Question: Let's forget about the Schrödinger picture and say we only know about the Heisenberg picture. Then we have a differential equation for the dynamic of the system with matrices (operators) as variables. If you solve this diff equation, you get an answer for a time varying matrix but you need an initial condition for that matrix as well. How do you determine the initial condition for the matrix when you are solving a problem? Also, after having the matrix, to calculate the statistics of that observable, you need a fixed vector (which is a fixed state in Schrodinger picture but we are assuming we don't know about the Schrodinger picture). How do you determine that fixed vector? In other word, every explanation I saw about the Heisenberg picture depends on the Schrodinger picture. Can you explain Heisenberg picture independently and how it can be used to solve a problem step by step?
We have been finding it difficult to upload regularly (this is only our side project), but we hope to have more time moving forward and go back to publish more often!
@@hp2594 Main job is that we are both professors at the University of Cambridge, which mostly involves research and teaching. The UA-cam channel is a side project :)
Hey guys, I just discovered your videos and I absolutely love them! Thanks for making such clear and helpful videos and I hope you make many more. I’ll definitely be recommending them to people
I’d love to chat with you guys and hear about your plans! If you’re interested then could you drop me an email? My email is on the about page of my channel :)
There are plenty of vids on "the tube" regarding QM. This channel is different as it does not only scratch the surface but is going into the interesting details as well. Thanks a lot for this!!! You are asking for topics, which might complete this series: I would highly appreciate some vids about the pertubation theory inside QM (e.g. Fermi's golden rule).
Glad you like the videos, and thanks for the suggestion! Perturbation theory is on our to-do list, probably after we cover spin-1/2 systems. We'll keep you informed!
glad to have you back! my thoughts on this one: 1. Does the "wavefunction" even make sense in the heisenberg picture? due to observables being the one who's time-evolving, we only have one constant state in time, doesn't seem like its waving at all 2. Can the matrix elements of hermitian operators in the heisenberg picture evolve from diagonal to non-diagonal or vice-versa?(because wavefunction can time-evolve from eigenstate to superposition) 3. After some moments thinking about several ways to describe time-evolution, i have this idea that *both* hermitian operators *and* quantum state are part of "the system" (rather than wavefunction containing everything). Because observables are obtained indirectly in quantum mechanics, you need hermitian operators and states working in tandem to produce the spectrum of experimental results. both of them are "the system", and the separation between observables and states are not really physical. 4. If hermitian operators are the one dynamically evolving in the heisenberg picture, what if we're trying to treat entanglement in heisenberg picture? do we superpose the operators instead? what do you think?
Glad to be back! :) Here are some thoughts in response: 1. The name "wavefunction" is indeed tricky; note that even in the Schrödinger picture we can have states that don't really change in time, for example the eigenstates of the Hamiltonian. So we should probably consider "wavefunction" as more general than just describing something that is waving. 2. I guess this is correct, just like in the Schrödinger picture you can have states oscillating between various eigenstates. 3. You are essentially correct that for a full description of a quantum system you need both observables and states. 4. This is an interesting question, and a full answer would require much more than a comment. But to get you started, note that to consider entanglement in the Schrödinger picture, the very first thing we do is to separate a state space V into two parts V1(x)V2 [where (x) represents tensor product]. You can then typically find operators that only act on V1, and operators that only act on V2. It turns out that specifying these operators is enough to define the separation V1(x)V2. In the Schrödinger picture, this separation V1(x)V2 does not change in time, and the states do, so you define entanglement and its time evolution through the states. In the Heisenberg picture, the operators associated with V1 (and V2) change in time, so the factorization V1(x)V2 also changes in time. This means that, even though the states are time-independent, the entanglement can change because of the change in the factorization of V with respect to which we define entanglement. I hope this helps!
@@ProfessorMdoesScience hmmm... on the last part... does changing pictures preserve the measurable effect of entanglement? oh! i think i know why i need time to grasp this. i've heard how entanglement is described as this bizarre nonlocal phenomenon, but never on how an entangled pair can evolve in time to be entangled in a *different* way, or even to eventually disentangle completely
Thank you all for these useful and beautiful videos that help me a lot while I'm studying QM. I really need now a video about the mathematics of Entangled States please 🥺. 🙏❤️❤️
Glad you like them! We do hope to go deeper into entangled states at some point, but for now we very briefly mention them in this video: ua-cam.com/video/kz3206S2B6Q/v-deo.html Hope it helps!
I love the two videos that I've watched, thank you so much! I have a question on the one above (Schrödinger vs. Heisenberg pictures of quantum mechanics): In some places, e.g. time stamp 26:56, you include a term (the A is actually A-hat:) i h-bar [d/dt A_S (t)]_H. Why do you not write, instead, a single subscript, on the A alone, i.e. i h-bar [d/dt A_H (t)]?
The notation is meant to make it clear what we are doing: first calculate the time derivative of A when written in the Schrödinger picture, and then transform the result to the Heisenberg picture. Note that, if the operator A is time-independent in the Scrhödinger picture (as many important operators are), then the derivative vanishes. Things become more interesting if the operator is time dependent. I hope this helps!
We currently don't have them available, but we are working on creating additional content to go with the videos (including notes and problems+solutions). It may take a while though...
At 20:33, I will still not try to put time dependence on observable at Schrodinger picture, as for some beginners it may confuse them. Don't you think the same?
sir ,i have a question when you derive the dynamics of observables in heisenberg picture ,why you put the observable of schrodinger picture sandwiched between the unitory operators as time dependent ,where we see the obsvbles in S picture are time independent
Observables in the Schrödinger picture can be time dependent, this is why I have a time dependence on those expressions. Standard observables like position, momentum, or angular momentum are indeed time independent in the Schrödinger picture. But you could, for example, have a time-dependent Hamiltonian describing, say, a short light pulse interacting with an atom. In this case, the interaction is time dependent in the Schrödinger picture: it is zero to start with, non-zero for a short time (the duration of the pulse) and zero again afterwards. I hope this helps!
Time dependence in quantum mechanics can be described in many different ways. For example, the time dependence can be captured by the state (as in the Schrödinger picture) or by the operators (as in the Heisenberg picture) or by any intermediate combination. Each one of these choices is called a "picture" in this context. I hope this helps!
@@ProfessorMdoesScience Thank you sir .I got your point but there is one confusion ."Each one of these choices is called picture it means if we talk about the time dependent Schrodinger equation so how can we represent it Schrodinger equation?that's called picture?I m right?if not please guide me:)
@@Physics_Walli The time dependent Schrödinger equation tells you how the quantum states evolve in time in the Schrödinger picture. By contrast, in the Heisenberg picture it is operators that capture the unitary time evolution, and the relevant equation that gives the corresponding time dependence is the one discussed in this video. I hope this helps!
This is a very interesting question! No space in the comments section for a full answer, but let me give you a few ideas. Both Schrödinger and Heisenberg pictures describe "unitary time evolution", that is, the time evolution that occurs in the absence of a measurement. When we perform a measurement, state collapse happens, and this is not described by the Schrödinger equation in the Schrödinger picture, nor the corresponding dynamical equation in the Heisenberg picture. Instead, at the point of state collapse we have "non-unitary time evolution". As described in our video on state collapse, this "non-unitary time evolution" is mediated by a projection operator applied to the quantum state in the Schrödinger picture (details here: ua-cam.com/video/UaC-gLZ0Zvc/v-deo.html). You can view this as, when going from a state at t0 to a state at t with a measurement in-between at ti (t0
@@ProfessorMdoesScience Prof. does the state really "collapse" or just its representation in the space we are measuring? I think, it is better to see this in terms of Fourier transform. Because when the position-space wave-function "collapses" (or more correctly, in my view, narrows or even better "is made to be narrower") under the external influence that we call "measurement of position", the momentum-space wave-function (its Fourier transform) actually widens (=uncertainty principle") and there is no collapse in that space or representation. In other words, "collapse" is a RELATIVE term. Relative to the observable and representation that we (as measuring agents) are working with and trying to measure. Similarly "superposition" is also a relative term: Every state (e.g. a position eigen-state) is a superposition in terms of eigen-states of some other observable. It is usually said that "superpositions cannot be observed". To me, this seems a lack of deep understanding of the mathematics of QM. When we observe a particle localized to a small region (after a "position measurement") we are simultaneously also observing a very wide superposition in terms of momentum eigen-states. I think, to talk about "collapse" and "superposition" in absolute terms, is a bias of the physicist's mind. Also the postulate or proposition that "measurement causes collapse of the wave-function" to me it seems that needs to be re-formulated backwards. That is, we want to make the wave-function narrow in some space (position, momentum, etc.), and then we orchestrate a physical process (that we call measurement) which exactly causes this narrowing in that particular representation or space (which also translates to a widening in the conjugate space and representation). In other words, this intentional narrowing of wave-function in some space (and the unavoidable, simultaneous widening in some other space) is the very definition of "measurement". Also don't you think the evolution is "non-unitary upon measurement" only because we have not incorporated the measurement sub-system itself into the state. That is, time-evolution looks non-unitary (and we need to resort to projection, etc.) ONLY because we have chosen a smaller window to view only the particle (not the bigger system), but choosing a larger window (larger system's state) to incorporate both "the measured" and the "measuring agent" will restore the unitarity (thus for example, position measurement could be defined as intentional unitary narrowing [by whatever resolution we might want] of the joint position-space wave-function). To me it seems, this view and recognition of biases (and subtle points) in understanding QM (and interpreting its math) makes QM much more intuitive and rational. Just some of my thoughts as a careful student of physics. I would be honored if any of my possible misconceptions are pointed out, Sir. Thank you.
@@itsawonderfullife4802 If I understand your points correctly, I think I agree with you in that "collapse" refers specifically to the state of the system in the basis of the eigenstates of the observable we are measuring. From this point of view, a term like "change" may be more appropriate, as would take into account the "collapse" in this basis, but also whatever change happens in another basis. Superposition is again basis-dependent indeed. Regarding measurements, this is a tricky topic that many have discussed, an example of an interesting discussion is here: physics.stackexchange.com/questions/35648/if-i-go-to-the-church-of-the-greater-hilbert-space-can-i-have-unitary-collapse
@@itsawonderfullife4802*Because when the position-space wavefunction "collapses" (or more correctly, in my view, "narrows," or even better, "is made to be narrower") under the external influence that we call "measurement of position," the momentum-space wavefunction (its Fourier transform) actually widens (=uncertainty principle) and there is no collapse in that space or representation.* Your view is incorrect, and the very conclusion that you draw concerning position-space versus momentum-space proves that your view is incorrect. You called the wavefunction collapse a "narrowing," failing to realize that this naming scheme is inappropriate, precisely because a "widening" also occurs in the Fourier transform of the representation used. This underpins a lack of understanding of the mathematics of wavefunction collapse. What you are failing to understand is that wavefunction collapse is representation-independent. For a given observable A, there is the eigenvalue equation Â|ψ) = λ·|ψ), where |ψ) denotes the state vector. Consider |ψ[α]) for all α to be the eigenbasis with respect to Â, so that Â|ψ[α]) = λ[α]·|ψ[α]), and |ψ) = Σ{c[α]·|ψ[α])}, where Σ denotes a Lebesgue sum. This gives us the decomposition of |ψ) as a superposition of the eigenbasis |ψ[α]). Wavefunction collapse simply refers to the non-unitary time evolution from |ψ) to |ψ[β]) for some β. It is improper to call this a narrowing or widening, since it is representation independent: all it amounts to is projecting the state vector to one of the basis vectors in the eigenbasis of Â, and this collapse happens when measuring the observable A. *In other words, "collapse" is a RELATIVE term. Relative to the observable and representation that we (as measuring agents) are working with and trying to measure.* No, this is incorrect. Collapse is representation-independent. When we measure the observable energy of a system, the system does not care whether our mathematical description of it is in the position space or momentum space. The collapse works the same way. Also, to say that it is relative to observable is nonsensical. Observables do not, in general, commute, so observables do not, in general, even share an eigenbasis with which we can speak of collapse. You can only consider collapse in the context of the measurement of a single observable, and this is the only context in which the notion is mathematically meaningful. *Similarly, "superposition" is also a relative term: every state (e.g. position eigenstate) is a superposition of eigen-states of some other observable.* You are mistaken. Position is not a fundamental observable to speak of, here, and the term "superposition" does not refer to any particular observable in particular. Also, what you are describing in your comment is a change of basis, the existence of a Jacobian between any two bases, but that is not what the word "superposition" means. What the word superposition refers to is the fact that any arbitrary state vector, whether an eigenstate of an observable or not, can be written as a superposition of _some_ basis, whether an eigenbasis or not. In this case, the observables are irrelevant for the notion of superposition, which is independently well-defined. *It is usually said that "superpositions cannot be observed."* No physicist has ever said this. The only people who say this are enthusiastic students who are nonetheless completely new to the study of quantum mechanics, and have no had any exposure to the topics beyond an extremely introductory-level discussion of certain intuitions. *To me, this seems a lack of deep understanding of the mathematics of QM.* You have no privilege to be saying this, considering your own ignorance of the mathematics. To be clear, I know that later in your comment, you admit to just being a student, which justifies the ignorance, but it is extremely arrogant and stupid to pretend that you understand the mathematics of QM better than expert physicists do, which is what you are doing here, especially when you also dare to attribute to them a saying that originated on the Internet among students, not physicists. *When we observe a particle localized to a small region (after a "position measurement"), we are simultaneously also observing a very wide superposition in terms of momentum eigenstates.* This is not simultaneous, because as you said, it occurs _after_ the position measurement. This scenario looks very different from a simultaneous measurement of two observables. *I think, to talk about "collapse" and "superposition" in absolute terms, is a bias of the physicist's mind.* Once again, this is extremely arrogant and insulting. A physicist understands the mathematics of QM better than you or I do, and a physicist understands that you have a bias towards a particular representation of the wavefunction, one which is not actually warranted, and which is obscuring and blockading your understanding of representation-independent quantum states. *Also, the postulate or proposition that "measurement causes collapse of the wavefunction," to me it seems that needs to be reformulated backwards. That is, we want to make the wavefunction narrow in some space (position, momentum, etc.), and then we orchestrate a physical process (that we call measurement) which causes exactly this narrowing in that particular representation or space (which also translates to a widening in the conjugate space and representation).* No, this is completely wrong. You are conflating "space" and "representation." The space of quantum states is a unique Hilbert state, characterized by the postulates, or dare I say, axioms, of quantum mechanics. A representation refers simply to the projection of the quantum state into the eigenstates of a particular observable. It is one thing to talk about |ψ), and it is another completely different thing to talk about (ψ[A, α]|ψ), which is the representation of the quantum state in the span of eigenstates of the observable A. The position representation is (r|ψ), not |ψ) itself. There is a video on the channel already explaining all of this. The way it works is that Â|ψ[A, α]) = λ[A, α]·|ψ[Α, α]), so the dual equation is the equation (ψ[A, α]|·λ[A, α]* = (ψ[A, α]|Â^t, where Â^t is the adjoint of Â. Therefore, λ[A, α]*·(ψ[Α, α]|ψ) = (ψ[A, α]|Â^t|ψ). Αnyway, when we perform a measurement, we are not seeking to narrow anything on a particular representation, since collapse is representation-independent. Also, collapse is not something that we seek. Collapse is something that happens when we perform a measurement we seek. So the reformulation you speak of already makes no sense. *In other words, this intentional narrowing of wavefunction in some space is the very definition of "measurement."* No, because the collapse of the wavefunction is not something we intend for. It is merely an unavoidable consequence of what we are looking for. The idea behind collapse is simple. For a simplified example, let us consider the simplest non-trivial vector space every person is familiar with on some intuitive level or other: the 2-dimensional vector space R^2. Here, we can choose orthornormal vectors (with respect to the Euclidean inner product, the dot product), so that one basis vector is called "horizontal," which I shall denote |r), with "r" for "right," and the other is called "vertical," which I shall denote |u), with "u" for "up." Every vector |ψ) in this vector space can be written as a·|r) + b·|u) for some a, b in R. Now, suppose we have an observable A whose operator  has eigenstates |r) and |u). Then, according to the postulate, an inevitable consequence of trying to measure A is that we can only ever measure the eigenvalues r and u respectively. And if the eigenvalue r is measured, then this must be equivalent to the projection a·|r) + b·|u) -> |r), while if the eigenvalue u is measured, then this must be equivalent to the projection a·|r) + b·|u) -> |u). This is what collapse refers. This is what collapse is. But this collapse is not intended. The only thing that is intended is the measurement itself. We measure A, without a preconceived expectation that only r or u will ever come out. It is not as if there is anything that, in principle, is stopping (|r| + |u|)/2 from being the valid outcome of a measurement, or sqrt[(|r|^2 + |u|^2)/2] from being the outcome. But the reason it is not a valid outcome is due to this unintended yet inevitable collapse that occurs. That is what the postulates of QM are encoding.
@@angelmendez-rivera351 First you need to cool down, my friend. I wasn't trying to insult any physicists. Many practicing QM experts (particularly in the foundations of QM and QM optics) share these views, implicitly or explicitly. Not understanding other people's meaning and being aggressive in dismissal harms only oneself, especially in the long run. "...superposition" does not refer to any particular observable in particular. " "Collapse is representation-independent" These very sentences suggest you have a serious and underlying problem with the very basics, especially when it comes to making cross connections in QM. And (in contrast to Prof M.) you neither understood the meaning nor intention behind any of my original comments, my dogmatic friend (Even my "student" reference above: We are all "students" ;) ). Also some Fourier analysis bg. wouldn't hurt either. So I would rather respectfully skip going through the rest of your comment.
It is still not available. We are working on a number of videos, so it will hopefullly arrive soon... And yes, the interaction picture is useful when we have, for example, a time-dependent perturbation.
Thank you so much. Your explanation of Heisenberg and Schrdinger's pictures in quantum mechanics was excellent. I really enjoyed it. Good luck
Glad you liked it! :)
Question: Let's forget about the Schrödinger picture and say we only know about the Heisenberg picture. Then we have a differential equation for the dynamic of the system with matrices (operators) as variables. If you solve this diff equation, you get an answer for a time varying matrix but you need an initial condition for that matrix as well. How do you determine the initial condition for the matrix when you are solving a problem? Also, after having the matrix, to calculate the statistics of that observable, you need a fixed vector (which is a fixed state in Schrodinger picture but we are assuming we don't know about the Schrodinger picture). How do you determine that fixed vector?
In other word, every explanation I saw about the Heisenberg picture depends on the Schrodinger picture. Can you explain Heisenberg picture independently and how it can be used to solve a problem step by step?
Thank you so much, professor! I was afraid that maybe you'd stopped uploading, but I'm so glad I was wrong!
We have been finding it difficult to upload regularly (this is only our side project), but we hope to have more time moving forward and go back to publish more often!
@@hp2594 Main job is that we are both professors at the University of Cambridge, which mostly involves research and teaching. The UA-cam channel is a side project :)
@@ProfessorMdoesScience Thank you for sharing your level of understanding with people around the globe
Love your brisk and energetic approach. Thank you very much.
Glad to hear this, thanks for watching!
Hey guys, I just discovered your videos and I absolutely love them! Thanks for making such clear and helpful videos and I hope you make many more. I’ll definitely be recommending them to people
Thanks for the kind words! We've been watching your videos for a while, and find them really inspiring :)
I’d love to chat with you guys and hear about your plans! If you’re interested then could you drop me an email? My email is on the about page of my channel :)
There are plenty of vids on "the tube" regarding QM. This channel is different as it does not only scratch the surface but is going into the interesting details as well. Thanks a lot for this!!! You are asking for topics, which might complete this series: I would highly appreciate some vids about the pertubation theory inside QM (e.g. Fermi's golden rule).
Glad you like the videos, and thanks for the suggestion! Perturbation theory is on our to-do list, probably after we cover spin-1/2 systems. We'll keep you informed!
@@ProfessorMdoesScience YEAH! Thanks a lot. I've an active subscription as well. Don't want to miss any ...
Thank you for your videos, they are awesome! Looking forward to the video on the Interaction Picture :D
Glad you like the videos! We hope to be able to publish it soon!
Please continue with this lovely work.
Thanks for your support! :)
Love your videos guys! Just in time for my exams this Easter term !
Glad you like them! Are you studying at Cambridge?
@@ProfessorMdoesScience yes!
Thank you so much.Excellent presentation.All the ideas are conveyed in a simple and clear way .
Glad you like it!
Thank you so much! Eagerly awaiting your video on the interaction picture. ❤
Thanks for watching, and it is in our soon-to-do list!
Yes please 👍🙏🙏
Brilliant video. So clear and well structured
Glad you like it!
great video !!! waiting for the Interaction picture! 💌
On our to-do list! :)
glad to have you back! my thoughts on this one:
1. Does the "wavefunction" even make sense in the heisenberg picture? due to observables being the one who's time-evolving, we only have one constant state in time, doesn't seem like its waving at all
2. Can the matrix elements of hermitian operators in the heisenberg picture evolve from diagonal to non-diagonal or vice-versa?(because wavefunction can time-evolve from eigenstate to superposition)
3. After some moments thinking about several ways to describe time-evolution, i have this idea that *both* hermitian operators *and* quantum state are part of "the system" (rather than wavefunction containing everything). Because observables are obtained indirectly in quantum mechanics, you need hermitian operators and states working in tandem to produce the spectrum of experimental results. both of them are "the system", and the separation between observables and states are not really physical.
4. If hermitian operators are the one dynamically evolving in the heisenberg picture, what if we're trying to treat entanglement in heisenberg picture? do we superpose the operators instead?
what do you think?
Glad to be back! :) Here are some thoughts in response:
1. The name "wavefunction" is indeed tricky; note that even in the Schrödinger picture we can have states that don't really change in time, for example the eigenstates of the Hamiltonian. So we should probably consider "wavefunction" as more general than just describing something that is waving.
2. I guess this is correct, just like in the Schrödinger picture you can have states oscillating between various eigenstates.
3. You are essentially correct that for a full description of a quantum system you need both observables and states.
4. This is an interesting question, and a full answer would require much more than a comment. But to get you started, note that to consider entanglement in the Schrödinger picture, the very first thing we do is to separate a state space V into two parts V1(x)V2 [where (x) represents tensor product]. You can then typically find operators that only act on V1, and operators that only act on V2. It turns out that specifying these operators is enough to define the separation V1(x)V2. In the Schrödinger picture, this separation V1(x)V2 does not change in time, and the states do, so you define entanglement and its time evolution through the states. In the Heisenberg picture, the operators associated with V1 (and V2) change in time, so the factorization V1(x)V2 also changes in time. This means that, even though the states are time-independent, the entanglement can change because of the change in the factorization of V with respect to which we define entanglement.
I hope this helps!
@@ProfessorMdoesScience hmmm... on the last part... does changing pictures preserve the measurable effect of entanglement? oh! i think i know why i need time to grasp this. i've heard how entanglement is described as this bizarre nonlocal phenomenon, but never on how an entangled pair can evolve in time to be entangled in a *different* way, or even to eventually disentangle completely
Minor correction: 23:29 is an application of the product rule, not the chain rule
Otherwise, great job! Keep up the good work
Thanks for this, and glad you like the video!
this is very good and in perfect timing for my advanced quantum mechanics exam, good job.
Glad it is helpful and the timing works! May we ask where you study?
Another GREAT video Professor M team 👌, thank you soooo much ❤. We are waiting for the "Interaction picture" video!
Glad you like it! We'll get to the interaction picture as soon as possible! :)
Thank you all for these useful and beautiful videos that help me a lot while I'm studying QM. I really need now a video about the mathematics of Entangled States please 🥺.
🙏❤️❤️
Glad you like them! We do hope to go deeper into entangled states at some point, but for now we very briefly mention them in this video: ua-cam.com/video/kz3206S2B6Q/v-deo.html
Hope it helps!
great video on clarifying this, please do more in quantum field
Glad you like it, and thanks for the suggestion!
Thank you very much! The video is wonderful!
Glad you like it!
Epic, finally I understood some QM! :) Thank you Prof. M!
This is great to hear! :)
Hi professor M
Thank you for this awesome video.
Can you please make a video on the Liouville picture of QM?
Thanks for the suggestion!
I love the two videos that I've watched, thank you so much! I have a question on the one above (Schrödinger vs. Heisenberg pictures of quantum mechanics): In some places, e.g. time stamp 26:56, you include a term (the A is actually A-hat:) i h-bar [d/dt A_S (t)]_H. Why do you not write, instead, a single subscript, on the A alone, i.e. i h-bar [d/dt A_H (t)]?
The notation is meant to make it clear what we are doing: first calculate the time derivative of A when written in the Schrödinger picture, and then transform the result to the Heisenberg picture. Note that, if the operator A is time-independent in the Scrhödinger picture (as many important operators are), then the derivative vanishes. Things become more interesting if the operator is time dependent. I hope this helps!
Great video!
A video on perturbation Theory would be awesome. Keep up the good work! :)
Thanks for the suggestion, it is on our to-do list :)
Thank you so much it's helps me a lot
Glad this is helpful!
Thanks a lot for these videos! I just wanted to know, would it be possible to access the notes you write on the video? It would help me a lot!
We currently don't have them available, but we are working on creating additional content to go with the videos (including notes and problems+solutions). It may take a while though...
At 20:33, I will still not try to put time dependence on observable at Schrodinger picture, as for some beginners it may confuse them. Don't you think the same?
Hi Professor M, is it possible for you to talk about the interaction picture also?
It is in our list indeed, we hope to get there soon!
@@ProfessorMdoesScience thank you so much! I do look forward to it!
Thank you so much😊
Thanks for watching! :)
Where is the video on interaction picture, I'm unable to find it
We haven't published the video on the interaction picture yet, hope to do this in the future!
@@ProfessorMdoesScience Yes please publish a video on interaction picture as soon as possible, this videos are too much helpful.
I miss you. where are you? Why not post a new video?
Been very busy lately, but we have the next 6 videos in the pipeline, so hopefully we'll start publishing again soon!
@@ProfessorMdoesScience
Thanks ❤❤
sir ,i have a question when you derive the dynamics of observables in heisenberg picture ,why you put the observable of schrodinger picture sandwiched between the unitory operators as time dependent ,where we see the obsvbles in S picture are time independent
Observables in the Schrödinger picture can be time dependent, this is why I have a time dependence on those expressions. Standard observables like position, momentum, or angular momentum are indeed time independent in the Schrödinger picture. But you could, for example, have a time-dependent Hamiltonian describing, say, a short light pulse interacting with an atom. In this case, the interaction is time dependent in the Schrödinger picture: it is zero to start with, non-zero for a short time (the duration of the pulse) and zero again afterwards. I hope this helps!
@@ProfessorMdoesScience thank you sir
I just realized that the handwriting is the same for both professors M and M'.
Indeed! :)
Very helpful
Glad you think so!
What does mean of "Picture" in quantum mechanics.Please give me answer clearly
Time dependence in quantum mechanics can be described in many different ways. For example, the time dependence can be captured by the state (as in the Schrödinger picture) or by the operators (as in the Heisenberg picture) or by any intermediate combination. Each one of these choices is called a "picture" in this context. I hope this helps!
@@ProfessorMdoesScience Thank you sir .I got your point but there is one confusion ."Each one of these choices is called picture it means if we talk about the time dependent Schrodinger equation so how can we represent it Schrodinger equation?that's called picture?I m right?if not please guide me:)
@@Physics_Walli The time dependent Schrödinger equation tells you how the quantum states evolve in time in the Schrödinger picture. By contrast, in the Heisenberg picture it is operators that capture the unitary time evolution, and the relevant equation that gives the corresponding time dependence is the one discussed in this video. I hope this helps!
@@ProfessorMdoesScience Thank you so much respected Sir😊
Can you do the legget-garg inequalities?
Thanks for the suggestion! They were not in our list, but will add them
How about measurement collapse in Heisenberg picture? 'Non-Hamiltonian'?
This is a very interesting question! No space in the comments section for a full answer, but let me give you a few ideas. Both Schrödinger and Heisenberg pictures describe "unitary time evolution", that is, the time evolution that occurs in the absence of a measurement. When we perform a measurement, state collapse happens, and this is not described by the Schrödinger equation in the Schrödinger picture, nor the corresponding dynamical equation in the Heisenberg picture. Instead, at the point of state collapse we have "non-unitary time evolution". As described in our video on state collapse, this "non-unitary time evolution" is mediated by a projection operator applied to the quantum state in the Schrödinger picture (details here: ua-cam.com/video/UaC-gLZ0Zvc/v-deo.html). You can view this as, when going from a state at t0 to a state at t with a measurement in-between at ti (t0
@@ProfessorMdoesScience Prof. does the state really "collapse" or just its representation in the space we are measuring?
I think, it is better to see this in terms of Fourier transform. Because when the position-space wave-function "collapses" (or more correctly, in my view, narrows or even better "is made to be narrower") under the external influence that we call "measurement of position", the momentum-space wave-function (its Fourier transform) actually widens (=uncertainty principle") and there is no collapse in that space or representation. In other words, "collapse" is a RELATIVE term. Relative to the observable and representation that we (as measuring agents) are working with and trying to measure.
Similarly "superposition" is also a relative term: Every state (e.g. a position eigen-state) is a superposition in terms of eigen-states of some other observable. It is usually said that "superpositions cannot be observed". To me, this seems a lack of deep understanding of the mathematics of QM. When we observe a particle localized to a small region (after a "position measurement") we are simultaneously also observing a very wide superposition in terms of momentum eigen-states. I think, to talk about "collapse" and "superposition" in absolute terms, is a bias of the physicist's mind.
Also the postulate or proposition that "measurement causes collapse of the wave-function" to me it seems that needs to be re-formulated backwards. That is, we want to make the wave-function narrow in some space (position, momentum, etc.), and then we orchestrate a physical process (that we call measurement) which exactly causes this narrowing in that particular representation or space (which also translates to a widening in the conjugate space and representation). In other words, this intentional narrowing of wave-function in some space (and the unavoidable, simultaneous widening in some other space) is the very definition of "measurement".
Also don't you think the evolution is "non-unitary upon measurement" only because we have not incorporated the measurement sub-system itself into the state. That is, time-evolution looks non-unitary (and we need to resort to projection, etc.) ONLY because we have chosen a smaller window to view only the particle (not the bigger system), but choosing a larger window (larger system's state) to incorporate both "the measured" and the "measuring agent" will restore the unitarity (thus for example, position measurement could be defined as intentional unitary narrowing [by whatever resolution we might want] of the joint position-space wave-function).
To me it seems, this view and recognition of biases (and subtle points) in understanding QM (and interpreting its math) makes QM much more intuitive and rational.
Just some of my thoughts as a careful student of physics. I would be honored if any of my possible misconceptions are pointed out, Sir. Thank you.
@@itsawonderfullife4802 If I understand your points correctly, I think I agree with you in that "collapse" refers specifically to the state of the system in the basis of the eigenstates of the observable we are measuring. From this point of view, a term like "change" may be more appropriate, as would take into account the "collapse" in this basis, but also whatever change happens in another basis. Superposition is again basis-dependent indeed. Regarding measurements, this is a tricky topic that many have discussed, an example of an interesting discussion is here: physics.stackexchange.com/questions/35648/if-i-go-to-the-church-of-the-greater-hilbert-space-can-i-have-unitary-collapse
@@itsawonderfullife4802*Because when the position-space wavefunction "collapses" (or more correctly, in my view, "narrows," or even better, "is made to be narrower") under the external influence that we call "measurement of position," the momentum-space wavefunction (its Fourier transform) actually widens (=uncertainty principle) and there is no collapse in that space or representation.*
Your view is incorrect, and the very conclusion that you draw concerning position-space versus momentum-space proves that your view is incorrect. You called the wavefunction collapse a "narrowing," failing to realize that this naming scheme is inappropriate, precisely because a "widening" also occurs in the Fourier transform of the representation used. This underpins a lack of understanding of the mathematics of wavefunction collapse. What you are failing to understand is that wavefunction collapse is representation-independent. For a given observable A, there is the eigenvalue equation Â|ψ) = λ·|ψ), where |ψ) denotes the state vector. Consider |ψ[α]) for all α to be the eigenbasis with respect to Â, so that Â|ψ[α]) = λ[α]·|ψ[α]), and |ψ) = Σ{c[α]·|ψ[α])}, where Σ denotes a Lebesgue sum. This gives us the decomposition of |ψ) as a superposition of the eigenbasis |ψ[α]). Wavefunction collapse simply refers to the non-unitary time evolution from |ψ) to |ψ[β]) for some β. It is improper to call this a narrowing or widening, since it is representation independent: all it amounts to is projecting the state vector to one of the basis vectors in the eigenbasis of Â, and this collapse happens when measuring the observable A.
*In other words, "collapse" is a RELATIVE term. Relative to the observable and representation that we (as measuring agents) are working with and trying to measure.*
No, this is incorrect. Collapse is representation-independent. When we measure the observable energy of a system, the system does not care whether our mathematical description of it is in the position space or momentum space. The collapse works the same way. Also, to say that it is relative to observable is nonsensical. Observables do not, in general, commute, so observables do not, in general, even share an eigenbasis with which we can speak of collapse. You can only consider collapse in the context of the measurement of a single observable, and this is the only context in which the notion is mathematically meaningful.
*Similarly, "superposition" is also a relative term: every state (e.g. position eigenstate) is a superposition of eigen-states of some other observable.*
You are mistaken. Position is not a fundamental observable to speak of, here, and the term "superposition" does not refer to any particular observable in particular. Also, what you are describing in your comment is a change of basis, the existence of a Jacobian between any two bases, but that is not what the word "superposition" means. What the word superposition refers to is the fact that any arbitrary state vector, whether an eigenstate of an observable or not, can be written as a superposition of _some_ basis, whether an eigenbasis or not. In this case, the observables are irrelevant for the notion of superposition, which is independently well-defined.
*It is usually said that "superpositions cannot be observed."*
No physicist has ever said this. The only people who say this are enthusiastic students who are nonetheless completely new to the study of quantum mechanics, and have no had any exposure to the topics beyond an extremely introductory-level discussion of certain intuitions.
*To me, this seems a lack of deep understanding of the mathematics of QM.*
You have no privilege to be saying this, considering your own ignorance of the mathematics. To be clear, I know that later in your comment, you admit to just being a student, which justifies the ignorance, but it is extremely arrogant and stupid to pretend that you understand the mathematics of QM better than expert physicists do, which is what you are doing here, especially when you also dare to attribute to them a saying that originated on the Internet among students, not physicists.
*When we observe a particle localized to a small region (after a "position measurement"), we are simultaneously also observing a very wide superposition in terms of momentum eigenstates.*
This is not simultaneous, because as you said, it occurs _after_ the position measurement. This scenario looks very different from a simultaneous measurement of two observables.
*I think, to talk about "collapse" and "superposition" in absolute terms, is a bias of the physicist's mind.*
Once again, this is extremely arrogant and insulting. A physicist understands the mathematics of QM better than you or I do, and a physicist understands that you have a bias towards a particular representation of the wavefunction, one which is not actually warranted, and which is obscuring and blockading your understanding of representation-independent quantum states.
*Also, the postulate or proposition that "measurement causes collapse of the wavefunction," to me it seems that needs to be reformulated backwards. That is, we want to make the wavefunction narrow in some space (position, momentum, etc.), and then we orchestrate a physical process (that we call measurement) which causes exactly this narrowing in that particular representation or space (which also translates to a widening in the conjugate space and representation).*
No, this is completely wrong. You are conflating "space" and "representation." The space of quantum states is a unique Hilbert state, characterized by the postulates, or dare I say, axioms, of quantum mechanics. A representation refers simply to the projection of the quantum state into the eigenstates of a particular observable. It is one thing to talk about |ψ), and it is another completely different thing to talk about (ψ[A, α]|ψ), which is the representation of the quantum state in the span of eigenstates of the observable A. The position representation is (r|ψ), not |ψ) itself. There is a video on the channel already explaining all of this. The way it works is that Â|ψ[A, α]) = λ[A, α]·|ψ[Α, α]), so the dual equation is the equation (ψ[A, α]|·λ[A, α]* = (ψ[A, α]|Â^t, where Â^t is the adjoint of Â. Therefore, λ[A, α]*·(ψ[Α, α]|ψ) = (ψ[A, α]|Â^t|ψ).
Αnyway, when we perform a measurement, we are not seeking to narrow anything on a particular representation, since collapse is representation-independent. Also, collapse is not something that we seek. Collapse is something that happens when we perform a measurement we seek. So the reformulation you speak of already makes no sense.
*In other words, this intentional narrowing of wavefunction in some space is the very definition of "measurement."*
No, because the collapse of the wavefunction is not something we intend for. It is merely an unavoidable consequence of what we are looking for. The idea behind collapse is simple. For a simplified example, let us consider the simplest non-trivial vector space every person is familiar with on some intuitive level or other: the 2-dimensional vector space R^2. Here, we can choose orthornormal vectors (with respect to the Euclidean inner product, the dot product), so that one basis vector is called "horizontal," which I shall denote |r), with "r" for "right," and the other is called "vertical," which I shall denote |u), with "u" for "up." Every vector |ψ) in this vector space can be written as a·|r) + b·|u) for some a, b in R. Now, suppose we have an observable A whose operator  has eigenstates |r) and |u). Then, according to the postulate, an inevitable consequence of trying to measure A is that we can only ever measure the eigenvalues r and u respectively. And if the eigenvalue r is measured, then this must be equivalent to the projection a·|r) + b·|u) -> |r), while if the eigenvalue u is measured, then this must be equivalent to the projection a·|r) + b·|u) -> |u). This is what collapse refers. This is what collapse is. But this collapse is not intended. The only thing that is intended is the measurement itself. We measure A, without a preconceived expectation that only r or u will ever come out. It is not as if there is anything that, in principle, is stopping (|r| + |u|)/2 from being the valid outcome of a measurement, or sqrt[(|r|^2 + |u|^2)/2] from being the outcome. But the reason it is not a valid outcome is due to this unintended yet inevitable collapse that occurs. That is what the postulates of QM are encoding.
@@angelmendez-rivera351 First you need to cool down, my friend. I wasn't trying to insult any physicists. Many practicing QM experts (particularly in the foundations of QM and QM optics) share these views, implicitly or explicitly. Not understanding other people's meaning and being aggressive in dismissal harms only oneself, especially in the long run.
"...superposition" does not refer to any particular observable in particular. "
"Collapse is representation-independent"
These very sentences suggest you have a serious and underlying problem with the very basics, especially when it comes to making cross connections in QM.
And (in contrast to Prof M.) you neither understood the meaning nor intention behind any of my original comments, my dogmatic friend (Even my "student" reference above: We are all "students" ;) ). Also some Fourier analysis bg. wouldn't hurt either.
So I would rather respectfully skip going through the rest of your comment.
Hopefully the 'Interaction picture' is coming soon.
Waiting for the interaction picture video...
We are hoping to get to it... but you may have noticed we've been struggling recently to keep a regular schedule, too busy with our "real" jobs...
Thabk you sir, very bery helpful
Where is the video for interaction picture? They say it is useful for perturbation is that true?
It is still not available. We are working on a number of videos, so it will hopefullly arrive soon... And yes, the interaction picture is useful when we have, for example, a time-dependent perturbation.
@@ProfessorMdoesScience Thank you, I will be waiting for it.
just commenting for the algorithm: 👍🏼👍🏼👍🏼
Thanks for your support! :)