Quantum harmonic oscillator via ladder operators
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- Опубликовано: 29 июн 2024
- A solution to the quantum harmonic oscillator time independent Schrodinger equation by cleverness, factoring the Hamiltonian, introduction of ladder operators, expression of the commutator, basic operator algebra, demonstration of the effect of the ladder operators on a hypothetical stationary state, and the resulting algebraic structure of the solutions. (This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at ruclips.net/user/playlist?list=...)
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The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.
- Sidney Coleman -
Has to be the best physics tutorial I've seen on youtube and it's way better than most profs do in class.
🙋
this is an old video but brooo you're saving my life I have a quantum mechanics midterm and this makes so much sense yaaay
I passed it!!! wOOP WOOP
@@carororororoCongratulations
The commutator of two operators is defined [A,B] = AB - BA, so -[A,B] = [B,A], so whether you get [x,p] or [p,x] with a negative somewhere else doesn't really change the result, just the way it's expressed.
Excellent series!!! Thank you Prof. Carlson. Extremely well explained. I'm glued to this series!!!!
This is by far, one if the best quantum mechanics explanations I have come across on youtube
QM by Griffiths uses a similar style to this lecture.
This whole video is very professionally done, I love how neat your writing is, and how clearly you relay the information. Thank you!
Excellent!! How logical and clear this lecture is. Appreciate it a lot!
Superb quantum mechanics videos. Your hard work is appreciated.
Thank you. It's really a pretty good explanation. It helps me figure out lots of questions.
I really wish you were my teacher and not who I have now... this was a question on the test and I was completely lost... anyways, I'm very grateful this video exists.
This tutorial was incredibly good!!!
Thanks for this amazing lecture!
Thank you for existing!
You're a lifesaver, that's all I have to say.
Thanks! Saved my exam.
Very timely! I need to see this! :)
pretty good explanation, thank you!
Thank you for this 🙏
Much appreciated.
Thank you so much sir ..now it is clear to me
That's awesome!! THANK YOU.
good and brief explanation!!
This is amazing thank you!
good lecture. thanks for upload
Thank you so much sir..It helps me a lot
thank you very clear and helpful.
excellent lecture
sick vid!!! very infromative
still the BEST video
I have a quantum mechanics test tomorrow and you just saved my life
me too
And one year later, I'm in the same boat :)
me too lmao
At 32:18 you say that the denominator is equal to one, so we can ignore it.
You say its because sqrt(n+1) where n is zero so the denominator is 1.
But actually it's because the denominator would be sqrt(1!) from the formula for PSI_ n.
جزاك الله خير
Brant Carlson
Thanks for the video. I think I am somewhat understanding. However you give (n + 1)^1/2 as the coefficient for psi(n+1) when doing a+psi. Could you explain/ show what that really looks like in terms of the actually numbers/variables? I'm trying to do this and make the n = 2 wave function from the n=0 wave function using the raising operator twice.
I've calculated several of the psi[n], and drawn a graph of each psi[n]^2 (no complex numbers here, to get the probability density rho(x)). Said graphs are VERY interesting (I don't know if it's possible to put them here).
Do conmutators have something to do with Poisson brackets?
Brilliant!
good explanation, perfect job!!!!; U subscribed to your channel))
@ Brant carlson : can you please explain schimidt orthogonalization process?
Dear Brant, could not you please tell me what kind of tablet/pen do you have and what is the software you use? I like that you do have a cursor on your videos and wanted to buy similar one)
at 32.26 example, may I know why is (n+1)^1/2 instead of (n)^1/2 as the formula as shown at 31.54?
It's very very usefull! Thank you! I work on the possible applications of quantum entanglement like a circuit QED model. If you know something about that I really like to see that.
How was the normalisation constant calculated at 28.49? Can anyone be kind enough to explain the calculation. I thought it would solving the integral from -infinity to infinity of psi*psi = 1. Are those our integral limits here? Not sure what domain this has been in.
when you factored out 1/2m how can you still have another m in the equation left???
What happened to the imaginary number in the last derivation of psi(1)?
if you had factored out +imw instead of -imw you would end up with [p,x] and get a different results, how did you know that you want to get [x,p]?
Nice class
Hello, You can give me the software that you use it to write on the screen? Thanks alot.
When you write p-hat squared psi, does that mean p-hat(p-hat(psi)) or does it mean (p-hat(psi))(p-hat(psi)), i.e. squared in the traditional sense. For that matter, it looks like you're using (x-hat)(p-hat)psi = x-hat(p-hat(psi)), otherwise they would commute. But earlier, you definitely treated (p-hat)(p-hat) as p-hat squared. Can someone clarify please?
Thank you very much
Does anyone know where to find an explanation of how to find the integral when calculating the coefficient of the e term for the equation of the ground state wave function? Thank you! These videos are so helpful!!!
I know I am late but it is just applying normalization Condition in Which you have to solve for Normalization coefficient.
my homework is complete
Same :)
why can you write a+ in the left side?
isn't that implying that the ladder and the hamiltonian commute? (which they seem not to)
I don't get why the ladder operator is quantized when omega is a continuous variable. If I choose a different omega, then my ladder is totally different, so that for any energy, I can find an omega that allows it. What am I missing here?
While calculating the lowest energy psi(0), where is "i" of the a- ladder operator.
Abhishek Ghosh original a operator has -i*p hat , but p hat operator is equal to -ih d/dx thus -i*i =1 and you get h d/dx
Hey Brant (Dr. Carlson), Can you provide the solutions to the test you knowledge problem? Thanks.
Thanks 😌
18:40. 'Now you notice I have an A+ here and an A+ here', that sounds so smug!
pls someone explain the logic of what he said from 25.00, esp the curving away dialogue at 25.25
Consider a coordinate plane considering psi(x) on the vertical axis and x on the horizontal axis. For this problem with a potential v(x) =mw^2x^2, the most obvious location to construct the origin is where v(x) = 0 and v(0) = 0. The only valid solutions are physical solutions which force the conditions for psi(x) and x to be greater than 0.
Brant, just for this time, I don't fully understand the whole concept of ladder operator.
Is ladder operator used to reconstruct the Schrodinger solution or just simplify it ?
From 21:50 onwards, if psi is a solution then a+ ladder operator acting on psi is another solution with higher enegy = h-bar*omega.
You can again apply a+ ladder operator on this new solution to get yet another solution of higher energy level with energy difference of h-bar*omega again.
The same process can repeated with a- ladder operators to get soltuions with lower energy levels & energy difference being h-bar*omega again.
very useful thank u
any non-physics students here , learning out of sheer interest here like me??
yeah :)
@Renzo Scriber lol yes!!
If you are learning physics you are a physics student.
What makes you a physics student is that you are studying physics, not that you get assigned homework.
@@menoetius8182 yeah haha, but I think they meant physics majors
i respect you guys so much
Can someone tell me that how at point 5:10 he solve that (m)???...When he take out (m) from the equation than how can he write it below again with the omega and x...
Samavia Rafiq he writes it as m^2 , so the m taken out in denominator will cancel one of the m in m^2 to give original m
[x, T] comes out to be (i h_bar p/m) ?
18:08 why not +1/2 a_+ ψ?
So it turns out that I had a choice between reading the same section 100 times or watching this video once.
I regret not choosing the ladder 😉
Can anyone confirm if I got the right answer at the end of the video? I got (ħ²/m)dΨ/dx... should I make this simpler?
i got the same result, i don't think you can simplify it though
'kissing the axis' but your making it kiss the 'E' line... Was this a mistake? If it tends to a positive value E, it's not normalizable
There's so much cleverness in this video that I was left completely confused :-/
2clever4me
Sir which book do u follow
respect
I don't see how a+a- gives you a different result. When you just switch the position of the negative sign, doesn't product foil out to p^2 + mwx^2 -imw[x,p] just like in the derivation here?
That is, a+a- = (-ip + mwx)(ip + mwx), which factors the same way, right?
[a+,a-]=[a-,a+]=1
(a+a-) is not equal to (a-a+)
In slide 7 fourth line, you cannot just add one without changing the order of a-a+ because a-a+ = a+a- +1, otherwise really helpful video!
+John Doe Shouldn't he then also subtract 1?
He wasn't changing the order of the operators because he was just rewriting +1/2 = -1/2 + 1
I still prefer the power-series solution. It just seems more intuitive to me than abstract 'ladder operators'!!!
Same. We covered the QHO in my quantum 1 course over the summer, and now my quantum 2 course at a different college is covering it to start the quarter and I still can’t fully wrap my head around the purpose of ladder operators other than to present a seemingly more “elegant” solution to the problem
@@SS-tu6kc Actually, being slightly smarter now, I understand that Ladder Operators play a CRUCIAL role in Quantum Field theory, you can essentially construct canonical quantization based off of ladder operators (plus some abstract algebra).
Why is there no minus sign in front of p at 5:37.
Look at the definition of momentum operator : -i*h-bar*(d/dx)
this is called algebraic method of TISE
you mixed up minus and plus signs in the opperators, but ill let it slide ;)
This is what I got for the answer for the check your understanding: www.physicsforums.com/showthread.php?p=4784187#post4784187
@@RyPangi5 I got hbar²/m( d/dx)
Ans plz of question at the end 36:57
36:42 For Check Your Understanding, do we have [x, T] = h^2/m * d/dx as our answer? Or am I wrong in some parts
I got the same, think it’s correct.( Acting on some “psi”wave function
Hey the answer is zero
@pixelberrychoicespodcast on instagram you can ask me for the solution
I got the same
10\10
m i the only who doesnt know how he get E value at the end?
Can anyone explain 25:40 in a little detail...??
If E < V(x), then the signs of the gradients in Schrodinger's equation indicate that the wave function blows up to infinity. But this is not allowed, for a wave function ought to be normalizable (i.e., the function should approach 0 as x tends to +/- infinity), in order to ensure that the area below that wave function, or the probability of finding the particle in a given state, remains finite. Therefore, there must be some minimum energy, i.e., at the ground state, below which we can no longer apply the lowering operator (a_) to generate meaningful (or normalizable) wave functions. It is interesting to think about this constraint from the perspective that no particle can physically exist with an energy below some minimum threshold energy, determined in this case by V(x). In other words, particles should have some little bit of energy at least to maintain its mass.
Hey Brant, this video is very informative, and it is easy to see the moving parts! I mentioned your video in my lecture on the "Mathematical Structure of Quantum Theory". To introduce some novelty, I did not use the standard method (i.e. the Frobenius method or the algebraic method), but instead I generated the Hermite polynomials using Gram-Schmidt orthogonalization. Please check it out!
Skip to 1:18:21: ruclips.net/video/1vMthGqUcr4/видео.html
Hey,i've watched you solve but i think you did a small mixed up at 5:37,the expression of your p operator is not correct,the p operator is equal to the momentum p divided by square root of mass times omega times h bar...
Verify that please...
But later on your expressions are correct for the anihilation operator and so on!!
i didn't know eric forman did quantum mechanics
"All of the energy and matter that existed still exist. Matter does not create energy of itself. The actions of matter enable energy to become manifest".
Heheh... pee hat
Toy Story 2 was ok.