The bound state solution to the delta function potential TISE

  Рет қаралды 47,307

Brant Carlson

Brant Carlson

Күн бұрын

Пікірлер: 41
@Oreoboy101
@Oreoboy101 9 жыл бұрын
Thanks for your videos. I'm taking a course in quantum physics right now and they are super helpful for studying
@stevenkay7269
@stevenkay7269 10 жыл бұрын
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 ?
@bol3r0
@bol3r0 3 ай бұрын
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).
@bakirev
@bakirev 5 жыл бұрын
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.
@BPHSadayappanAlagappan
@BPHSadayappanAlagappan 3 жыл бұрын
Awesome man 🧠
@cadetkhan133
@cadetkhan133 2 жыл бұрын
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?
@vishwaashegde4987
@vishwaashegde4987 9 жыл бұрын
At 6:20 - 6:38.. Shoudn't it be psi_1 = Be^kx and psi_2 = Ae^-kx ??
@solethunkosi6624
@solethunkosi6624 8 жыл бұрын
You are right.
@ManojTiwari-ky3fl
@ManojTiwari-ky3fl 8 жыл бұрын
Yes, you are Right. But it doesnt matter. Because A and B are just constants and you can replace them. ;)
@orchoose
@orchoose 6 жыл бұрын
yes but it makes it confusing it looks like solutions are zero on graph
@akhilmaru6999
@akhilmaru6999 2 жыл бұрын
was gonna comment the same thing
@alexiahartzell4055
@alexiahartzell4055 Жыл бұрын
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
@yahyahamdan147
@yahyahamdan147 10 жыл бұрын
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.
@manishsingh-vk8if
@manishsingh-vk8if 5 жыл бұрын
No. It will only vanish if the function was odd and not even.
@MiguelGarcia-zx1qj
@MiguelGarcia-zx1qj 3 жыл бұрын
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
@weizhou3928
@weizhou3928 2 жыл бұрын
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..
@MrEzystreet
@MrEzystreet 8 жыл бұрын
@Brant Carlson at 15:05 how did you get two B's as constants since the initial wave functions have A and B?
@philandros3195
@philandros3195 8 жыл бұрын
For psi to be continuous he found that A must be equal to B
@DargiShameer
@DargiShameer 2 жыл бұрын
Good explanation 😍😍
@fadiliqbal8329
@fadiliqbal8329 6 жыл бұрын
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.
@emillytabara9410
@emillytabara9410 4 жыл бұрын
Great explanation. What software do you use to write?
@thequazzman
@thequazzman 9 жыл бұрын
I love you mysterious Quantum Mechanic
@蕭宏鎮-e2y
@蕭宏鎮-e2y 2 жыл бұрын
thank you for you awesome video
@waqarmirza8372
@waqarmirza8372 6 жыл бұрын
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
@AkshitSharma0
@AkshitSharma0 2 жыл бұрын
Sir I have the same doubt, did you ever get answer to your ques?
@iyziejane
@iyziejane Жыл бұрын
​@@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.
@waqarmirza8372
@waqarmirza8372 6 жыл бұрын
Sorry for asking the previous question. I forgot that the coefficients can be complex, since u were only solving for real coefficients.
@Crusherdub
@Crusherdub 9 жыл бұрын
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?
@商君-l4y
@商君-l4y 8 жыл бұрын
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.
@moniadixit28
@moniadixit28 7 жыл бұрын
商君 try solving again with normalisation condition. I got the correct one.i.e the root of k.
@matrixate
@matrixate 4 жыл бұрын
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.
@matrixate
@matrixate 4 жыл бұрын
@@moniadixit28 Not sure about this anymore because I too got root 2k which yields 2ma in the numerator.
@ifrazali3052
@ifrazali3052 6 ай бұрын
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?
@Petarat94
@Petarat94 10 жыл бұрын
THANK YOU!
@ReynaLuviano-f4f
@ReynaLuviano-f4f Жыл бұрын
Explain why A=B.
@ilayws272
@ilayws272 Жыл бұрын
The wave function must be continuous. Therefore at x=0, the two functions (the one for x>0 and the one for x
@neallasta
@neallasta 10 жыл бұрын
Hi Brant. What happens if the E = 0? Thanks.!!
@Spaceman585
@Spaceman585 8 жыл бұрын
Neal Alfie Lasta this condition is not allowed in QM.
@sagarrawal8332
@sagarrawal8332 7 жыл бұрын
What actually is TISE?
@jonatan8973
@jonatan8973 7 жыл бұрын
time independent schrödinger equation
@MiguelGarcia-zx1qj
@MiguelGarcia-zx1qj 3 жыл бұрын
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.
Scattering state solutions to the delta function potential TISE
24:15
Potential functions in the Schrodinger equation
22:34
Brant Carlson
Рет қаралды 64 М.
BAYGUYSTAN | 1 СЕРИЯ | bayGUYS
36:55
bayGUYS
Рет қаралды 1,9 МЛН
Une nouvelle voiture pour Noël 🥹
00:28
Nicocapone
Рет қаралды 9 МЛН
Support each other🤝
00:31
ISSEI / いっせい
Рет қаралды 81 МЛН
Bound states, scattering states, and tunneling
16:39
Brant Carlson
Рет қаралды 59 М.
Infinite square well (particle in a box)
21:31
Brant Carlson
Рет қаралды 179 М.
Finite square well bound states
24:32
Brant Carlson
Рет қаралды 88 М.
The Dirac delta function
27:01
Brant Carlson
Рет қаралды 117 М.
Boundary conditions in the time independent Schrodinger equation
14:36
Separation of variables and the Schrodinger equation
32:11
Brant Carlson
Рет қаралды 67 М.
Free particles and the Schrodinger equation
14:19
Brant Carlson
Рет қаралды 119 М.
Schrodinger equation in 3d
18:02
Brant Carlson
Рет қаралды 71 М.
Quantum harmonic oscillator via power series
48:11
Brant Carlson
Рет қаралды 54 М.
Superposition of stationary states
21:37
Brant Carlson
Рет қаралды 36 М.
BAYGUYSTAN | 1 СЕРИЯ | bayGUYS
36:55
bayGUYS
Рет қаралды 1,9 МЛН