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 :)
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!
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!
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.
Thanks for the suggestion, we'll add to our list! In the meantime, we are still going to use the displacement operator in the next two upcoming videos, which look at the wave functions of coherent states, first mathematically and next on a more conceptual level.
Some we like include: Quantum Mechanics by Merzbacher, Quantum Mechanics by Cohen-Tannoudji, Modern Quantum Mechanics by Sakurai, or Principles of Quantum Mechanics by Shankar. We are hoping to do some videos reviewing books, so hopefully that will also be helpful!
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?
It is on our list! For now we have a playlist on time dependence in quantum mechanics (ua-cam.com/play/PL8W2boV7eVflUqUY3dLhQdYuZjlbXi0mU.html), which will serve as a starting point for the topics you want :)
As always, a great video. The displacement operator D(alpha) operates on the HO ground state |0> to generate a coherent state |alpha> , but isn't it true that you do not need the whole displacement operator to do that? Looking at the form of D(alpha) which equals exp(-1/2|alpha|^2) x exp(alpha a-hat-dagger) x exp (alpha* a-hat) -- the last term operating on |0> just gives |0>, I think. This is because exp(alpha* a-hat) = 1 + various powers of (alpha*) x (a-hat), which powers of a-hat all kill the ground state |0>. Since this is the right-most part of D(alpha), it operates first on |0> and basically does nothing. And in fact if you check exp(alpha a-hat-dagger)|0>, it seems to generate |alpha>, with the remaining first term from D(alpha) -- exp(-1/2|alpha|^2) -- just being the appropriate normalization term. So in a sense isn't it really exp (alpha x a-hat-dagger) acting on |0> that generates |alpha>, and the final term in D(alpha) -- exp (alpha* x a-hat) -- is just sort of along for the ride? So then why do we use D(alpha) instead of just exp(-1/2|alpha|^2) x exp(alpha a-hat-dagger), which is sufficient? I think it is because the shorter operator is not unitary and so does not produce a lot of the cool stuff we get from the full displacement operator (such as transforming operators a-hat and a-hat-dagger and having helpful commutation relations with them -- and eventually leading to a very intuitive form of the coherent state wave function. Am I totally missing the boat here? No need to answer if you do not want to -- I really appreciate all the work you do for your viewers, and ultimately we need to be able to work this stuff out for ourselves!
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. :)
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
We do not prove this expression in that video, but we do present the result as the Baker-Campbell-Hausdorff formula, of which this result is a special case when A and B commute with their commutator, that is, when [A,[A,B]]=0 and [B,[A,B]]=0. I hope this helps!
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?
Glad you like the videos! Note that the relation you are trying to prove is not quite what I claim in the video. My claim is specifically for the expressions in the video in terms of raising and lowering operators a and a^dagger. I would therefore encourage you to try it again in this specific case. In the general case of operators A and B, then the answer will depend on the commutator of A and B, so your results do make sense in this more general context. I hope this helps!
That is a cool thought. My math background is not strong. With that caveat, I guess usually I think of the situation where operators act on kets in a Hilbert space, and an eigenvalue equation for an operator identifies those kets which the specified operator leaves unchanged except for a scaling factor. This has great value; e.g., if the operator is Hermitian, we can use its eigenkets as a basis of the Hilbert space. In particular, for example, we can use energy eigenkets to expand states and analyze time dependence easily, as Prof. M has pointed out in various videos. Here you are thinking of operators as vectors in a vector space (which I think they are). Then the object [a^, …] would be a map (operator) mapping one vector in that (operator) vector space to another vector in that vector space. But would the usual benefits of eigenvalue equations be available here? For example, is there an analog of Hermitian operators here, where we could use the idea of completeness to treat the eigenvalues of some set of maps [a^, …] as bases for our operator vector space? Also, I do not see immediately that this operator vector space is a Hilbert space (does it have an inner product?). If it is not, maybe this would limit the utility of this very cool idea? I guess we could also think more generally of the object [… , …] as a bilinear map of two operators into a third operator; e.g. {a, a-dagger} maps to the identity operator, and {a, D(alpha)} maps to alpha*D(alpha). But I do not see immediately that this gets us anything significant . . . . Thanks for the interesting comment!
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 !!
Glad you like it! No resource comes immediately to mind, I guess you have tried the standard quantum optics textbooks? After we complete our series on the fundamentals of quantum mechanics, we hope to cover topics like condensed matter or quantum optics, but unfortunately it will take a while to get there...
This is an interesting point. The translation operator in real space by an amount omega is given by exp(-ip*omega/hbar), while the translation operator in momentum space by an amount omega is given by exp(ix*omega)/hbar). The displacement operator involves both position and momentum operators in the exponent, and describes a translation in "phase space". Interestingly, position and momentum operators don't commute, so we need to be careful when working with the displacement operator. An example of its use is in the video on coherent state wave functions: ua-cam.com/video/0p9pH85SLIU/v-deo.html I hope this helps!
The only difference from doing the position and momentum translations separately or at the same time as in the displacement operator is a net global complex phase. Since quantum states are indistinguishable when they differ by a constant global phase, they really are the same states. Nevertheless, the convention is to use the displacement operator as the fundamental operator as opposed to translation of position followed by translation of momentum or vice versa. If you like, it is halfway between the other two options.
A coherent state is an eigenstate of the lowering operator, so the number of photons in a coherent state is not fixed. However, you can calculate the expectation value of the number of photons in a coherent state, and for coherent state |alpha> it is |alpha|^2. You can show this easily by calculating , where n=a^dagger a is the number operator. If you have a different coherent state |alpha'>, then the expectation value of the number operator changes to |alpha'|^2. I hope this helps!
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.
I did the calculations properly and there are 2 mistake that cancel each other. With D(alpha)^dagger we should still have the minus sign in front of |alpha|²/2 but you should also swapp the exponential with a and a^dagger. In the end you should respect D(alpha)^dagger = D(-alpha) (D is a unitary operator (check with the initial expression before splitting the exponentials)). In the final calculation (D(alpha)D(alpha)^dagger) you should use BKH one more time to swapp e^(-alpha* a) and e^(-alpha a^dagger) resulting in an additional e^|alpha|² canceling the product of the two initial e^|alpha|²/2 from D and D^dagger. In the end you end up with the identity as expected. Hope this helps somewhat
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 :)
Best qauntum mechanics video on yt not time consuming, fun and also academic and rigorous...
Glad you like our approach! :)
Passed my quantum mechanics exam today (Bachelor in physics KIT).
With your help and your great videos!
Thank you very much.
Congratulations! Glad to hear our videos have been helpful! :)
Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!
Glad you find our videos 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!
Thank you so much for making QM that intuitive. I'm taking Engineering Physics course and it's really helpful!
Glad you like this, and nice to see the enthusiasm! :)
And as allways thank you so much
:) Thank you!
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!
Glad to hear that you like our videos! Where are you studying if I may ask?
thankyou so much....impressive explanation....
Glad you like it! :)
Thank you .
Thanks for watching!
Thank you ma'am, it's really helpful!
Glad you find it helpful!
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.
Thanks for the suggestion, we'll add to our list! In the meantime, we are still going to use the displacement operator in the next two upcoming videos, which look at the wave functions of coherent states, first mathematically and next on a more conceptual level.
Thank you!! It help me a lot!
Glad it helped!
thanku so so muchhh
Thanks for watching! :)
Can you please refer us a book or couple of books to study thoroughly Quantum mechanics...
Great content keep it up🙂🙂
Some we like include: Quantum Mechanics by Merzbacher, Quantum Mechanics by Cohen-Tannoudji, Modern Quantum Mechanics by Sakurai, or Principles of Quantum Mechanics by Shankar. We are hoping to do some videos reviewing books, so hopefully that will also be helpful!
Really helpful thank you
Glad you found it helpful!
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?
Please make a video on SCHRONDIGER and Hisenburg picture annd on Quantum dynamics
It is on our list! For now we have a playlist on time dependence in quantum mechanics (ua-cam.com/play/PL8W2boV7eVflUqUY3dLhQdYuZjlbXi0mU.html), which will serve as a starting point for the topics you want :)
@@ProfessorMdoesScience Wow! You spoke the truth -- it looks like this video dropped today!
@@richardthomas3577 We do literally have tens of videos mapped out that we are planning on publishing, and we just go down the list as time allows! :)
As always, a great video. The displacement operator D(alpha) operates on the HO ground state |0> to generate a coherent state |alpha> , but isn't it true that you do not need the whole displacement operator to do that? Looking at the form of D(alpha) which equals exp(-1/2|alpha|^2) x exp(alpha a-hat-dagger) x exp (alpha* a-hat) -- the last term operating on |0> just gives |0>, I think. This is because exp(alpha* a-hat) = 1 + various powers of (alpha*) x (a-hat), which powers of a-hat all kill the ground state |0>. Since this is the right-most part of D(alpha), it operates first on |0> and basically does nothing. And in fact if you check exp(alpha a-hat-dagger)|0>, it seems to generate |alpha>, with the remaining first term from D(alpha) -- exp(-1/2|alpha|^2) -- just being the appropriate normalization term. So in a sense isn't it really exp (alpha x a-hat-dagger) acting on |0> that generates |alpha>, and the final term in D(alpha) -- exp (alpha* x a-hat) -- is just sort of along for the ride? So then why do we use D(alpha) instead of just exp(-1/2|alpha|^2) x exp(alpha a-hat-dagger), which is sufficient? I think it is because the shorter operator is not unitary and so does not produce a lot of the cool stuff we get from the full displacement operator (such as transforming operators a-hat and a-hat-dagger and having helpful commutation relations with them -- and eventually leading to a very intuitive form of the coherent state wave function. Am I totally missing the boat here? No need to answer if you do not want to -- I really appreciate all the work you do for your viewers, and ultimately we need to be able to work this stuff out for ourselves!
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. :)
Thanks for the suggestion, we'll look into this. In the meantime, I guess taking screenshots would be a temporary solution...
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
I don't think e^A e^Be^-[A,B]/2 = e(A+B) is in the video on functions of operators
We do not prove this expression in that video, but we do present the result as the Baker-Campbell-Hausdorff formula, of which this result is a special case when A and B commute with their commutator, that is, when [A,[A,B]]=0 and [B,[A,B]]=0. I hope this helps!
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?
Glad you like the videos! Note that the relation you are trying to prove is not quite what I claim in the video. My claim is specifically for the expressions in the video in terms of raising and lowering operators a and a^dagger. I would therefore encourage you to try it again in this specific case. In the general case of operators A and B, then the answer will depend on the commutator of A and B, so your results do make sense in this more general context. I hope this helps!
Great video ! Could we see the relation [a,D]=alpha D as an eigenvalue equation for the map [a,.] ?
I had not thought about this, interesting thought! I wonder whether this may be useful in some context?
That is a cool thought. My math background is not strong. With that caveat, I guess usually I think of the situation where operators act on kets in a Hilbert space, and an eigenvalue equation for an operator identifies those kets which the specified operator leaves unchanged except for a scaling factor. This has great value; e.g., if the operator is Hermitian, we can use its eigenkets as a basis of the Hilbert space. In particular, for example, we can use energy eigenkets to expand states and analyze time dependence easily, as Prof. M has pointed out in various videos.
Here you are thinking of operators as vectors in a vector space (which I think they are). Then the object [a^, …] would be a map (operator) mapping one vector in that (operator) vector space to another vector in that vector space. But would the usual benefits of eigenvalue equations be available here? For example, is there an analog of Hermitian operators here, where we could use the idea of completeness to treat the eigenvalues of some set of maps [a^, …] as bases for our operator vector space?
Also, I do not see immediately that this operator vector space is a Hilbert space (does it have an inner product?). If it is not, maybe this would limit the utility of this very cool idea?
I guess we could also think more generally of the object [… , …] as a bilinear map of two operators into a third operator; e.g. {a, a-dagger} maps to the identity operator, and {a, D(alpha)} maps to alpha*D(alpha). But I do not see immediately that this gets us anything significant . . . .
Thanks for the interesting comment!
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 !!
Glad you like it! No resource comes immediately to mind, I guess you have tried the standard quantum optics textbooks? After we complete our series on the fundamentals of quantum mechanics, we hope to cover topics like condensed matter or quantum optics, but unfortunately it will take a while to get there...
why we can't use translational operator instead of displacement operator?
This is an interesting point. The translation operator in real space by an amount omega is given by exp(-ip*omega/hbar), while the translation operator in momentum space by an amount omega is given by exp(ix*omega)/hbar). The displacement operator involves both position and momentum operators in the exponent, and describes a translation in "phase space". Interestingly, position and momentum operators don't commute, so we need to be careful when working with the displacement operator. An example of its use is in the video on coherent state wave functions: ua-cam.com/video/0p9pH85SLIU/v-deo.html
I hope this helps!
The only difference from doing the position and momentum translations separately or at the same time as in the displacement operator is a net global complex phase. Since quantum states are indistinguishable when they differ by a constant global phase, they really are the same states. Nevertheless, the convention is to use the displacement operator as the fundamental operator as opposed to translation of position followed by translation of momentum or vice versa. If you like, it is halfway between the other two options.
can we define the inverse operation: |n>=D^-1|alpha>? is this bijective?
We can certainly define the inverse of the displacement operator (which also happens to be equal to its adjoint).
thanks, what is the displacement operator for two mode light
Could you please clarify what you mean?
does the number of photons in a displaced coherent state is same as the original coherent state?
A coherent state is an eigenstate of the lowering operator, so the number of photons in a coherent state is not fixed. However, you can calculate the expectation value of the number of photons in a coherent state, and for coherent state |alpha> it is |alpha|^2. You can show this easily by calculating , where n=a^dagger a is the number operator. If you have a different coherent state |alpha'>, then the expectation value of the number operator changes to |alpha'|^2. I hope this helps!
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.
I did the calculations properly and there are 2 mistake that cancel each other. With D(alpha)^dagger we should still have the minus sign in front of |alpha|²/2 but you should also swapp the exponential with a and a^dagger. In the end you should respect D(alpha)^dagger = D(-alpha) (D is a unitary operator (check with the initial expression before splitting the exponentials)). In the final calculation (D(alpha)D(alpha)^dagger) you should use BKH one more time to swapp e^(-alpha* a) and e^(-alpha a^dagger) resulting in an additional e^|alpha|² canceling the product of the two initial e^|alpha|²/2 from D and D^dagger. In the end you end up with the identity as expected. Hope this helps somewhat