What an excellent series of minilectures! My probability density curve for grokking QM integrated over all of your videos is equal to one (understanding!) thanks to your insightful presentations.
That relationship came from simplifying the original differential equation, so if that equality is false then X(x)•T(t) would not be a solution to the differential equation (Six years ago, hopefully this is still relevant for you 😅)
25:30 So this is basically an *eigenvalue problem* for the differential operator :J The eigenfunction of the differential operator is the exponential function, and the eigenvalue is -i·E/ℏ, or from de Broglie's equation E = h·f = ℏ·ω → E/ℏ = ω we get that the eigenvalue has a form -i·ω .
If the schrodinger equation comes from he classical wave equation, why in the first term in the time dependent schrodinger equation appears the psi function in the first derivative and not in the second derivative form? Hope i make miself clear, im not that good at english. thanks
It's because the Schrödinger's equation has been derived by taking the formula for the wave function of a "free particle" (plane wave) and replacing the ω and k with E/ℏ and p/ℏ, and then looking for the time and space derivatives, which are needed for the Hamiltonian. The wave equation is not a Hamiltonian - it expresses the dependence between the curvature in space and the curvature in time of the wave function, which are both second-order derivatives. The Schrödinger's equation, on the other hand, expresses the dependence between the kinetic energy (related to changes in time) and the potential energy (related to changes in space), which are first-order derivatives. There's a nice analogy you can take a look at, regarding harmonic motion: The equation for the Simple Harmonic Oscillator (SHO) i s d²x/dt² = -ω²·x, so it is has a second-order derivative, and the solution is that nice sinusoidal oscillation you're probably well familiar with. But the harmonic oscillator can be also expressed in the Hamiltonian way, in terms of energies: the total energy E consists of kinetic energy m·v²/2 and potential energy k·x²/2, so the Hamiltonian is E = m·v²/2 + k·x²/2. Now you can replace the `v` in this formula by the first derivative of position with respect to time, dx/dt, obtaining: E = m·(dx/dt)²/2 + k·x²/2 It is just a different way to express the conditions on the function which describes the motion of the oscillator: one is in terms of curvatures, the other is in terms of energies. But the solution is the same for both, because they both describe the same harmonic oscillator.
Can someone explain why both solutions to the 2nd Order ODE was not included. Should we not get the solution T = Aexp[-sqroot(a)t] + Bexp[sqroot(a)t]? Why has the negative one been ignored?
No! The ODE has been solved by rearranging the variables and integrating, it is not a seond order ODE with the general solution of Aexp[-wt] + Bexp[wt]
Hmm so what is it that makes an equation separable? How can we tell if a given equation can be solved by separation of variables or not before actually trying? (since the separation process can be messy sometimes, and it would be cool if we could tell if it will work before going into all those gory details :q )
Sci Twi I've been wondering myself. From what I can gather (also from the book), if there is another function in the PDE that makes it so that the variables can't be seperated easily, solving the PDE will become a whole lot harder. The example they gave was if V was V(x,t) instead of just V(x), then it will not be possible to solve with SOV if V's variables aren't seperable themselves...? I don't know. Maybe there exists some weird PDE that has has the exact opposite property.
WOAH! I wish you where my professor! You explain things so thoroughly it makes so much sense.
Neil Volk Our professor gave this as a homework .. this video was my savior
What an excellent series of minilectures! My probability density curve for grokking QM integrated over all of your videos is equal to one (understanding!) thanks to your insightful presentations.
I envy your students Dr. Calson. I wish that my University had you in their Physics Department. I've learned so much from just your videos alone.
Great lecture for self independent learners!
At around 12:15, you explain that the equation f(t) = g(x) must hold for every t and every x. Could you explain why that must be the case? Thanks!
That relationship came from simplifying the original differential equation, so if that equality is false then X(x)•T(t) would not be a solution to the differential equation
(Six years ago, hopefully this is still relevant for you 😅)
@@faielgila7375 hopefully he hasnt bene stuck on this for 6 years lol
@@pedramnoohi2715 or dead
Lovely insight into the wave equation being a relationship between acceleration and curvature.
Yeah. That point really is a lovely idea.
Thank you for your clear explanations. I think you are a great teacher.
I understand many concepts with you :D I wish that you share more videos about undergraduate physics like 'electromagnetic theory' :)
25:30 So this is basically an *eigenvalue problem* for the differential operator :J The eigenfunction of the differential operator is the exponential function, and the eigenvalue is -i·E/ℏ, or from de Broglie's equation E = h·f = ℏ·ω → E/ℏ = ω we get that the eigenvalue has a form -i·ω .
Very cool and insightful video. Looking forward to the rest. Thanks.
What a fantastic video!! Thanks a lot!
excellently explained and simplified
Amazing videos! Very clear.
If the schrodinger equation comes from he classical wave equation, why in the first term in the time dependent schrodinger equation appears the psi function in the first derivative and not in the second derivative form? Hope i make miself clear, im not that good at english. thanks
It's because the Schrödinger's equation has been derived by taking the formula for the wave function of a "free particle" (plane wave) and replacing the ω and k with E/ℏ and p/ℏ, and then looking for the time and space derivatives, which are needed for the Hamiltonian.
The wave equation is not a Hamiltonian - it expresses the dependence between the curvature in space and the curvature in time of the wave function, which are both second-order derivatives.
The Schrödinger's equation, on the other hand, expresses the dependence between the kinetic energy (related to changes in time) and the potential energy (related to changes in space), which are first-order derivatives.
There's a nice analogy you can take a look at, regarding harmonic motion:
The equation for the Simple Harmonic Oscillator (SHO) i s d²x/dt² = -ω²·x, so it is has a second-order derivative, and the solution is that nice sinusoidal oscillation you're probably well familiar with.
But the harmonic oscillator can be also expressed in the Hamiltonian way, in terms of energies: the total energy E consists of kinetic energy m·v²/2 and potential energy k·x²/2, so the Hamiltonian is E = m·v²/2 + k·x²/2.
Now you can replace the `v` in this formula by the first derivative of position with respect to time, dx/dt, obtaining:
E = m·(dx/dt)²/2 + k·x²/2
It is just a different way to express the conditions on the function which describes the motion of the oscillator: one is in terms of curvatures, the other is in terms of energies. But the solution is the same for both, because they both describe the same harmonic oscillator.
At 24:36 how do you know E is real and not complex?
I needed this video. Thank you!
Can someone explain why both solutions to the 2nd Order ODE was not included. Should we not get the solution T = Aexp[-sqroot(a)t] + Bexp[sqroot(a)t]? Why has the negative one been ignored?
No! The ODE has been solved by rearranging the variables and integrating, it is not a seond order ODE with the general solution of Aexp[-wt] + Bexp[wt]
this is fantastic and so clear thank you!
hi mister and for the solution of the shrodinger equation in polar cordinates by separation of variables ?
thanks. You are a good teacher
Sir you are just amazingg, thank you so much!
Hmm so what is it that makes an equation separable? How can we tell if a given equation can be solved by separation of variables or not before actually trying? (since the separation process can be messy sometimes, and it would be cool if we could tell if it will work before going into all those gory details :q )
Sci Twi
I've been wondering myself.
From what I can gather (also from the book), if there is another function in the PDE that makes it so that the variables can't be seperated easily, solving the PDE will become a whole lot harder.
The example they gave was if V was V(x,t) instead of just V(x), then it will not be possible to solve with SOV if V's variables aren't seperable themselves...?
I don't know.
Maybe there exists some weird PDE that has has the exact opposite property.
Excellent sir
You are the best!
Can V depend on time? Can solve the equation then?
Sr great explaination
thank you so much sir
Perfect
Jesus Christ is God.
*\0(*_*)0/* ᕦ(ಠ_ಠ)ᕤ😡🥴😤
fuck yeah baby
Amen brother
Thank you very very much