Yes! Thank you very much. I did look at a few videos on the momentum operator but none were as clear as this one. A bit cranky and untidy but that’s part of the fun 🎉 . Dr-J
there are several different ways you can axiomatize QM. I'm choosing the SE as an axiom and going from this. The point of this video is to help people in an introductory quantum class, and at least in my first QM class that's how we started. But you raise a great point
First of, THANK YOU! I was also wondering about the appearance of the momentum operator as part of the Schrodinger equation. Is there a way to derive the operator without utilizing it in advance?
@@jonathanericson4066 yes there is a way i don't know about any videos teaching this method but you may look jj sakurai modern quantum mechanics in its first chapter on page 54 eqn (1.7.15) under the topic momentum operator in position basis
We don't drive it. We define it. Just like we have defined the energy operator, right in the Schrödinger equation. The easiest way to get to the definition is to take a wave functions ψ=Ae ^i(xp-Εt) and differentiate it wrt x. Then, you'll immediately see how the momentum operator pops out.
Ah thank you! I had to derive this for my midterm yesterday and i ended up trying to do it from memory from the way Griffiths does it in Intro to QM 2nd ed. But he skips that integration by parts step at 7:00. So when i did it i ended up with a negative sign that i couldn't account. Grifffiths just skips all that part and goes from one step to the next getting the exact same equation but with a sign switch. NOW i see where that sign switches. Thank you!
For the step involving the fundamental theorem of calculus - the derivative you use is a partial with respect to x, while the dv is not a partial, but independent of time. Do you know where how i can resolve this difference? I.e. Using the partial of x instead of dx and then equating the two
One thing that always stroke me as weird, is that mass is always treated like a constant, but never as an operator. I mean isn't mass an observable as well?
Great question. In non-relativistic QM the mass behaves like a fixed parameter because NR-QM obeys Galilean symmetries. If you really want to dive in, look into the Bargmann mass superselection rule. Or if you like, just pretend we're doing a toy model of QM, like how when we're learning classical mechanics, we first assume mass is a constant.
@@mathforphysics2444 thanks! Unfortunately all the materials I can find are too rigorous and not very intuitive, they start talking SU(3) and other simmetry groups and Hilbert space and my mind blows up😂 so that's too bad. But thanks!
theleastcreative yes i've made some thoughts but im not sure if im correct is it because this is actually the place opperator which shouldn't change with time
How the hell does the integral of psi*psi goes from negative infinity to positive infinity goes to zero when for a normalised psi when it's literally defined as equal to 1 for a normalised function as the probability of finding a particle from positive to negative infinity is one .. this makes no sense and its driving me crazy ughh 🤬🤬
Mostly this video is for helping undergrads with their homework. But there are several ways you can axiomatize QM, and the text I'm following takes the SE as an axiom
God bless for this man. My modern physics prof couldn't teach this to save his life.
You're too kind. Need anything else explained?
After numerous google searches, I finally found this among the google picture search results. Nicely done and easy to follow! Keep up!
Thanks! I’m going through Griffiths and this has been really helpful.
commodorekitty is there any other topics you'd like to see covered?
Same story here, he seems to like cutting out some of the important maths
Thank you so much for explaining this black magic you absolute legend.
I'm happy I can help
You are, to put it short, a life saver... Thank you so much for having made this video!!!!!!!
I'm really glad I could help, are there any other topics or derivations you'd like to see?
Yes! Thank you very much. I did look at a few videos on the momentum operator but none were as clear as this one. A bit cranky and untidy but that’s part of the fun 🎉 . Dr-J
However, in the Schrodinger equation we already have the momentum operator explicitly given, and then you used it to eventually derive the same thing.
If you start by assuming the plane wave solution to SE, then using De Broglie equation you can show this in 2 lines
there are several different ways you can axiomatize QM. I'm choosing the SE as an axiom and going from this. The point of this video is to help people in an introductory quantum class, and at least in my first QM class that's how we started.
But you raise a great point
First of, THANK YOU!
I was also wondering about the appearance of the momentum operator as part of the Schrodinger equation. Is there a way to derive the operator without utilizing it in advance?
Hi Sam, could you refer me to a vid explaining this option?
@@jonathanericson4066 yes there is a way i don't know about any videos teaching this method but you may look jj sakurai modern quantum mechanics in its first chapter on page 54 eqn (1.7.15) under the topic momentum operator in position basis
This is simply a consistency check.
We don't drive it. We define it. Just like we have defined the energy operator, right in the Schrödinger equation.
The easiest way to get to the definition is to take a wave functions ψ=Ae ^i(xp-Εt) and differentiate it wrt x. Then, you'll immediately see how the momentum operator pops out.
thanks.. very explicit, very useful
Dude, wonderful video. Thanks so much
Your h-bars look like symbol for the British Pound lol
Ah thank you! I had to derive this for my midterm yesterday and i ended up trying to do it from memory from the way Griffiths does it in Intro to QM 2nd ed. But he skips that integration by parts step at 7:00. So when i did it i ended up with a negative sign that i couldn't account. Grifffiths just skips all that part and goes from one step to the next getting the exact same equation but with a sign switch. NOW i see where that sign switches. Thank you!
oh
have you full mathematics ???
nice work dude
Great. Maybe You would show us how to derive Conservation laws from symmetry (space, time translation and rotating)? (Noether's theorem)
Excellent!! thank you very much.
You sir are a hero :) thank you so much
Thank you.
What is the difference between velocity and speed in this context?
For the step involving the fundamental theorem of calculus - the derivative you use is a partial with respect to x, while the dv is not a partial, but independent of time. Do you know where how i can resolve this difference?
I.e. Using the partial of x instead of dx and then equating the two
One thing that always stroke me as weird, is that mass is always treated like a constant, but never as an operator. I mean isn't mass an observable as well?
Great question. In non-relativistic QM the mass behaves like a fixed parameter because NR-QM obeys Galilean symmetries. If you really want to dive in, look into the Bargmann mass superselection rule.
Or if you like, just pretend we're doing a toy model of QM, like how when we're learning classical mechanics, we first assume mass is a constant.
@@mathforphysics2444 thanks! Unfortunately all the materials I can find are too rigorous and not very intuitive, they start talking SU(3) and other simmetry groups and Hilbert space and my mind blows up😂 so that's too bad. But thanks!
awesome vid!
Oh, Thanks alot !
THANK YOU!
You're welcome
perfect
How do you know x isn't a function of time? So that pulling it out of the derivative around 1:55 isn't a mistake?
Never mind! I figured it all out! Thanks for the vid!
+CrushOfSiel I just saw this now, sorry for the slow reply, but I'm glad you figured it out
+theleastcreative can you help me figure it out?
Do you have the same question?
theleastcreative
yes
i've made some thoughts but im not sure if im correct
is it because this is actually the place opperator which shouldn't change with time
I wish you reply p,x’s poisson relation with ih.
How the hell does the integral of psi*psi goes from negative infinity to positive infinity goes to zero when for a normalised psi when it's literally defined as equal to 1 for a normalised function as the probability of finding a particle from positive to negative infinity is one .. this makes no sense and its driving me crazy ughh 🤬🤬
can anyone help to explain the derivation at 13.23
I'm repeating integration by parts
Your derivation for momentum uses the shrodinger equation which in itself requires momentum to derive… so it's a cyclical derivation. Not so useful
Mostly this video is for helping undergrads with their homework. But there are several ways you can axiomatize QM, and the text I'm following takes the SE as an axiom
@@theleastcreative Oh okay, thanks!