Quantum mechanics with mixed states 2:01 Phase space introduction 1:14:44 Properties of Wigner function 1:36:06 Gaussian states 2:07:54 Gaussian moment theorem 2:18:56
Why are your bounds of integration dp' rather than d^2r at 1:39:25, second line of calculation? Wouldn't the dirac delta function consume the dp' on the RHS, and you are left with the LHS bounds of integration? (sorry this be my lack of experience with the dirac delta function in this context). Thank you for clarification !
hi, no worries, let me see if this helps you: d^2r’ = dx’ dp’, right? then, the dirac delta(x’) consumes the integration over dx’ and sets x’ to zero, leaving only the integration over dp’. Does that make sense to you?
I had a few questions. I will be glad if you could answer them. Thanks in advance! (i) I have found in the literature that for a pure Gaussian state, the Wigner function is always non-negative (Weedbrook, 2011). On the other hand, you proved the possibility for a Wigner function to go negative on the basis of the orthogonality of pure states. My question is does the non-negativity of the pure Gaussian states have anything to do with the fact that two pure Gaussian states are never orthogonal (there is always a finite overlap)? (ii) Do the quadrature eigenstates only make sense as the 'post-measurement states'? Since their eigenvalues are continuous, it is not possible to prepare quadrature eigenstates physically with exactly well-defined quadrature eigenvalues. But upon homodyne measurement one collapses a quantum state of radiation to a quadrature eigenstate. The measurement reading provides the eigenvalue of the collapsed state. Of course, the post-measurement state is just a mathematical construct, not anything physical. (iii) The literature also states that for a real, symmetric matrix to qualify as a covariance matrix the condition is $V + \imath\Omega \geq 0$. Based on your derivation, it means $\expval{\delta \hat{\vec{R}} \delta \hat{\vec{R}}^T}$ has to be greater than 0. Is it because of Robertson's inequality? Thanks a lot for your time!
(i) Yeah, good question. First, note that, by definition, Gaussian states have a non-negative Wigner, no matter whether they are pure or mixed states. I think the message of what you might have heard is that pure states with a positive Wigner function must necessarily be Gaussian. And as you say, because the overlap of quantum states it's equal to the overlap of their Wigners in phase space, this implies that two pure Gaussian states cannot be orthogonal. (ii) Just to be clear: quadrature eigenstates NEVER make sense as physical states. They are OK as a theoretical idealization of the outcomes of a quadrature measurement, yeah, but in practice, creating one would require perfect resolution in an experiment, which never happens. Experiments always have a finite precision with which they can measure any continuous observable. Hence, in reality, you never collapse the state to an eigenstate of the observable, but to a superposition o f several of them, spanning around a "bandwidth" given by the resolution of the experiment. (iii) Note that this condition you mention precisely converges to det{V}>=1 for a single mode. This is because a 2x2 real, symmetric matrix has only two invariants (under change of reference quadratures in phase space, knwon as "symplectic transformations"), the trace and the determinant. The trace of the condition you mention (bringing V to the other side of the inequality) produces a trivial identity (sum of the variances of the position and momentum quadratures larger or equal than 0), so you are left with just the determinant of the expression, which is precisely det{V}>=1. For many modes it's more complicated, because the covariance matrix is higher-dimensional, which means that it has several symplectic invariants. In fact, the formulation I like the most for this general inequality is based on symplectic eigenvalues, the eigenvalues of i*Omega*V, which have to be larger than or equal to one. These eigenvalues, in turn, have a very physical explanation: they fix the entropy of the state; in particular, any Gaussian state can be written as a tensor product of thermal states for each mode, onto where you apply Gaussian unitaries (Williamson's theorem). The symplectic eigenvalues are related to the number of thermal excitations on each mode, and since the unitaries cannot change the entropy, they fixed the mixedness of the state. I'm not sure if this answers your question, but it's the most accurate description I can give (in a few lines) of what's going on for many modes. To be honest, I don't recall what the Robertson inequality is, but I hope what I explained puts you on the right track :-D Most of these things I mentioned are in my notes of arxiv.org/abs/1504.05270 (which I turned into a book in 2015, and I can share on demand if you don't have access to it) and in the various reviews and books of quantum information with continuous variables that I mention there.
@@CarlosNBvs5 Thank you for your detailed answers. Thank you very much for taking the time. I will definitely go through the relevant materials from your book. Incidentally, is there an email I can get you in touch with?
Quantum mechanics with mixed states 2:01
Phase space introduction 1:14:44
Properties of Wigner function 1:36:06
Gaussian states 2:07:54
Gaussian moment theorem 2:18:56
Gracias profesor Carlos son las clases mas completas en optica cuantica que he encontrado!
muchas gracias!! 😄
Great lecture. The notes are helpful too. Cool prog music as well!
thanks for taking the time to tell me this! :-D I really appreciate it and it really motivates me!
@@CarlosNBvs5 Fancy Strandberg and full TesseracT album cover. Damn. If things go well, I might just become you in 10 years lol
@@lilaitch705 hahaha, all the best finding your way!
awesome lecture sir
I want to know about Q and P probability distribution function. and can you can share me material of quantum optics, I am undergraduate student.
Awesome lecture! Hi Carlos, since the lectures are long, I think, it might help segmenting them.
yeah, I thought about it, but I am so busy… I might do it some time… I still even have to edit the last 7 videos, cutting the irrelevant parts… 😅
Why are your bounds of integration dp' rather than d^2r at 1:39:25, second line of calculation? Wouldn't the dirac delta function consume the dp' on the RHS, and you are left with the LHS bounds of integration? (sorry this be my lack of experience with the dirac delta function in this context). Thank you for clarification !
hi, no worries, let me see if this helps you: d^2r’ = dx’ dp’, right? then, the dirac delta(x’) consumes the integration over dx’ and sets x’ to zero, leaving only the integration over dp’. Does that make sense to you?
@@CarlosNBvs5 I see, thank you for the quick response Carlos!
¡Gran clase Carlos! Saludos
muchas gracias!!
I had a few questions. I will be glad if you could answer them. Thanks in advance!
(i) I have found in the literature that for a pure Gaussian state, the Wigner function is always non-negative (Weedbrook, 2011). On the other hand, you proved the possibility for a Wigner function to go negative on the basis of the orthogonality of pure states. My question is does the non-negativity of the pure Gaussian states have anything to do with the fact that two pure Gaussian states are never orthogonal (there is always a finite overlap)?
(ii) Do the quadrature eigenstates only make sense as the 'post-measurement states'? Since their eigenvalues are continuous, it is not possible to prepare quadrature eigenstates physically with exactly well-defined quadrature eigenvalues. But upon homodyne measurement one collapses a quantum state of radiation to a quadrature eigenstate. The measurement reading provides the eigenvalue of the collapsed state. Of course, the post-measurement state is just a mathematical construct, not anything physical.
(iii) The literature also states that for a real, symmetric matrix to qualify as a covariance matrix the condition is $V + \imath\Omega \geq 0$. Based on your derivation, it means $\expval{\delta \hat{\vec{R}} \delta \hat{\vec{R}}^T}$ has to be greater than 0. Is it because of Robertson's inequality?
Thanks a lot for your time!
(i) Yeah, good question. First, note that, by definition, Gaussian states have a non-negative Wigner, no matter whether they are pure or mixed states. I think the message of what you might have heard is that pure states with a positive Wigner function must necessarily be Gaussian. And as you say, because the overlap of quantum states it's equal to the overlap of their Wigners in phase space, this implies that two pure Gaussian states cannot be orthogonal.
(ii) Just to be clear: quadrature eigenstates NEVER make sense as physical states. They are OK as a theoretical idealization of the outcomes of a quadrature measurement, yeah, but in practice, creating one would require perfect resolution in an experiment, which never happens. Experiments always have a finite precision with which they can measure any continuous observable. Hence, in reality, you never collapse the state to an eigenstate of the observable, but to a superposition o f several of them, spanning around a "bandwidth" given by the resolution of the experiment.
(iii) Note that this condition you mention precisely converges to det{V}>=1 for a single mode. This is because a 2x2 real, symmetric matrix has only two invariants (under change of reference quadratures in phase space, knwon as "symplectic transformations"), the trace and the determinant. The trace of the condition you mention (bringing V to the other side of the inequality) produces a trivial identity (sum of the variances of the position and momentum quadratures larger or equal than 0), so you are left with just the determinant of the expression, which is precisely det{V}>=1.
For many modes it's more complicated, because the covariance matrix is higher-dimensional, which means that it has several symplectic invariants. In fact, the formulation I like the most for this general inequality is based on symplectic eigenvalues, the eigenvalues of i*Omega*V, which have to be larger than or equal to one. These eigenvalues, in turn, have a very physical explanation: they fix the entropy of the state; in particular, any Gaussian state can be written as a tensor product of thermal states for each mode, onto where you apply Gaussian unitaries (Williamson's theorem). The symplectic eigenvalues are related to the number of thermal excitations on each mode, and since the unitaries cannot change the entropy, they fixed the mixedness of the state.
I'm not sure if this answers your question, but it's the most accurate description I can give (in a few lines) of what's going on for many modes. To be honest, I don't recall what the Robertson inequality is, but I hope what I explained puts you on the right track :-D
Most of these things I mentioned are in my notes of arxiv.org/abs/1504.05270 (which I turned into a book in 2015, and I can share on demand if you don't have access to it) and in the various reviews and books of quantum information with continuous variables that I mention there.
@@CarlosNBvs5 Thank you for your detailed answers. Thank you very much for taking the time. I will definitely go through the relevant materials from your book. Incidentally, is there an email I can get you in touch with?
where is the link to lecture notes? I want to download.....
hi, you should be able to find it in the description of the video
nice video and good lectures!
Thanks! I appreciate you watching them and commenting!
@@CarlosNBvs5 Thanks for uploading