Tensor product state spaces

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📚 State spaces allow us to describe each one of the degrees of freedom of a quantum particle. But what happens when we need to represent several degrees of freedom at the same time, for example three physical dimensions, orbital and spin degrees of freedom, or a system of multiple particles? Tensor product state spaces provide the mathematical tools to study these more complex systems, and in this video we learn how to extend the standard quantities of state spaces to tensor product state spaces. Some systems with multiple degrees of freedom can be described by a trivial combination of the individual components making up the whole system. But quantum mechanics is more interesting than this: some systems cannot be described by a trivial combination of the individual components, and this leads to entangled states.
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⏮️ BACKGROUND
Dirac notation in state space: • Dirac notation: state ...
Operators: • Operators in quantum m...
Representations: • Representations in qua...
⏭️ WHAT NEXT?
Eigenvalues and eigenstates in tensor product state spaces: • Eigenvalues and eigens...
Identical particles: • Identical quantum part...
Entangled states: [COMING SOON]
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Director and writer: BM
Producer and designer: MC

[@PaulWintz]

I've been struggling with understanding tensor products for a while because many sources I've been reading give them very abstractly. This video feels like it made them "click" by making it more concrete. Thank you!

[@ProfessorMdoesScience]

Glad it helped!

[@sedication]

The explanation of entangled states with the system of equations was so clear and well done and made it super clear just with basic Linear Algebra. Also, I liked how you hinted with the l x , y ,z > = l r > ket that when we're working in 3D, we were actually using tensor product without knowing it. Super well made, thanks!

[@ProfessorMdoesScience]

Thanks for the feedback, really glad you like it! :)

[@Impatient_Ape]

Very nicely done! Glad to see that you also addressed the notational variations since it's what most students will see outside of the classroom.

[@ProfessorMdoesScience]

Yes, that is a topic that often leads to confusion, so thought it important to include :)

[@adaircampos4240]

Your videos are helping me a lot in building knowledge towards my bachelor thesis. Thank you very much!

[@ProfessorMdoesScience]

Great to hear you find them helpful!

[@yonedelcastillo5206]

Amazing quality of the content of this video!! Thank you so much!!

[@ProfessorMdoesScience]

Glad you liked it! :)

[@syedrumman3920]

Really like these videos!! 😊

[@clasicus]

Genius lecturer! I am very proud that I watch his videos!

[@ProfessorMdoesScience]

Thanks for your support!

[@dutonic]

This was so clear and well done. When I dig into the pure math of tensor products, it's rather overwhelming. But this video really helped bring things together for the purposes of physics. Thanks :)

[@ProfessorMdoesScience]

Glad it was helpful!

[@_BhagavadGita]

You do a wonderful job. Thank you so much.

[@ProfessorMdoesScience]

Glad you like it!

[@abhishekjangid8953]

Your videos are very helpful in understanding the concepts very clearly. Thanks a lot for making videos on these topics of QM.

[@ProfessorMdoesScience]

Glad you like them! The next video we'll publish will actually be a follow-up of this one on how to calculate eigenvalues and eigenstates in tensor product states, mentioning it just in case you are interested! :)

[@abhishekjangid8953]

@@ProfessorMdoesScience Thanks, I will watch them also...😄

[@4youtubezgz533]

Very good video, thank you very much!

[@ProfessorMdoesScience]

Thanks, glad you liked it! :)

[@user-rz1bx9ic6l]

Very useful for a beginner!

[@ProfessorMdoesScience]

Glad you like it!

[@AnsImran]

Thanks a lot for this video! Impatiently waiting for a video on "Entangled States?" & "How to quantity entanglement"

[@ProfessorMdoesScience]

Glad you like it! Entanglement is definitely on our list, but not sure yet when we'll get there as the list is very long!

[@AnsImran]

@@ProfessorMdoesScience Ok thanks!

[@lorismopsas9398]

Amazing work!

[@ProfessorMdoesScience]

Thanks for your support! :)

[@cappy_bara]

Great video as always! Waiting for the video about entangled states ;)

[@ProfessorMdoesScience]

We hope to get to it soon!

[@richardthomas3577]

excellent video -- thanks!

[@ProfessorMdoesScience]

Glad you liked it! :)

[@josephusmiller2427]

Great video!

[@ProfessorMdoesScience]

Glad you like it!

[@Mhoikano]

Thank you !!

[@ProfessorMdoesScience]

Thanks for watching! :)

[@sayanmaji2845]

Thank you so much Sir. I understand very well. I actually was struggling with the concept of tensor product space. My physics professor often write the symbol without introducing us, what the tensor product is? When he tells adding of angular momenta, I was not properly able to understand what he tries to explain me. But , now I understand that, what he actually tried to tell me. Thank you once again sir to make my life simple. ❤❤Take love sir. Thank you

[@ProfessorMdoesScience]

Really glad to hear our approach helped you! :)

[@workerpowernow]

excellent explanation! thanks. Your videos are really helping refresh QM for me before graduate level classes. One question: are all your videos sorted into playlists? I'm knocking out the playlists one by one and want to make sure I don't miss anything

[@ProfessorMdoesScience]

In principle we do sort all our videos into playlists. Some videos may feature in more than one playlist. Others may temporarily be in no playlist (if we are just starting a new "topic") but will eventually join one. Moving forward we hope to create a website with the "recommended" way of following the videos (and with problems+solutions etc), but this will take some time...

[@workerpowernow]

@@ProfessorMdoesScience that sounds like a great system, thanks

[@arashedrisi40]

great videos and very informative. Thank You! Would you have a plan to make a video on Entangled states?

[@ProfessorMdoesScience]

We are hoping to, but we have many other topics we want to cover too so not sure when we'll be able to get there...

[@orionexploding1112]

It's possible to provide some motivation for representing the state space of two particles by their tensor product. (Although you could argue that no such motivation is required since the tensor product is the obvious choice.) To do this, we have to see the state vector of a single particle as consisting of the "square roots" of the corresponding entries in a probability distribution. (I'm using "square roots" extremely loosely here to mean any number multiplied by its conjugate. The difference between classical and quantum probability mainly arises from the fact that classical probability only allows a positive square root with a complex phase of zero.) The joint probability distribution of two independent random variables, A and B, P(a,b) = P(a)P(b), where "a" is a possible value of A and "b" is a possible value of B. Taking "square roots" of both sides, this is the rule for tensor multiplication of two state vectors. Entangled states correspond to A and B being dependent variables. It might also be worth pointing out that the inner product of two tensors is just the sum of the corresponding products of bra and ket components of the tensors. (Although maybe you did that and I missed it - these videos take several viewings!)

[@ProfessorMdoesScience]

Indeed I would say that the motivation is simply that the tensor product is the mathematical operation that allows us to describe what we need. There are various ways to see this, and to add another one, we can consider the sizes of the state spaces. Let us consider N1-dimensional state space V1 (with N1 basis vectors) and N2-dimensional state space V2 (with N2 basis vectors). To describe the combined quantum system, we need N1*N2 basis vectors, and this is given by the tensor product state space.

[@146fallon9]

Thank you very much. 🎉🎉

[@ProfessorMdoesScience]

Thanks for watching!

[@cochaviz]

I think the problem of entanglement would've been easier to understand if you came from a 'factorization' angle. While totally equivalent, explicitly mentioning not being able to factor out the tensor into the bases of the original state spaces makes the word 'entangled' way more appropriate. The complete system can then not be described by a simple linear combination of any number of states, and the 'particles' cannot be described separately from each other, entangling the system and its parts with one another. (This is at least my physical interpretation of what's happening, so if it's not -- or not very -- correct, please let me know!) Other than that, absolutely love the videos, I can't believe not more people are watching them! I provided the different playlists to my professor, so I hope he refers to them in future editions of the course! :)

[@ProfessorMdoesScience]

Thanks for the suggestion! We hope to look at entanglement and related topics in more detail in the future, and will certainly take in your suggestion about the best way to approach it. And thanks for your support! :)

[@angelmendez-rivera351]

You explained it incorrectly. Every tensor in the tensor product space can be written as a linear combination of the tensor products of the basis vectors. What entanglement is about is that it cannot be written as the tensor product of two linear combinations from the vector spaces in the tensor product space.

[@Upgradezz]

Why do we write the ket |r> as tensor product of x, y, z?

[@ProfessorMdoesScience]

Tensor products are the mathematical operation that allows us to combine state spaces (vector spaces) to build more complex state spaces. In this context, the ket that describes a particle moving in 3D can be built by combining kets living in the three dimensions making up the 3D space, and such a ket in general is the tensor product of x,y,z. Calling this tensor product |r> is just a shorthand, but for mathematical manipulations I would recommend sticking with the explicit tensor product of x,y,z. Hope this helps!

[@dutonic]

I'm working through Griffith's QM this semester and finding by experience that letting the tensor product be commutative is not producing the results I am expecting. I googled it and it appears that the tensor product is not commutative, even when we specify that |φ⟩ and |ψ⟩ are from different hilbert spaces. The only exception I can think of here is if |ψ⟩ and |φ⟩ are symmetric states (thus they can be swapped without changing the physics of the system). So I'm a bit lost at what 1:40 is communicating. Is this just saying that |φ⟩_2 ⊗ |ψ⟩_1 should be read mathematically as |ψ⟩_1 ⊗ |φ⟩_2, but we physicists are just too lazy to put them in the correct order every time? Would you be willing to clarify a bit here? Thanks so much!

[@ProfessorMdoesScience]

Essentially yes, the addition of the subindex identifying the subspace gives us this extra flexibility to move things around. And more generally, when using tensor product states we as physicists very often use many simplifying conventions in how to write them to make the formulas neater, with some discussion on this in our series on identical particles and second quantization. I hope this helps!

[@angelmendez-rivera351]

If the tensor product space you are working with is the space V ⊗ V for a vector space V, then in general, the tensors v ⊗ w and w ⊗ v are not equal, for v, w in V. However, if the tensor product space you are working with is the space V ⊗ W, where V ≠ W, then order does not matter. It does not matter, because when you write v ⊗ w or w ⊗ v, there is no confusion as to which of v and w is from V, and which of v and w is from W.

[@TheWingEmpire]

The linearity and distributive property you mentioned in 2:34 , are those similar to the property of vector spaces?

[@ProfessorMdoesScience]

There are many mathematical objects that obey linearity and distributivity, and indeed vectors in state space do :)

[@TheWingEmpire]

@@ProfessorMdoesScience thank you very much :)))

[@marcrindermann9482]

7:38 Since I've got a recommendation for this channel on another channel yesterday I've watched quite a lot of your videos. Consequently, I stumbled upon this little gem and naturally, I would like to see more about entangled states and looked in the description of the video for more information. The Information I got is that entangled states are coming soon. Since this video was released three years ago, I was wondering if you are still working on the video in particular or this channel in general?

[@ProfessorMdoesScience]

Thanks for watching! May we ask which other channel recommended us? And we are definitely still active, but unfortunately we cannot publish as regularly as we'd like to because we are both full time professors at Cambridge and this is only a side project... But we are exploring ways of making the channel more sustainable, so we hope to be able to increase our output in the near future!

[@marcrindermann9482]

@@ProfessorMdoesScience sure, @ParthGChannel recommended you guys in his video on Breaking Quantum Physics (But Not Really): Mixed States + Density Operators

[@qbtc]

Great video as always. Out of pure curiosity, are the presenters actors reading from a script or actual physicists? My guess is the latter. I ask because some channels use actors.

[@ProfessorMdoesScience]

We are definitely not actors :) We are both professor at the University of Cambridge.

[@zeusilver]

Do I have a entagled state only if the two single states can overlap each other?

[@ProfessorMdoesScience]

Do you mean spatial overlap? You can definitely think about entangled states without any notion of spatial overlap, for example considering entanglement between spin degrees of freedom. I hope this helps!

[@prathamhullamballi837]

Hey, wonderful video! I was just wondering if there's any rule on how operators (that cannot be written as a tensor product) act on product states; like if a product state evolves via a joint unitary, say CNOT gate, how does this operator act on a product state?

[@angelmendez-rivera351]

You can always write the operator as a linear combination of tensor products.

[@Maxwells_demon_0031]

Thanks a lot for all your amazing videos. This channel is a great resource for me. I think that the matrix analogue of the tensor product is the Kronecker product. Kronecker products do not commute. But the video mentions that in tensor products, order doesn't matter (i.e., tensor products commute). I am inclined to think that the order in tensor products matter. What's the gap in my understanding here? Please advise.

[@ProfessorMdoesScience]

I believe I may have already responded to your email, but for completeness: We consider two state spaces V1 and V2, and the tensor product space V=V1(x)V2. For every pair of states |phi>_1 in V1 and |chi>_2 in V2, then the tensor product |phi>_1(x)|chi>_2 belongs to V. The key is that, in our notation, we use a subindex “1” or “2” for the kets to identify which original state space they belong to. This makes order unnecessary, so we can write: |phi>_1(x) |chi>_2 or equivalently |chi>_2 (x) |phi>_1 Note that the “1” is always with the “phi” and the “2” is always with the “chi”. From this point of view, the order is included in the subindex. I hope this helps!

[@angelmendez-rivera351]

The order only matters when you tensor a vector space V with itself, but generally speaking, you never do this in the context of quantum theory. This only has relevance when working geometric vector spaces.

[@TheWingEmpire]

does the tensor product has something to do with "rank" of the matrix vectors? Is the tensor product similar or analogous to the term "outer product" ? and about the entangles states you mentioned...does is have to do something with "Quantum Entanglement" ? that Schrodinger's cat sort of thing? Very helpful video though! :)

[@ProfessorMdoesScience]

This is an interesting point. Essentially they are the same, but we make a convenient distinction in quantum mechanics to use them in different contexts. Roughly speaking, the outer product takes two kets |psi>, |phi> in state space V and generates an operator |psi>

[@TheWingEmpire]

@@ProfessorMdoesScience thank you very much. :)

[@narfwhals7843]

How is this distinct from the outer product? I had thought the terms were basically interchangeable. And it looks like ki>=psi> x phi> is the same as ki>=psi> is also an operator?

[@ProfessorMdoesScience]

The outer product of two kets is an operator, whereas the tensor product of two kets is another ket. So they are fundamentally different quantitites. I hope this helps!

[@narfwhals7843]

@@ProfessorMdoesScience but aren't they calculated the same way? I think the wiki article says "the outer product of two tensors is called their tensor product". Is the difference just one of interpretation?

[@ProfessorMdoesScience]

@@narfwhals7843 Mathematically you can think of both as arising from products of two Hilbert spaces, with the outer product being a product of a Hilbert space with itself and the tensor product of two distinct Hilbert spaces. But from a practical point of view in quantum physics, these two operations represent very different things.

[@narfwhals7843]

@@ProfessorMdoesScience Thank you very much for your continued answers :)

[@angelmendez-rivera351]

@@narfwhals7843​​⁠​​⁠ *How is this distinct from the outer product?* The outer product is an operation you perform on vectors and operators, not on vector spaces. The tensor product applies to vector spaces, and as a consequence, also the vectors in the vector spaces. *I had thought the terms were basically interchangeable.* They are not. They are very similar, and in many contexts, the distinction is not relevant, but in general, they are not interchangeable. Names in mathematics are not like names in the English language. If concepts have two different names in mathematics, then that is because the two concepts are different, even if the difference is imperceptible at your level of education. *And it looks like ki> = psi> ⊗ phi> is the same as ki> = psi> is also an operator?* No. ki> is a ket, as stated in the video. *but aren't they calculated the same way?* No. Where are you getting that they are calculated the same way? *I think the wiki article says "the outer product of two tensors is called their tensor product."* The Wiki is wrong. The Wiki is very confused, and in my experience, I find that Wikipedia is more of an obstacle than a tool when teaching about any topic that involves the word "tensor." You are better off relying on proper educational resources instead.

[@dien6]

Very good video, really well explained. I have two questions; you said the 3D particle state space is just the tensor product of its 1D components, this means (?) technically there exist entangled states in this space (i just assume this because of how you described tensor product). Do these states actually exist or not, and if not, is there a way to derive/intuitively see why they cannot exist or are they just assumed to be nonphysical? The second question relates to angular momentum; does it just not exist in this tensor product space? (Since angular momentum operator components do not commute, you cannot create a simultaneous eigenbasis so its impossible to measure the whole thing). Or is there a way of implementing angular momentum in this tensor product state space? Thanks in advance!

[@ProfessorMdoesScience]

Glad you like it! For your first question, I cannot see of a reason why such states could not exist. For the second question, when we study angular momentum we typically use the set of two commuting operators J^2 and Jz (or L^2 and Lz for orbital angular momentum). These do not form a complete set of commuting observables in 3D space, and in general you need to add more operators to fully characterize the space. A good example of this is the hydrogen atom, in which we add the Hamiltonian H and then the set of three operators (H,L^2,Lz) does form a complete set of commuting observables and is sufficient to fully characterize every state. We have actually just started a series on hydrogen, so you can learn more about how this comes about in those videos (we will add more videos to the series over the next few weeks/months): kzbin.info/aero/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa Hope this helps!

[@dien6]

Thank you for the reply! I should maybe clarify; for the first question, my confusion arises from what entangled states actually mean for measurements; in my understanding entangled states correspond to states of which you cannot independently measure obvervables of the original system. So say a 3D particle is in an entangled state in the position basis ( tensor product of coordinate position bases) and you measure the x coordinate, you would automatically know its y and z coordinates. Can this actually occur or am I missing something in the interpretation? The second answer is very clear! I know these questions are pretty nonstandard but i really appreciate you taking the time to answer them!

[@quantum4everyone]

I will take a stab at answering your second question. It probably is not the best thing to formulate as you want in terms of position space eigenstates, because they are very hard if not impossible to maintain in a superposition (they are not the eigenstates of any Hamiltonian, for example). But, what you are asking is if you create an entangled state, that has unique values for one quantity in the tensor product and you measure it then you can immediately infer the other quantities in the tensor product, the answer is yes, you can and do. But, in quantum mechanics, the real challenge is figuring out how to prepare interesting states and how to measure them. For example, just because an operator is self-adjoint does not mean that we have a way to actually measure it. Instead, we have to try to work out a method to do so, and we are quite limited in the things we can measure. Think about how one measures momentum without measuring position at the same time. There are a few ways, such as via the Doppler effect, but it is pretty hard to do that unless the object is moving at relativistic velocities. So, usually position is measured at the same time as momentum.

[@richardthomas3577]

Hi Dien, You asked if “3D particle state space is just the tensor product of its 1D components, this means (?) technically there exist entangled states in this space . . . Do these states actually exist or not, and if not, is there a way to derive/intuitively see why they cannot exist or are they just assumed to be nonphysical?” This is a really interesting question and made me think about a lot of stuff. Quite possibly all these thoughts are off target! I mention them to seek corrections. First, I do not think the tensor product space (|x> cross |y> cross |z>) is the same as what we usually think of a three-dimensional space. For one thing, each individual space here is one dimensional, N=1, and the dimension of the TPS |x> cross |y> cross |z> should equal N_1 x N_2 x N_3 =1x1x1=1. In other words, the tensor product space V is one dimensional, and has only one basis vector. Similarly, the TPS V must be linear with respect to scalar multiplication (see lecture at 2:00), so for example (2|x>) cross |y> cross |z> = |x> cross (2|y>) cross |z> = 2(|x>cross |y> cross |z>). This does not really match how x, y and z act in regular 3-D space. That said, I think in this particular case entangled states cannot exist, because all the individual state spaces are 1-D. Of course I could be wrong! Here is my reasoning. We have three individual state spaces here, each of dimension 1, so each has only one basis vector. We can write the general states in these individual spaces - call them V_1, V_2 and V_3, -- as follows: V_1: |psi> = Sum over i of c_i |u_i> = c_1|u_1> since there is only one value of i for a 1-D space. V_2: |phi> = Sum over j of d_j |v_j> = d_1|v_1> and V_3: |theta> = Sum over k of f_k |w_k> = f_1|w_1> An entangled state exists if a state in V = V_1 cross V_2 cross V_3 cannot be written as a tensor product of states in the individual spaces V_1, V_2, and V_3. I.e., we ask if there can be a state in V: = Sum over ijk of a_ijk {|u_i> cross |v_j> cross |w_k>} that cannot be written as a tensor product of individual space states: |TP form> = sum over ijk of c_i |u_i> cross d_j |v_j> cross f_k |w_k> But since i, j and k each can only equal 1, |chi> -- the state in V - can be written: = a_111 {|u_1> cross |v_1> cross |w_1>} Likewise, the general tensor product of individual space states can be written as: |TP form> = c_1 |u_1> cross d_1 |v_1> cross f_1 |w_1> By linearity with respect to scalar multiplication, |TP form> = (c_1)(d_1)(f_1) |u_1> cross |v_1> cross |w_1> So as long as we can find c_1, d_1 and f_1 such that their product = a_111, we do not have an entangled state. And this seems easy - e.g., pick c_1 = a_111 and d_1=f_1=1. Interestingly, it seems that there are an infinite number of ways that any general state in V can be expressed as a tensor product of individual space states. Sorry for the long response. It helps me to write things out sometimes - I share them in case it might help others, or at least elicit corrections that show my mistakes!

[@quantum4everyone]

@@richardthomas3577 Unfortunately, I believe you have a lot of things wrong. First off the space of solutions of the position eigenstates is uncountably infinite dimensional, not one dimensional. Each |x> is an independent basis vector in this space, and x can take any value from -infinity to infinity. So, the position space eigenstates for one dimension and for three dimensions are actually all uncountably infinite dimensional spaces. As for making entanglements, a simple entangled state would be integral dx |x>|x>|x>, corresponding to the states that live only on the diagonal subspace of the coordinate space. They cannot be constructed as a product state due to the integration, which can be thought of as a summation over the different states. The challenge, as I mentioned before, is that we do not know how to easily make such states. The mathematical details for how to work with these states can be quite complex, as they do not live in the L2 Hilbert space, but instead in the space of tempered distributions, which includes, but is larger than the L2 Hilbert space. It is easy to make mistakes when working with such states, so they need to be handled very carefully.

[@nomanahmadkhan7791]

Very good video indeed, but a difficult concept for me to digest. How can one write the tensor product in matrix notation? For example if V1 and V2 are two dimensional state spaces each and |psi> and |phi> belong to V1 and V2 respectively, then how can one write the tensor product |psi>*|phi> in matrix notation?

[@ProfessorMdoesScience]

In your example, the tensor product space of V1 and V2 would be 4-dimensional, and the matrix representation of states would then be 4-component vectors, and operators would correspond to 4x4 matrices. I hope this helps!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience Ok, that partially answers the question, let me rephrase. In V1 and V2, |psi> and |phi> are column vectors each having two matrix elements (e.g. [a1 a2] and [b1 b2], it is hard to write column vector in this text but hope you understand). How can I combine these column vectors to obtain another column vector having four matrix elements (say [c1 c2 c3 c4]) which represents tensor product of |psi> and |phi>?

[@ProfessorMdoesScience]

@@nomanahmadkhan7791 Thanks for the clarification! Let's say that the basis vectors in V1 are |u1> and |u2> (so that your state is a1|u1>+a2|u2>) and the basis vectors in V2 are |v1> and |v2> (so that your state is b1|v1>+b2|v2>). The basis states in the tensor product space V1(x)V2 are four: |u1>(x)|v1>, |u1>(x)|v2>, |u2>(x)|v1>, |u2>(x)|v2>. The tensor product state has components a1*b1 for the first basis state, a1*b2 for the second, and so on, so c1=a1*b1, c2=a1*b2, etc. I hope this helps!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience great! got it, thank you so much!

[@_BhagavadGita]

2:48

[@sandippaul468]

How come the tensor product of two (ket) vectors is another (ket) vector? Or am I saying it wrong? Is it producing another tensor?

[@ProfessorMdoesScience]

The outcome is indeed another ket, but the ket lives in a larger state space compared to the two original kets. I hope this helps!

[@sandippaul468]

@@ProfessorMdoesScience Thank you sir

[@alexanderbuchreiter5128]

Thanks a lot for your videos, they are really helpfull for my quantum mechanics exam. you do a great job. nevertheless, at Minute 1:40 you say that the tensor product of the two states is commutative, which doesn't hold when i'm multiplying it out. Are you sure of that, or am i making a mistake? Assume they are both two-dimensional: |psi> (x) |phi> = (psi_1*phi_1 , psi_1*phi_2 , psi_2*phi_1 , psi_2*phi_2) =! |phi> (x) |psi>

[@ProfessorMdoesScience]

Thanks for watching! To be precise, what I say is that the notation of the tensor product I use is flexible, in that the subindex outside the kets tells us to which of the original state spaces that ket belongs to. And this is simply convention. For example, let |psi> be in V1 and |phi> be in V2. Then, when we write the tensor product we write |psi>_1, where the subindex is reminding us that it is associated with V1, and we also write |phi>_2, where again the subindex reminds us that the state is associated with V2. As the subindices already identify the original state space, then what I say is that we can write the tensor product as |psi>_1(x)|phi>_2 or as |phi>_2(x)|psi>_1, in both cases it is clear that psi goes with V1 and phi with V2. An alternative convention is to omit the subindices, and then the *order* of the states in the tensor product is what identifies the original state spaces. For example, we could assume that the first entry always refers to states from V1 and the second to states from V2. In this notation, |psi>(x)|phi> would mean that psi comes from V1 and phi from V2. But in this notation, this is different to |phi>(x)|psi>, which means that phi comes from V1 and phi from V2. As to your expression, I am not sure I understand your notation. How are psi_1 and psi_2 related to psi? I realise that it is hard to write this clearly in a comment, but I hope this helps!

[@alexanderbuchreiter5128]

that doesn't really answer my question i think. what i meant is: assume: |psi>_1 is a state (vector) in space 1 |phi>_2 is a state (vector) in space 2 Both spaces 1 and 2 are each 2-dimensional, that means the states can be written as linear combination of the basis vectors of each space. |psi>_1 = (psi_1, psi_2) (psi_1 and psi_2 are the elements of the vector) |phi>_2 = (phi_1, phi_2 (phi_1 and phi_2 are the elements of the vector) then, the tensor product of these states (which gives us a 4-dimensional vector) is not commutative: |psi>_1 (x) |phi>_2 = (psi_1*phi_1 , psi_1*phi_2 , psi_2*phi_1, psi_2*phi_2) which is not equal to: |phi>_2 (x) |psi>_1 = (phi_1*psi_1 , phi_1*psi_2 , phi_2*psi_1, phi_2*psi_2) i don't really see how the tensorproduct can be flexible. Thank you for answering.

[@ProfessorMdoesScience]

The problem with your caculation is that when you are writing the different components in the 4-dimensional tensor product state space, you are not writing the basis states in the same order when you calculate the two tensor products. To be absolute clear, let me use a basis |u_i> for space 1, and |v_j> for space 2. Then, the two original vectors are: |psi>_1 = psi_1|u_1>_1 + psi_2|u_2>_1 |phi>_2 = phi_1|v_1>_2 + phi_2|v_2>_2 (in your case, you are not writing a basis explicitly, you only write the components with respect to a basis, but the basis is implicit). We then get: |psi>_1 (x) |phi>_2 = psi_1*phi_1|u_1>_1(x)|v_1>_2 +psi_1*phi_2|u_1>_1(x)|v_2>_2 +psi_2*phi_1|u_2>_1(x)|v_1>_2 +psi_2*phi_2|u_2>_1(x)|v_2>_2 and: |phi>_2 (x) |psi>_1 = phi_1*psi_1|v_1>_2(x)|u_1>_1 +phi_1*psi_2|v_1>_2(x)|u_2>_1 +phi_2*psi_1|v_2>_2(x)|u_1>_1 +phi_2*psi_2|v_2>_2(x)|u_2>_1 You can now see that in the 4-dimensional tensor product state space, the basis state which has u_1 and v_2 is always associated with the coefficient psi_1*phi_2, and the basis state which has u_2 and v_1 is always associated with the coefficient psi_2*phi_1. I hope this helps!

[@alexanderbuchreiter5128]

okay, thank you very much, i think i found my mistake. Thanks again for your videos, they are really helpful. I will definitely recommend them to my colleagues.

[@theopdiamond6226]

Do you have a source to read on the abstract tensor product ? It seems to me that everyone explains it differently on the internet and idk a source which gives an introduction to the "analysis" of tensor product. Atm it seems to me that all we are doing is overcomplicating our lives by adding a circle with a cross in between our kets and call that "a tensor product" however we don't explain the properties or what that circle-cross thing actually means mathematically which then brings up the question, why introduce it if we don't understand it mathematically?

[@ProfessorMdoesScience]

I am not sure exactly what you are looking for. In the video we provide the definition of a tensor product vector space. Specifically, a vector space V is called a tensor product state space of V1 and V2, written as V=V1xV2, if for every pair of vectors psi_1 in V1 and phi_2 in V2, there is a corresponding vector in V, which we write as psi_1xphi_2. Additionally, this state in V has to obey a list of properties (e.g. linear with respect to scalar multiplication). We then go on to explore some of the properties that arise from this definition, such as scalar products or basis vectors. I realise I have mostly repeated what is in the video, but I am not sure what you are looking for? We have the definition and a few properties that are useful for us. Are you perhaps interested in learning more properties that arise from the definition? More generally, this is also how we would work with a standard vector space: we first define it and then explore its properties. I hope this helps!

[@DivisionbyZer0]

Diedrik Aerts wrote extensively about it from a conceptual and logical point of view.

[@ProfessorMdoesScience]

@@DivisionbyZer0 Thanks for the pointer!

[@quantum4everyone]

You might want to look at some linear algebra texts that discuss this, because for finite sized vector spaces you really insert a vector from one space into the position of the element of a vector for another space and multiply them together. The end result is a vector whose dimension is the product of the dimensions of the original vectors, so the product notion is very useful to keep in one’s mind. But, because it isn’t a simple product but a generalization of the products of scalars, it needs a new notation for it. Hence the otimes.

[@angelmendez-rivera351]

*It seems to me that everyone explains it differently on the Internet and IDK a source which gives an introduction to the "analysis" of tensor product.* Can you give an example? As far as I can find, there really is only one description of the tensor product. Any discrepancies are entirely a matter of the language used, but this true for any concept you want explained. *ATM it seems to me that all we are doing is overcomplicating our lives by adding a circle with a cross in-between our kets and call that "a tensor product,"...* If we are overcomplicating it, then what alternatives would you propose for doing the work everyone does on this topic?Think about it carefully before replying. *...however we don't explain the properties or what that circle-cross thing actually means mathematically...* Did you watch the video? Because the properties were explained in the video. The product is bilinear, associative, and pseudo-commutative (see explanation in video), and there are no further restrictions applied: if two distinct tensor products are not related by the above properties, then the resulting tensors are not equal to each other. *...why introduce it if we don't understand it mathematically?* What makes you think we do not understand it?

[@shayanmoosavi9139]

Amazing video. I have a question, first you said that the order doesn't matter in a tensor product and at the end of the video you said order is important. Also when I looked up tensor product on the internet it seems that tensor product is not commutative. Am I missing something here? Is the tensor product in physics different from mathematics?

[@ProfessorMdoesScience]

Glad you like it! Mathematically, the tensor product may be commutative or non-commutative depending on the commutativity of the elements. However, what I say in the video is to do with notation. In the notation I use throughout most of the video, then the kets have a subindex (outside them) that indicates to which original state space they belong to. For example, the tensor product |u_i>_1(x)|v_j>_2 is telling us that the state u_i belongs to the original state space V1 (as the ket |u_i> has a subindex "1" outside) and that state v_j belongs to the original state space V2 (for the same reason). With this very explicit notation, then |u_i>_1(x)|v_j>_2 and |v_j>_2(x)|u_i>_1 are equivalent: in both cases state u_i is associated with V1 and state v_j is associated with V2. This notation is always clear, but it is a little cumbersome to use. Therefore, at the end of the video I go over alternative simpler notations that people frequently use to describe tensor products. One of these alternative notations is one in which the subindex outside the kets that identifies the original state space to which the states belong is omitted. When that happens, then we need a different convention to identify the state spaces to which the kets belong. What is done then, it that it is agreed that the *order* of the terms implicitly indicates to which state space the ket belongs. For example, in the notation |u_i>|v_j>, we would implicitly understand that u_i belongs to V1 because it is in the first position, and v_j belongs to V2 because it is in the second position. Put another way, this notation would be equivalent to |u_i>_1(x)|v_j>_2 (and to |v_j>_2(x)|u_i>_1). However, in this simplified notation, |v_j>|u_i> is a *different* state because state v_j belongs to V1 as it is in the first position, and states u_i belongs to V2 as it is in the second position. Put another way, this notation would be equivalent to |u_i>_2(x)|v_j>_1 (and to |v_j>_1(x)|u_i>_2). Overall, it all boils down to notation and convention. I hope this helps!

[@shayanmoosavi9139]

@@ProfessorMdoesScience thanks for the clarification. I definitely need to dig a little deeper in this subject. however the formal mathematics of it is intimidating to me as it's too abstract for me. Do you have some tips on how not to get intimidated by very abstract math?

[@ProfessorMdoesScience]

I would say that a good way to approach very abstract maths is to look at many examples of that math that relate to topics with which you are more comfortable with. This typically provides an easier route in. For the specific example of tensor products in quantum mechanics, they are used, for example, for systems of many identical particles, but that is a rather advanced topic. An intermediate step you could take would be to look at simple 2 and 3-dimensional systems, such as the 2 and 3-dimensional infinite square well, but re-written in the language of tensor products. We actually hope to publish a few videos on these simpler examples, precisely for the reason to provide more accessible examples of using tensor products.

[@shayanmoosavi9139]

@@ProfessorMdoesScience thank you so much for your answer. I'm looking forward to those videos.

[@mayaiveljic7975]

Hi, Do you give Private Instruction?

[@ProfessorMdoesScience]

We don't do this at the moment, although we are preparing extra material to go with the channel. For more details, please email us using our contact email that you can find in the "about" page of the channel.

[@user-os9np1fg7j]

Hi, Thanks a lot. I'm still find it a bit difficult to understand the connection between examples and math. let's imagine the x,y,z tensor product. The dimensions of each of them is 1 so according to math the tensor product dimension should also be 1, but it doesn't make sense...

[@ProfessorMdoesScience]

What dimension are you talking about here? If |x> is a state describing a position eigenstate, then it actually lives in an infinite dimensional state space. Its tensor product with |y> and |z> would lead to an overall infinite dimensional state space as each of |y> and |z> also live in such a space. I hope this helps!

[@user-os9np1fg7j]

Thanks. But the dimensions of the new space x,y,z is 3, no?

[@ProfessorMdoesScience]

@@user-os9np1fg7j Ok, these are two different spaces. (x,y,z) are indeed coordinates in 3D space, and for example we may describe the position of a particle in this 3D space. But |x>, |y>, and |z> (and their tensor product) represent quantum states that live in an infinite dimensional vector space (that is also complex). In quantum mechanics you can measure the coordinates (x,y,z) of a particle in 3D space (or its probability) from an arbitrary quantum state |psi>, with the latter living in an infinte dimensional state space. If these concepts are a bit unclear, I recommend you first take a look at our series on "the postulates of quantum mechanics": kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb I hope this helps!

[@user-os9np1fg7j]

@@ProfessorMdoesScience Thanks a lot. Now it's much clearer ☺

[@user-cr4yh9yw2s]

where is the video about the entangled state XD

[@ProfessorMdoesScience]

Not ready yet! We have many many things we want to cover and not enough time to do them as soon as we'd like! We are really hoping to cover entanglement at some point in the nearish future :)

[@thomasandersen4009]

Speaks way too fast.

[@ProfessorMdoesScience]

Other people have also mentioned this, and in the more recent videos we try to speak slower. However, note that KZbin has a "speed" functionality which allows you to slow down the video speed, hopefully that should help!