Density operator for mixed quantum states

[@shreyaspradhan8546]

This video was so great. Super intuitive and clear. I love this channel.

[@ProfessorMdoesScience]

Thanks for your support, glad you like the videos!

[@physicsfundamentalsbyhasan]

I tried Nielsen & chuang ,but it was your video that cleared so many things about density operator formalism for me ,I hope this channel grows more and more,thank you.

[@ProfessorMdoesScience]

Glad to hear that our approach was helpful!! :)

[@usamathakur-math]

I was in search of channel like this. I am so glad I found you I just wanna get across my concepts and mathematics of quantum mechanics to dig deeper into quantum computation and information these video's are really freaking helpful thank you so much for this can't say enough ❤️

[@ProfessorMdoesScience]

Really glad to hear you find our videos useful! :)

[@workinuka6110]

the world needs more of these, fast!

[@ProfessorMdoesScience]

Thanks for your support!

[@guoxinxin693]

Your video is short and so informative that I carefully studied every scentence indeed (maybe partially due to my poor english). They are valuable for my self studying. :)

[@ProfessorMdoesScience]

Happy to help! Something useful may be to turn on the automatic subtitles, which tend to be rather good :)

[@sandippaul468]

no not due to your English. I usually watch lectures from other channels in 2x speed. In this case I'm compelled to watch in normal speed with pauses bc every sentence is so so important and makes things clearer.

[@A.Shafei]

Extremely extremely neat and well presented! Even for those who disagree on which concepts are results and which are definitions, this video makes a reasonable choice and sticks with it with impressive consistency and clarity!

[@ProfessorMdoesScience]

Glad you like it! :)

[@nickknight5373]

A measly 18K views for this wonderfully lucid, focused and concise presentation? It deserves a couple of orders of magnitude better at least. Would love more videos on many-body systems, but thank you for the excellent tutorials that you have had time to make.

[@ProfessorMdoesScience]

Thanks for your kind words and for the suggestion!

[@miriamstudyaccount8735]

Great video! :) It would be cool to see this mini series on density operators continued to reduced density operators for composite systems

[@ProfessorMdoesScience]

Thanks for the suggestion! We do hope to continue along this direction, but quite a few things to cover so not sure when we'll get there!

[@fernandojimenezmotte2024]

Thank You for the very clear, neat, elegant explanation. Thank You for the motivation in the field of Quantum Mechanics that I love a lot. I am following your channel.

[@ProfessorMdoesScience]

Glad you like the videos, and thanks for following us! :)

[@學習中的哈密瓜]

Finally understand the density matrix, love the video and thanks!!!!

[@ProfessorMdoesScience]

Glad it helped!

[@quantum4everyone]

It might be useful to mention mixed states and incomplete information also arise when one takes a partial trace of an entangled state as often occurs in quantum computation and can be illustrated easily with Bohm’s version of EPR. So, mixed states do not really require large systems, but can occur much more broadly, especially in quantum information applications.

[@ProfessorMdoesScience]

Thanks for the insights! We do hope to move on from our current series on the basics of quantum mechanics to other topics including statistical mechanics and possibly quantum computing, which would give us a chance to expand on these topics.

[@panosthunder643]

Very nice analyze. I would like to ask, how the lindbland equation, arises from the form you wrote for the time evolution of the mixed state density matrix?

[@samuelsitina8558]

Sir, thank you, the language is so clear and the derivations and proofs very easy to understand and you probably saved my a$$, because I was not able to wrap my head around it by myslelf.

[@ProfessorMdoesScience]

Glad to be helpful!

[@pew6126]

Excellent video as usual. Just 2 points. Could have pointed out that density operators are positive ie >= 0 for all states |a>? Also Tr(AB) = Tr(BA) to show d/dt Tr(rho) = 0. May puzzle some viewers given evolution equation for rho: d/dt rho = /(i hbar) [H, rho]. Also, different mixtures can lead to same density operators. 😎

[@ProfessorMdoesScience]

Thanks for your insights! :)

[@vickymar3836]

Hi, at 16:19 you assume the individual states are orthogonal. If we assume that then we can write the mixed state as a superposition of basis created by the individual states right? Does that not defeat the whole purpose of the exercise as now we again have a pure state, just in a different basis.

[@ProfessorMdoesScience]

Not quite, the pk are not the standard expansion coefficients, they are the weights of the different states in the mixed state. The pk obey: Sum_k pk = 1 but in this final expression they appear as pk^2. Another way to see this is that for a pure state, rho^2=rho. However, here we show explicitly that, even for an orthogonal set of states, rho^2/=rho for a mixed state. I hope this helps!

[@vickymar3836]

@@ProfessorMdoesScience Thanks a lot, Actually my doubt got cleared when I referred your pure states vs mixed states video. Thank you so much for putting out such quality content!!

[@ProfessorMdoesScience]

@@vickymar3836 Glad this is all clear now!

[@lilboihere]

🙏🙏🙏 Beautiful and clear explanation & comparisons

[@ProfessorMdoesScience]

Glad you like the video! :)

[@aniksingharoy59]

Thank You Professor 🤝. It was soo much helpful.

[@ProfessorMdoesScience]

Glad you liked it! :)

[@bobdorsett6572]

Thank you! Now if you have time and inclination, could you please discuss the link between the mixed state density function and the partition function? These appear at the heart of recent work on the black hole information paradox and quantum gravity. You guys are great!

[@ProfessorMdoesScience]

Thanks for the suggestion! In short, the density operator of the canonical ensemble is rho=exp(-beta*H)/Z, where beta=1/(kB*T) is the inverse temperature, H the Hamiltonian, and Z the partition function. We have plans for new topics after we are done with the fundamentals of quantum mechanics, and statistical mechanics is one area we would like to cover, so we are noting down your suggestion, thanks!

[@bobdorsett6572]

@@ProfessorMdoesScience Hmmm. Z itself is the sum of e^(-beta*Ei), where the Ei are the energy states in the ensemble. And it appears in the denominator of the exponent in the density operator underneath (-beta*H)? Makes sense in unit analysis. Now I need to figure out what it implies for the physics. Any suggestions? Thanks!

[@ProfessorMdoesScience]

No, Z does not appear in the denominator of the exponent, it divides the exponent. I guess it is not the easiest to write mathematical formulas in the comments, so to be absolutely clear, let me rewrite it again in a slightly different way: rho=(1/Z)*exp(-beta*H). You can also understand why the density operator has this form from your expression for Z as the sum over exp(-beta*Ek). In the canonical ensemble, the probability of having a state of energy Ek, what we called pk in the video, is pk=exp(-beta*Ek)/Z, where Z is the normalization ensuring that the sum over all pk is 1. Then, following the video, rho=Sum_k (pk*|k> are the energy eigenstates. Plugging in the expression for pk into this sum, you arrive at the original expression for the density operator I wrote. [note I changed from your notation Ei to Ek to be consistent with the notation in the video]. As to understand this from a physical point of view, I would suggest a course on statistical mechanics (we will hopefully get there eventually ourselves). Introductory books I like include "Statistical Physics of Particles" by Kardar, or "Fundamentals of Statistical and Thermal Physics" by Reif. A more advanced book using the formalism of density operators is "Statistical Mechanics" by Feynman (not part of his 3-volume series). I hope this helps!

[@bobdorsett6572]

@@ProfessorMdoesScience Got it! Thanks a lot for taking time to explain, and thanks for the references.

[@JohnVKaravitis]

@@ProfessorMdoesScience I second the motion for Statistical Mechanics!

[@drdca8263]

This was helpful, thank you! It also brought a question to my mind, because it reminded me of something about spinors and I’m wondering if the connection I’m seeing is real/means anything. Aiui, in at least one sense of the term(I’m unsure how the different senses fit together) “spinor”, one can take a 2d complex vector (the spinor), and taking the outer product of it with itself (with the appropriate conjugation) one gets a 2x2 hermitian matrix, where the space of these matrices as a vector space over R correspond to spacetime vectors, where applying certain linear operations on the spinors corresponds to (orientation preserving) Lorentz transformations on the vector obtained from the outer product of the spinor with itself. (And like, angles where transforming the spinors are doubled when looking at the corresponding vector) Meanwhile, with the density matrix for a pure state being an outer product of the vector with itself, And for the vector to be normalized it needs the sum of the squares to be 1, while once we have it as a density matrix, this squaring kinda has already been done, so we just need to add up the diagonals, which, seems kind of like the angle doubling, if I squint really hard. So I guess I’m wondering, “does this similarity mean anything, or did I just draw a strained analogy that is probably mostly meaningless?” I guess the analogy extends to , for a unitary matrix U, “if you apply U^* rho U (or maybe I put the * on the wrong one, idr ) , this will correspond to just applying U to v, just as you do when conjugating [the outer product of the spinor with itself] by some 2x2 matrix corresponds to just applying the matrix to the spinor”.. But like, does that similarity “mean anything”? Or am I probably just making too big of a deal out of the tiny similarity of “they both involve the outer product of something with itself”?

[@ProfessorMdoesScience]

Interesting points. I would probably need to think more carefully about this, but my initial thought is that these are probably just mathematically analogous, not sure there is a deeper connection... A few thoughts that may or may not be relevant: unitary operators are those that preserve the scalar product (hence the norm), so they provide "rotations" in the complex vector space. This means that the density matrix should stay normalized under a unitary transformation.

[@Rational_Guy]

loved your explaination

[@ProfessorMdoesScience]

Glad to hear!

[@pujanajmera6651]

Would it be appropriate to say then that a mixed state is a more general form of a pure state, if we set p_k=1 for a specific k?

[@ProfessorMdoesScience]

Yes, you can view it like this :)

[@marcelorozas7516]

nice video! Could you make a video about Von Neumann Entropy, please?

[@ProfessorMdoesScience]

Thanks for the suggestion!

[@physicsandmath3290]

Much appreciation!!!

[@ProfessorMdoesScience]

Thanks for watching and glad you like it!

[@torqueRxF]

16:21 can someone please explain why the different states in the mixture are orthogonal?

[@ProfessorMdoesScience]

This is an assumption I have taken to simplify the maths, but the whole discussion can be generalized to non-orthogonal states. I hope this helps!

[@DrMarcoArmenta]

So clear! as always!

[@ProfessorMdoesScience]

Glad you like it! :)

[@subhajitsadhukhan8521]

Professor how to calculate the probabilities associated with the incomplete information about state? They are not same as probabilities associated with quantum mechanical measurements right?

[@ProfessorMdoesScience]

These two probabilites are different, we go over this in the following video: kzbin.info/www/bejne/goOYnJmep9hnecU I hope this helps!

[@Mark-IamNum1]

Excellent videos in general. How do you fancy doing spinors, the Dirac equation and its derivation? I think that would be very useful!!

[@ProfessorMdoesScience]

Thanks for the suggestion! Definitely on our to-do list!

[@subhajitsadhukhan8521]

What about the matrix element of density operator for mixed states? I computed the element and it leads to a sum over all pure states of the quantity pk Ci Cj*.looks like an ensemble average of Ci Cj*.is it ok? (¶ stands for density operator for mixed states) And Ci 's are expansion coefficients of the state |psi k>

[@ProfessorMdoesScience]

The density matrix for mixed states can indeed be written as a sum over the outer products of the individual pure states making it up with the appropriate weights pk, so if I understand your notation, I think you are correct.

[@subhajitsadhukhan8521]

Can you pls suggest a friendly book on quantum statistical mechanics( Introductary). It's in this semester and I'm struggling badly in it :(

[@ProfessorMdoesScience]

@@subhajitsadhukhan8521 Depending on the level, Feynman's is very nice.

[@mustufadosani4982]

If you can provide references then it would be lot helpful studying this in deeper.

[@ProfessorMdoesScience]

Most textbooks on quantum statistical mechanics would be a good reference for this topic. I hope this helps!

[@TszHoKwoK]

May you show how this is applied to PCA?

[@ProfessorMdoesScience]

Thanks for the suggestion, we'll add it to our list!

[@olieyin2313]

there is a little mistake at 3:58 that rho and operator A hat should be exchanged in the trace bracket

[@ProfessorMdoesScience]

I think the expression we have is correct? Also, in general the trace is invariant under cyclic permutations. I hope this helps!

[@olieyin2313]

@@ProfessorMdoesScience yeah it is. i just noticed the order in the video is different from my lecture notes but i didnt calculate that rho and observables commute. so both is ok! sry~~

[@ProfessorMdoesScience]

@@olieyin2313 No need to apologize, we are all here to learn together :)

[@nancy4186]

I have Eigenvalues of my density operator. One of the Eigen values are zero .So what does it mean??

[@ProfessorMdoesScience]

If you have a density matrix for a pure state, then as it is simply a projection operator, it always has the same two eigenvalues: 0 and 1. For the density matrix of a mixed state, the eigenvalues p_i and eigenstates |phi_i> provide a set of weights and states such that you can write it in the form: rho=sum_i p_i |phi_i>

[@5ty717]

Excellent

[@ProfessorMdoesScience]

Glad you like it!

[@canyadigit6274]

So a mixed state isn’t the state of the system, but a state of our knowledge?

[@ProfessorMdoesScience]

A mixed state is used to describe the state of the system as well as we can, which indeed is limited by our knowledge. You can find all the details in the video comparing pure and mixed states here: kzbin.info/www/bejne/goOYnJmep9hnecU

[@sadeghi73]

amazing

[@ProfessorMdoesScience]

Glad you like it!

[@2064899987]

Very nice!

[@ProfessorMdoesScience]

Glad you like it!

[@JohnSmall314]

very nice.

[@ProfessorMdoesScience]

Thanks!

[@ryan124007]

The pace of your videos is too fast. I have to play at 75% or 50% speed.

[@ProfessorMdoesScience]

Thanks for the feedback! Others have also mentioned this, while yet others think it is too slow... Luckily KZbin allows everyone to pick their preferred pace :)

[@iliasp4275]

[@ProfessorMdoesScience]

Glad you like it!

[@psylonmusic5264]

Amazing

[@ProfessorMdoesScience]

Thanks!