Generally when we talk about a solution to an equation we mean we can calculate what values the variables can take. For example if x=y then x and y must have the same value but x can have any value (as long as y has the same value). But if y=2x+1 and y=3x-1 then there is an absolute and unique solution. x=2 and y = 5.
At 2:41 where does the 1 minus come from? I understand that k squared must be minus because of where it is being substituted but not why there is a 1 before it. Thanks for your time
It's not a 1-k^2, just -k^2. He started to write the k before scribbling it out, so the 1 that you're seeing is actually just the scribbled out 'stick' from the k.
the solution to the differential equation could also just be written as psi = Acos(kx)+Bsin(kx) In quantum mechanics, does it matter which form it takes?
You can think of it like this: Assuming that the energy, E, in the system is positive and -h(bar) 2/2m is a negative term, it means that the second order differential equation must have a wave function that will become a negative number multiple of itself hence -K^2 I'm not sure if I explained it well but, I hope that helps.
I have a doubt, how could you have written E=p^2/2m ( it's the energy not the kinetic energy) the relationship between the Energy and K.e is E=2K.E, therefore E=p^2/m, and in the 2.53 from where the 1-k^2 Psi had come?
This is wrong, the free particle's energy is not quantized because you have no boundary conditions such as continuity for example. The particle is completely free and it can have any energy.
i did not understand the part at 4: 13 , (ik)^2 = -k^2 that is understood but how come (-ik)^2 is again gives -k^2 , -i^2 = 1, so it must be k^2 . Thanks in advance doctor .
In my text book it shows that p=nh/2L while in your video around 5:03 p=nh/L. Is my book looking at a different case perhaps? Regardless thank you sooooo much, these are great videos. By the way my Textbook is Young/Freeman 12th edition of University Physics pg 1377. Chapter 40.1 Again, thank you so much for these videos and please keep them coming.
I think in most textbooks, it shows that angular momentum is quantized. That is - mvr=nh/(2pi). Dividing both sides by r gives us mv=p=nh/(2pi*r) And 2pi*r is the circumference of the orbit, or in this case the length of the particle's path. Hope this clears your doubt!
Schrödinger eq. is no more than a "second order differential equation". Check any mathematics forum and they will tell you what the solution should be in specific cases.
@DrPhysicsA absolutely wrong video. energy of a free particle is NOT quantized. Quantization occurs only when you have BOUNDARY CONDITION like a particle trapped in a one dimensional box!!!!! PLEASE RECTIFY
Utter nonsense. While infinitely long wave functions may provide a valid solution to the wave equation, there are none in nature and if there were, they'd have infinite mass, momentum and energy (despite the nonsensical "L" length construct). Further, one can't assert quantized momentum as justification for quantized energy, it's a circular argument. For these and other reasons, the free-space solution presented here is nonsense.
Generally when we talk about a solution to an equation we mean we can calculate what values the variables can take. For example if x=y then x and y must have the same value but x can have any value (as long as y has the same value). But if y=2x+1 and y=3x-1 then there is an absolute and unique solution. x=2 and y = 5.
We would love to have you back and teach us more. You are simply brilliant.
I love your kitchen clock and dog barking as the avant-garde type background music.
I have a cuckoo type clock - but instead of a cuckoo, a cow comes out and moos on the hour. Somehow it managed to feature on many of my videos.
best and simplest theorical explainer! love your videos!
ok, i'm starting to watch your videos more and more often, so i subscribed and i'll let other people know of your amazingly useful lectures. thanks!
A & B are just coefficients associated with the wave function. Their values will depend on the location of the particle.
you are the best tutor in Mr Dr Physics
Thank you so much for all of your great videos! I just had a quick question: why is the wavefunction here not e^i(kx-wt)?
Thank you ^_^
👍👍 I think you belong to feynman's family. You are a very good explainer kudos!!
you are funny, I am sure!
Your explanation is very clear and organized.. Thanks
Just covered this in a lecture. Great clarification! Thanks.
Absolutely brilliant explanation, thank you so much!!
At 2:41 where does the 1 minus come from? I understand that k squared must be minus because of where it is being substituted but not why there is a 1 before it. Thanks for your time
It's not a 1-k^2, just -k^2. He started to write the k before scribbling it out, so the 1 that you're seeing is actually just the scribbled out 'stick' from the k.
Well i feel like an idiot! haha! thanks for pointing it out
Haha, not at all man, I had the exact same thought as you when I first saw it, took me a while to see what had happened.
the solution to the differential equation could also just be written as
psi = Acos(kx)+Bsin(kx)
In quantum mechanics, does it matter which form it takes?
yes, you have 2, not one with a copy. 2 mirror the same in mirror (*/-)
Nice kitchen clock u got there... XD!!!
@4:06 how can -k^2 be a coefficient of B, shouldn't it be positive k, because squaring -i is positive thus k^2??????
You can think of it like this: Assuming that the energy, E, in the system is positive and -h(bar) 2/2m is a negative term, it means that the second order differential equation must have a wave function that will become a negative number multiple of itself hence -K^2
I'm not sure if I explained it well but, I hope that helps.
I have a doubt, how could you have written E=p^2/2m ( it's the energy not the kinetic energy) the relationship between the Energy and K.e is E=2K.E, therefore E=p^2/m, and in the 2.53 from where the 1-k^2 Psi had come?
can you explain how the result of -k^2 psi gives a complex solution and also what it means to have a solution to the equation ? Thanks
Does this solution to Schrodinger equation obey the all the fundamental postulates of Q.M.?
2:45 ,instead of E=-k2???
Oh and also, why for the energy of the particle did you only consider the kinetic and not the potential energy too?
Thank you :)
DUDE! Ur Videos= Awesome
This is wrong, the free particle's energy is not quantized because you have no boundary conditions such as continuity for example. The particle is completely free and it can have any energy.
i found a 1d schrodinger problem which the box isn't rectangular ,its triangular shaped (the barrier has gradient) i get confused in solving it!
thanks but what i meant from my question was what does it mean to have a solution to an equation, any equation ?
love your videos!
what is A and B in the equation?
why so that in the case of E=V0. on the wave function solution (psi_2)=C+Dx
i did not understand the part at 4: 13 , (ik)^2 = -k^2 that is understood but how come (-ik)^2 is again gives -k^2 , -i^2 = 1, so it must be k^2 . Thanks in advance doctor .
(ik)^2 and (-ik)^2 are both equal to -k^2 - because (-i)^2 = i^2 = -1
This had me confused too until I thought of it this way;
(-ik)^2 = (-1*i*k)^2 = (1*i*k)^2
Why not Acos(kX)?
I wish I had seen this yesterday. >_
In my text book it shows that p=nh/2L while in your video around 5:03 p=nh/L. Is my book looking at a different case perhaps?
Regardless thank you sooooo much, these are great videos.
By the way my Textbook is Young/Freeman 12th edition of University Physics pg 1377. Chapter 40.1
Again, thank you so much for these videos and please keep them coming.
I suspect it depends on the precise example which has been used. The key point is that momentum is quantised in some units of planks constant.
I think in most textbooks, it shows that angular momentum is quantized. That is - mvr=nh/(2pi). Dividing both sides by r gives us mv=p=nh/(2pi*r) And 2pi*r is the circumference of the orbit, or in this case the length of the particle's path. Hope this clears your doubt!
Cow ringtone at @6:12 for the win.
+Aaron Wolbach i actually glanced outside for a second. i live in an urban area. ._. convincing ringtone haha!
Very understandable, thank you for sharing the knowledge ;)
Schrödinger eq. is no more than a "second order differential equation". Check any mathematics forum and they will tell you what the solution should be in specific cases.
How much is Ψ^2?
Can someone tell me where does the one come from?
is not a one, he had first put the k but then corrected saying it was "- k"
good progress on this
good work
sorry what is k?
your dog is just really excited about the psi function
ouside the box-- look at gravity and sound (Hz) you'r almost there
and magnetism, sorry forgot;;; it is like perpetuum mobile- SO:grav- mag - hertz
thanks .
thanks! :-)
@DrPhysicsA absolutely wrong video. energy of a free particle is NOT quantized. Quantization occurs only when you have BOUNDARY CONDITION like a particle trapped in a one dimensional box!!!!!
PLEASE RECTIFY
+aditya2004 - And what's the kinetic energy of a free object in space (far away from gravitational fields?)
nice.
I almost thought u owns a cow as well
90% of the stuff explained in class, wont get me points on the test :) i like science because of that
But a simple sinusoid is not normalized.
mooo!
Utter nonsense. While infinitely long wave functions may provide a valid solution to the wave equation, there are none in nature and if there were, they'd have infinite mass, momentum and energy (despite the nonsensical "L" length construct). Further, one can't assert quantized momentum as justification for quantized energy, it's a circular argument. For these and other reasons, the free-space solution presented here is nonsense.
Macroscopic wave-particle in a box: /watch?v=nmC0ygr08tE