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.
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.
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.
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?
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.
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?
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.
As far as I can tell, the domain is from -infinity to infinity, and then you get A^2 times the integral from -infinity to infinity of e^(-mw*x^2/h)dx (squaring the e^(-mw*x^2/2h) cancels out that two in the denominator), and then that becomes a gaussian integral (integral from -infinity to infinity of e^(-ax^2) dx is equal to sqrt(pi/a)), which in this case a is equal to mw/h, which gives that A^2*sqrt(pi*h/mw)=1 => A^2=sqrt(mw/pi*h) => A=(mw/pi*h)^1/4. Correct me if I made a mistake. Here is the source: en.wikipedia.org/wiki/Gaussian_integral
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.
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)
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).
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.
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!!!
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).
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...
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?
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.
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,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!!
"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".
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: ua-cam.com/video/1vMthGqUcr4/v-deo.html
Has to be the best physics tutorial I've seen on youtube and it's way better than most profs do in class.
🙋
The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.
- Sidney Coleman -
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.
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 series!!! Thank you Prof. Carlson. Extremely well explained. I'm glued to this series!!!!
Excellent!! How logical and clear this lecture is. Appreciate it a lot!
Superb quantum mechanics videos. Your hard work is appreciated.
You're a lifesaver, that's all I have to say.
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.
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.
Do conmutators have something to do with Poisson brackets?
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?
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)
Thank you. It's really a pretty good explanation. It helps me figure out lots of questions.
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.
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
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?
This tutorial was incredibly good!!!
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]?
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.
As far as I can tell, the domain is from -infinity to infinity, and then you get A^2 times the integral from -infinity to infinity of e^(-mw*x^2/h)dx (squaring the e^(-mw*x^2/2h) cancels out that two in the denominator), and then that becomes a gaussian integral (integral from -infinity to infinity of e^(-ax^2) dx is equal to sqrt(pi/a)), which in this case a is equal to mw/h, which gives that A^2*sqrt(pi*h/mw)=1 => A^2=sqrt(mw/pi*h) => A=(mw/pi*h)^1/4. Correct me if I made a mistake. Here is the source: en.wikipedia.org/wiki/Gaussian_integral
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.
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?
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)
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).
Thank you so much sir ..now it is clear to me
'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
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
Thank you for existing!
Hey Brant (Dr. Carlson), Can you provide the solutions to the test you knowledge problem? Thanks.
What happened to the imaginary number in the last derivation of psi(1)?
when you factored out 1/2m how can you still have another m in the equation left???
Hello, You can give me the software that you use it to write on the screen? Thanks alot.
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.
still the BEST video
[x, T] comes out to be (i h_bar p/m) ?
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
18:08 why not +1/2 a_+ ψ?
Thanks! Saved my exam.
my homework is complete
Same :)
@ Brant carlson : can you please explain schimidt orthogonalization process?
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).
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
Very timely! I need to see this! :)
جزاك الله خير
Thanks for this amazing lecture!
Thank you for this 🙏
Much appreciated.
good lecture. thanks for upload
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
Thank you so much sir..It helps me a lot
18:40. 'Now you notice I have an A+ here and an A+ here', that sounds so smug!
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)
good and brief explanation!!
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+)
Sir which book do u follow
pretty good explanation, thank you!
This is amazing thank you!
excellent lecture
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)
thank you very clear and helpful.
That's awesome!! THANK YOU.
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.
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
sick vid!!! very infromative
Nice class
good explanation, perfect job!!!!; U subscribed to your channel))
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 😉
Thanks 😌
Thank you very much
There's so much cleverness in this video that I was left completely confused :-/
2clever4me
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.
Ans plz of question at the end 36:57
Brilliant!
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!!
you mixed up minus and plus signs in the opperators, but ill let it slide ;)
this is called algebraic method of TISE
very useful thank u
respect
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".
m i the only who doesnt know how he get E value at the end?
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: ua-cam.com/video/1vMthGqUcr4/v-deo.html
10\10
Heheh... pee hat
Toy Story 2 was ok.