On one side you do a total derivative w.r.t. time while on the other you do a partial derivative. How does this work ? can both be a function of time and space, right ?
do you neglect the boundary terms in the integration by parts like d(wavefunction)/dx |+-infinty ? do they yield to zero because the wave function goes to zero
In quantum mechanics we often assume that all derivatives of the wave function go to zero at infinity. This is the mathematical way of saying that we can ignore what happens far away from our experiment. Of course, there are exceptions to this (crystal lattices for example)
Why exactly does the sign change at 4.50? Is there some property of the conjugate that makes it negative when differentiated? This seems straightforward but isn't entirely obvious to me..
google the solutions file related to the book and you will notice this guy simpley read out the asnwer. To answer your question tho: As the hint tells you this term should be zero, you know you can get rid of it somehow. As you have an integral over a product of two functions, integration by parts is the way to go. Doing so will show you that you get twice the integral as well as some conditions. The conditions turn out to be zero in this case. As half of zero is still zero, the integral will be zero as well ;)
Dude if you are going to copy the solutions...... simpley share the PDF on screen..... 10:40 you cant really tell by looking at it, this will be the case..... you just read out the solution manual....
@@johndoe-ow2ns You can, so long as your wavefunction has differential partial derivatives (which is an assumption we generally make): en.wikipedia.org/wiki/Symmetry_of_second_derivatives
Cadê os Brasileiros?
please speak clearly next time!!. I hardly hear what are you saying TT
On one side you do a total derivative w.r.t. time while on the other you do a partial derivative. How does this work ? can both be a function of time and space, right ?
Thank you so much. Awesome video.
Your voice was smooth pls start asmr physics 🙏 😻
do you neglect the boundary terms in the integration by parts like d(wavefunction)/dx |+-infinty ? do they yield to zero because the wave function goes to zero
In quantum mechanics we often assume that all derivatives of the wave function go to zero at infinity. This is the mathematical way of saying that we can ignore what happens far away from our experiment. Of course, there are exceptions to this (crystal lattices for example)
Thank you this is very helpful
Why exactly does the sign change at 4.50? Is there some property of the conjugate that makes it negative when differentiated? This seems straightforward but isn't entirely obvious to me..
ibb.co/dbxTjQ, i found it in the book page 16
Because he ran integration by parts uv-Integral(vdu)
What exactly are you doing when you say youre using integration by parts? i see it moves a derivative but i dont really understand why.
google the solutions file related to the book and you will notice this guy simpley read out the asnwer.
To answer your question tho: As the hint tells you this term should be zero, you know you can get rid of it somehow. As you have an integral over a product of two functions, integration by parts is the way to go. Doing so will show you that you get twice the integral as well as some conditions. The conditions turn out to be zero in this case. As half of zero is still zero, the integral will be zero as well ;)
Obrigado meu caro!! Física BSB
Very Good!
Dude if you are going to copy the solutions...... simpley share the PDF on screen..... 10:40 you cant really tell by looking at it, this will be the case..... you just read out the solution manual....
4:42 is incorrect
I agree, you can't just switch around the derivatives like that.
@@johndoe-ow2ns You can, so long as your wavefunction has differential partial derivatives (which is an assumption we generally make): en.wikipedia.org/wiki/Symmetry_of_second_derivatives
Sir, please increased your audio
his channel is literally Kinda Sorta ASMR Physics