The displacement operator in quantum mechanics

📝 Problems+solutions:

📚 In this video we learn about the mathematical properties of the displacement operator in the context of the quantum harmonic oscillator. We also learn that the application of the displacement operator on the ground state of the harmonic oscillator generates a coherent state.

0:00 Intro
2:28 Definition
5:54 Mathematical properties
15:36 Coherent states
19:45 Wrap-up

⏮️ BACKGROUND

⏭️ WHAT NEXT?

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Director and writer: BM
Producer and designer: MC

[pawejakubik]

Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!

[asmaa.ali]

One correction: at 8:34 actually we can change the order of the sum in the exponent because D operator is UNITARY as you nicely proved earlier NOT simply because addition of operators is commutative as here operators are the power of exponent. Thank you so much for these great videos :)

[dbf]

Best qauntum mechanics video on yt not time consuming, fun and also academic and rigorous...

[warmesuppe]

Passed my quantum mechanics exam today (Bachelor in physics KIT).
With your help and your great videos!
Thank you very much.

[SR]

Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!
Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!
Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!

[eeshgupta]

I ABSOLUTELY ADORE YOU!!!! Thanks for making this content and your accent is impeccable. I am currently trying to learn cavity QED and your videos are a blessing!!!! Gosh I wanna be a physicist just like you!

[rainsongeng]

Thank you so much. The original intuitive of the "displacement" is really helpful.

[danielborrero]

Thanks! it would have been interesting to see the displacement of vacuum state on the quadrature space as an application of the displacement operator. Coherent states are quite interesting, I am currently working with them in the light-matter interaction via stimulated and spontaneous emission. There is a lot of issues to be clarified in this topic. Anyway, nice explanation.

[Mooriza]

Hello. You talked about the displacement operator in this video, which was great. My question is, what is the form of this operator in the phase space in terms of X and P?

[siddhantak]

thankyou so much....impressive explanation....

[assassin_un]

another good video Thanks, I think one could build the coherent states from expression of displacement operators and decomposing of coherent stetes i mean mathematically i mean that could be included too to make it more complete aside from the reason that both are eigen state of lowering operator

[shamitavaroy]

Thank you ma'am, it's really helpful!

[dilanperez]

First of all. I really love ❤️ your videos : ). Second, I have not managed to convince myself that "this expression is consistent with the adjoint of this expression" (minute 6:50). It leads me to the conclusion that (e^A e^B)^dagger is not equal to e^(B^dagger) e^(A^dagger) but there is an extra factor that must appear. Could anyone explain to me more about this, please?

[garvitmakkar]

Can you please refer us a book or couple of books to study thoroughly Quantum mechanics...
Great content keep it up🙂🙂

[Badgamerrrr]

It's really great, is it possible to attach the file where you explained as pdf.? If so then please provide it as we can't see all videos again when preparing for exams. :)

[danielegiunchi]

I have a question related to minute 7:02 when we adjoint the three terms on the right of the top equation. From that operation why the exponential with |alpha|^2 change sign? It is ok if we start to adjoint the left side of the equation and then apply the formula that split those exponential, but I can not see where that minus can come out from the adjoint of the right side.

[bastiencasini]

Many thanks for this really clear video. I was wondering, do you have any advice to learn about the Squeezing operator, I am struggling to find good resources on the subject.
Thanks again for your amazing job !!

[romainmorleghem]

Great video ! Could we see the relation [a, D]=alpha D as an eigenvalue equation for the map [a, .] ?

[jozsefkele]

I don't think e^A e^Be^-[A, B]/2 = e(A+B) is in the video on functions of operators

[abdiduguma]

thanks, what is the displacement operator for two mode light