es impresionante tanta humildad de este hombre ... gana tanto dinero y no le interesa manchar su pantalón con tiza, para él lo importante es compartir el conocimiento. Héroe sin capa.
If you ever find yourself recording lecture for youtube, it is better to keep the instructor on screen at ALL TIMES instead of lingering on equations they have left behind on the board. We can pause to read equations. We can't recover the hand gestures, etc. that we miss while they are off screen.
If you check the time independent Schrodiger equation, you will notice that if psi=0, then psi'' must also be 0. Now, if you derive the two sides of the schrodinger equation, you will again notice that if you have psi'=0, then psi''' must also be 0. Keep deriving the SE and all higher derivatives will be 0. As a result, psi will be identically 0 whenever psi=psi'=0. 0 is not a valid value for a wavefunction
What I meant was that, when you derive the SE, you will have d^3(psi)/dx^3=constant*psi'. Here, if psi'=0, then d^3(psi)/dx^3=0. If you keep deriving, all higher derivatives will be 0. I hope I helped.
It's a late answer, but if someone else has this doubt: the Time Independent Schrodinger equation is a second-order ODE, which obeys the unicity theorem for its solution. Given that, if the boundary conditions (BC) psi(x0) = psi'(x0) = 0 and TISE are satisfied for a guessed psi(x) solution, this is a unique solution to the problem. A good solution guess will be psi(x) = 0 for all x (including x0), which satisfies both BC and TISE, so this must be the unique solution. However, this is a trivial solution and not physically acceptable to our problem. Therefore, given a continuous 1D potential that generates bounding states, those eigenstates will never be pis(x0) = psi'(x0) = 0 for any particular x0.
@@dorysouza1770 I'm a bit confused by the difference between boundary conditions and initial conditions here. Would you call psi(x0) = psi'(x0) = 0 the *inital conditions* instead?
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es impresionante tanta humildad de este hombre ... gana tanto dinero y no le interesa manchar su pantalón con tiza, para él lo importante es compartir el conocimiento. Héroe sin capa.
pero la tiza es elegante
Bién, el es Peruano, Universidad Nacional de Ingeniería.
If you ever find yourself recording lecture for youtube, it is better to keep the instructor on screen at ALL TIMES instead of lingering on equations they have left behind on the board. We can pause to read equations. We can't recover the hand gestures, etc. that we miss while they are off screen.
Beautiful explanation! Thanks professor Zwiebach.
el MEJOR ESTUDIANTE DE LA UNI PERU, promedio historico de 19!!!, orgullo peruano
Un alumno de la FIEE, grande Barton
Brilliant
Thanks ❤️🤍
Talento peruano
Momento peruano
Grande!
Realmente impresionante
This is helpful ❤️🤍
Great lecture, thanks a lot!
Peruvian teacher!!!!!!!
Why wavefunction must satisfy that at any point it can have a Taylor expansion or ψ = ψ′ couldn't equal 0 at same time?
If you check the time independent Schrodiger equation, you will notice that if psi=0, then psi'' must also be 0. Now, if you derive the two sides of the schrodinger equation, you will again notice that if you have psi'=0, then psi''' must also be 0. Keep deriving the SE and all higher derivatives will be 0. As a result, psi will be identically 0 whenever psi=psi'=0. 0 is not a valid value for a wavefunction
Thanks. I ignore "if psi'=0, then psi''' must also be 0."
What I meant was that, when you derive the SE, you will have d^3(psi)/dx^3=constant*psi'. Here, if psi'=0, then d^3(psi)/dx^3=0. If you keep deriving, all higher derivatives will be 0. I hope I helped.
Junjiro Obando Molina that did help. Thanks!
@@JunO-tq6td Thanks for this.
i do not understand the 3:20 , can any one please explain
It's a late answer, but if someone else has this doubt: the Time Independent Schrodinger equation is a second-order ODE, which obeys the unicity theorem for its solution. Given that, if the boundary conditions (BC) psi(x0) = psi'(x0) = 0 and TISE are satisfied for a guessed psi(x) solution, this is a unique solution to the problem.
A good solution guess will be psi(x) = 0 for all x (including x0), which satisfies both BC and TISE, so this must be the unique solution. However, this is a trivial solution and not physically acceptable to our problem. Therefore, given a continuous 1D potential that generates bounding states, those eigenstates will never be pis(x0) = psi'(x0) = 0 for any particular x0.
@@dorysouza1770 I'm a bit confused by the difference between boundary conditions and initial conditions here. Would you call psi(x0) = psi'(x0) = 0 the *inital conditions* instead?
@@jonathan3372 Yes, that would be more accurate. They serve the same purpose and with partial differential equations, the lines can become blurred.
gran peruano