A and B must be equal
@Brant Carlson at 15:05 how did you get two B's as constants since the initial wave functions have A and B?
For psi to be continuous he found that A must be equal to B
For normalized constant B, I got (2ma)^(1/2)/(h_bar)^2 instead of (ma)^(1/2)/(h_bar)^2 in the video. I double checked my answer. I am not sure about my answer.
商君 try solving again with normalisation condition.
I got the correct one.i.e the root of k.
I get 2ma on the numerator as well. I got it from multipling the e twice which yields 2kx in the exponent. I'm not sure which answer is right now.You do get the root of k but there's a 2 which comes from the squaring.
@@moniadixit28 Not sure about this anymore because I too got root 2k which yields 2ma in the numerator.
I love you mysterious Quantum Mechanic
My answers:
1) because only one side is bounded, unlike in the infinite square well case where we have two sides both bounded
2) I think so because initial conditions must also satisfy the boundary conditions
Answer to "Why do we not solve for the imaginary part of psi(x)":
By the theory of ordinary diff eqns, only real parts are enough.. If you want the imaginary part, you can simply add a complex number to it; nothing changes..
Hi Brant. What happens if the E = 0? Thanks.!!
Neal Alfie Lasta this condition is not allowed in QM.
What actually is TISE?
time independent schrödinger equation
At time 12:20 the write hand side is zero not because we are integrating on a very small area(There is a discontinuity!!).In fact, the reason is that wave function is even so the area under the curve to the right is equal to the area under the curve to the left so the term vanishes.
No. It will only vanish if the function was odd and not even.
The integral gives (considering psi(x) almost constant near x=0, i.e. for very small epsilon): E by 2 by epsilon by psi(0), that goes to zero as epsilon does. And there's no discontinuity in psi(x), because A=B
Explain why A=B.
The wave function must be continuous. Therefore at x=0, the two functions (the one for x>0 and the one for x
A tad sloppy at times. And I just don't understand why ask for the uniqueness of the bound state, because it should be obvious from the exposition ... or I'm missing something.