Projection operators in quantum mechanics

[@Lyxtwa]

I was reading Griffiths explanation of the projection operator and I had no idea how the inner product was moved in front of the ket. It was a very simple realization, but you helped me realize that any inner product is a scalar which can trivially be moved in front of the ket and probably saved me hours of confusion. Thank you so much.

[@ProfessorMdoesScience]

Glad it was helpful!

[@anthonycortez975]

What a gem!! Glad I found this channel! I'll be back.

[@ProfessorMdoesScience]

Thanks! Looking forward to having you back! :)

[@zonglangfrancis9794]

Thanks a lot for this vivid description of the projection operator.

[@ProfessorMdoesScience]

Glad you like it and thanks for your support!

[@supergravity66]

Another presentation that displays clarity, completeness and accuracy! Keep it up!

[@ProfessorMdoesScience]

Thanks for your support!

[@musictoonz]

This is an amazing video than others available on youtube. Thank you so much for this video

[@HankGussman]

It is an incredible series indeed. Best on youtube so far. Hope the series runs through full course of Quantum Mechanics.

[@hollywoodbanayad7755]

What a rich video, thank you so much for sharing

[@ProfessorMdoesScience]

Glad you like it!

[@mehanaziqbal5434]

You’re a lifesaver. Can’t thank you enough for these great videos😊. Can you please make videos on Electrodynamics?😢

[@ProfessorMdoesScience]

Glad you like the videos, and thanks for the suggestion! We are hoping to expand beyond quantum mechanics in future videos, but it will probably take a while before we get there...

[@attilauhljar3636]

Super clear and easy to follow explanation, very helpful, thanks. Keep up the god work!

[@attilauhljar3636]

good work, I mean.. :)

[@ProfessorMdoesScience]

Thanks, glad you like it! :)

[@jayjain1033]

Thank you so much, I didn't know QM could be taught this nicely

[@ProfessorMdoesScience]

Glad you like it!

[@nomanahmadkhan7791]

I have few questions. 1- I can see intuitively that eigenvalues of P are 0 and 1. But let me re-write the equation at 4:28 lambda|lambda>=c|psi> lambda=c lambda c=c lambda=1 How do we get lambda=0 exactly? 2- Projection operator is hermitian and it has two eigenvalues as well as two eigenstates. So its matrix form in the basis of its eigenstates is a 2*2 diagonal matrix with diagonal entries 1 and 0, am I right? 3- Spin also has two eigenvalues 0 and 1, so is it also a projection operator? (Anxiously waiting for your video on spin!) 4- Since basis of a state space is not unique so can there be more than one pairs of complementary spaces?

[@ProfessorMdoesScience]

Here are some thoughts: 1. Are you happy with the derivation up to c=0? If so, then from the equation c|psi>=lambda|lambda>, we must in general have that lambda=0. 2. First a clarification: a Hermitian operator is one for which its adjoint is equal to itself. For this to be true you need a stronger requirement than the operator having the same number of eigenvalues as eigenstates. Also note that the projection operator may have many more than two eigenstates. For example, if you have an N-dimensional state space, you can project on a single direction, and then you get N-1 perpendicular directions to it, all of which would be degenerate eigenstates of the projection operator with eigenvalue 0. Having said this, a matrix representation of the projection operator in its own basis would simply have 1s and 0s in the diagonal entries. 3. Spin for spin-1/2 particles is described in a 2-dimensional state space (video hopefully coming soon!), but note my answer above about the projection operator. Spin is an angular momentum operator, not a projection operator. 4. The complementary subspaces for a given projection operator are fixed, but your description of them can change depending on your basis choice. I hope this helps!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience Thank you so much, I missed two key concepts, first physical significance of eigenstate and second the degeneracy due to which questions arose, thanks again.

[@nourelhudazuraiki5400]

Thank you so much. This was very helpful. Just why is the projection operator for a ket is the outer product of the ket with itself?

[@ProfessorMdoesScience]

Take a look at the analogy with a Euclidean 2D vector which may help: kzbin.info/www/bejne/g2q5ZZuept6lgbM More generally, imagine you have a state |phi> and you want to project it on state |psi>. This means that after the projection your state will be in the "direction" of |psi>, and its length will be the component of |phi> along |psi>. The outer product |psi> direction, and the "

[@enricolucarelli816]

You have proved that the projection operator is Hermitian. You also proved in another video that any Hermitian operator has always n eigenvectors with all its eigenvalues real, and that this set of n eigenvectors form a full orthonormal base of the vector space.Thus, I thing that the interpretation I understand of your proof regarding the eigenvectors of a projection operator is to say that: the protection operator on |psi> has one eigenvector which is the actual vector |psi>, and the remaining n-1 eigenvectors are necessarily orthogonal to |psi>. The solution lambda = 0 represents the n-1 eigenvectors that are perpendicular to |psi>, but they are to be found by the classical procedure. Please correct me if I am mistaken.

[@ProfessorMdoesScience]

You are of course absolutely correct, and this is exactly what we mean to say in the video. Looking at the discussion again, I can see how our notation can lead to confusion. To summarise what you just wrote: 1. The projection operator |psi>. 2. The projection operator has n-1 eigenstates with eigenvalue 0, and the eigenstates are any set of n-1 states in the n-1 dimensional space perpendicular to |psi>. I think what we "say" in the video in consistent with this. However, in what we write we label the eigenstates |lambda> by the corresponding eigenvalue lambda. When lambda=0, then the eigenstate in this notation would be |0>. We never explicitly write this down, but it would certainly follow from our notation. This is indeed confusing. We should have been more careful and clarify that |lambda> when lambda=0 refers to the possible n-1 states perpendicular to |psi>. We do explicitly say this, but overall I agree that our notation could have been better. Thanks for pointing this out, and I hope this helps!

[@enricolucarelli816]

Thank you!👏👏👏👏👏

[@qubit597]

Good explanation, easy to understand. Thanks sir!👍👍👍

[@ProfessorMdoesScience]

Glad you like it!

[@paulbk2322]

Thanks for the lucid explanation

[@ProfessorMdoesScience]

Glad you ilke it! :)

[@badanjacoub4750]

Really helpful thank u

[@ProfessorMdoesScience]

Glad you like it!

[@sathirakatugaha974]

Incredible explanation

[@ProfessorMdoesScience]

Glad you liked it!

[@yashodabhattarai4206]

This is very helpful video 👏 Thank you very much 👌

[@ProfessorMdoesScience]

Glad it was helpful! :)

[@syedanaushabinzakirkhan20p50]

What about the operator (X+iY) where X and Y are Pauli spin operators. It's not a projection operator as (X+iY)²=0, it has real eigenvalue, but isn't a Hermitian operator as (X+iY)†≠(X+iY). I'm not sure if it's unitary (I don't think it is). Then what kind of operator is it? Is it even a valid operator?

[@ProfessorMdoesScience]

The combination X+iY looks proportional to the raising operator, which allows us you to change the spin m quantum number by +1. In the case of spin 1/2 particles, there are only two possible values of m, so you terminatethe series after applying the operator twice. We haven't yet covered this operator for spin 1/2 particles (although working on it and it should be out soon!). However, we have already covered the general case, of which spin 1/2 is just a particular example. You can find all the details about these operators for the general case here: kzbin.info/www/bejne/r3jZl6SIm9xkeKc And you can find the full series on angular momentum here: kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI I hope this helps!

[@paulbk2322]

Brilliant! No question about that💞

[@ProfessorMdoesScience]

Glad you like it!

[@tomaskubalik1952]

1) Do we know before measurement onto which subspace will the vector (quantum state) from V be projected or do we just know the probability that the vector will be projected onto subspace V1 and probability it will be projected onto V0? 2) Do I get identity operator if I put together all projection operators? 3) Can we use identity operator to project vector (quantum state) from V onto itself? Can V be its own subspace? Thanks

[@ProfessorMdoesScience]

All we can know before a measurement is the probability. We discuss the relation between projection operators and measurements in this video: kzbin.info/www/bejne/pZWvqIiOgL5jgNU For your second question, yes, the identity operator can also be thought of as a projection operator to the full space. I hope this helps!

[@tomaskubalik1952]

​@@ProfessorMdoesScience I understand that there is some probability that the measurement will result in a projection onto one of the subspaces. But what is the probability that the vector will be projected onto itself? Will it be alpha times aplha star plus beta times beta star? Thank you for your patience

[@ProfessorMdoesScience]

@@tomaskubalik1952 After a measurement, the state will only be projected onto itself if it was already in an eigenstate of the operator you are measuring before the measurement. And in this case, the probability of getting that state after measurement is 1. I hope this helps!

[@tomaskubalik1952]

@@ProfessorMdoesScience But if it wasn´t in an eigenstate and I would use an identity operator to project a vector onto itsef? (I mean if it was an eigenstate of identity operator). I'm wondering if there can ever be a projection into full space or is it always a projection onto one of the basis vectors in one of the subspaces? Then it is understandable that a vector in full space can be at the same time an eigenvector in one of the subspaces.

[@tomaskubalik1952]

@@ProfessorMdoesScience Is identity operator applied to |ψ 〉equal to (P+ (1−P)) applied to |ψ 〉?

[@williamberquist3923]

at 9:30, is the projective operator P_m or P_n? It seems like you are saying P_n and it makes sense it would be P_n given we are working on an n-dimensional subspace. However, it looks like you wrote P_m. Is that just a fancy cursive n?

[@ProfessorMdoesScience]

It is P_n, I write "n" with two "bridges" and "m" with three bridges... I hope this is clear!

[@vladimirkolovrat2846]

my question too, thanks for asking!!

[@aidenigelson9826]

Greetings. May I ask how we can project the system state from a 3 basis subspace to a 5 basis subspace and vice versa? I am trying to understand how mathematically we can (if possible) do the projection of the system state in a sequential incompatible observable projection, where we have two incompatible states, one is in 3 basis, the other in 5. How can we project the state sequentially from one onto the other? I understand the mathematics for the same dimension scenario, but wish to be sure regarding the generalized approach of any two arbitrary dimension subspace observable. I would be immensely grateful for any guidance

[@ProfessorMdoesScience]

When you project a state on a particular subspace, you only keep the components of that state along the subspace of interest, and remove all other components. If I understand your question correctly, you cannot project a state that is spanned by three basis vectors into one that is spanned by five if the basis is the same in both cases. I hope this helps!

[@jsconventschooldallaobraso6222]

Sir explain sign used in this method easily or make another video

[@ProfessorMdoesScience]

What do you refer to specifically when you say "sign used in this method"?

[@rishabhkumarsrivastav8169]

Great , but your speed is very fast .. Plz slow your speed... Other all thing perfect 👏👏👏

[@ProfessorMdoesScience]

Thanks for the feedback!

[@enricolucarelli816]

I have the same problem, but I don’t see it as your speed being to fast, I see it as my brain being to slow! 😁 Seriously, what I do is I adjust the video speed to 0,75, and this is perfect pace for me the first time a watch them. Then, for later reviews, I watch them at normal speed. So no worries! I think your speed is no problem at all, at least while seen through KZbin! 🤗

[@gowrissshanker9109]

Respected sir 🌟, can classical electromagnetic theory exist without imaginary number?,is complex number just an convenient tool for classical em theory? Can quantum mechanics exist without imaginary number?? Thank you sir 🌟

[@ProfessorMdoesScience]

Quantum mechanics needs complex numbers: the state space is a complex vector space. For classical physics, what exactly do you have in mind?

[@sayanjitb]

Can the projection operator be a Unitary operator?

[@ProfessorMdoesScience]

In general no. A projection operator P is idempotent, which means P^2=P. It is also Hermitian, which means P^dagger = P. If P was unitary, then by definition it would imply that P^dagger = P-1. As P is Hermitian this would imply P-1=P. This in turn would imply that P^2=1, so that the only projection operator that is unitary is the identity operator P=1.

[@sayanjitb]

@@ProfessorMdoesScience I understood. Thank you so much indeed!

[@ProfessorMdoesScience]

We have now published a video on unitary operators, you can check it out here: kzbin.info/www/bejne/mJKshWl-lsaMq7M

[@sayanjitb]

@@ProfessorMdoesScience Thank you so much indeed

[@sayanjitb]

I am a subscriber to your channel from the first place. So I get every update as your team posts. I am very privileged that you personally notified me of the latest video update. Thank you from the bottom of my heart.

[@jzl4646]

This was an amazing video, thank you so much! I understood a lot more leaving than coming =] I feel like this might be a basic question, but intuitively, why can any arbitrary vector be represented as a sum of parallel/perpedicular vectors to another vector?

[@ProfessorMdoesScience]

This is the same in standard vector spaces that you may be more familiar with, say a 2D or a 3D real vector space. For a given vector, you can always define the direction along it, and then build all other orthogonal vectors to it, of which you'll have as many as are needed to fill up the full dimension of the space. I hope this helps!

[@prikarsartam]

Wow, your explanations are very nice. Why don't you make a lecture note of yourself, maybe it'd help thousands.

[@ProfessorMdoesScience]

Thanks! We do plan to expand the type of material we share (problems, etc), but it will take some time to get there...

[@prikarsartam]

@@ProfessorMdoesScience that'll be great, waiting for that!

[@schmetterling4477]

It is amazing how you can go through all the math without saying anything about the actual physics of it. Do you really assume that students, let alone laymen, can gather from the math that a projection operator is the mathematical description of an actual physical process that removes (or adds) energy, momentum, angular momentum etc. from (to) the system? That is not my experience. I don't know how many times I came across people who told me that the Copenhagen Interpretation is bullshit because of this part of the formalism. They really think that these are just some magical "assumptions" that can not be actually found in nature. I would suggest discuss with students that each such operator is the physical description of an actual external system that can act as source or sink of these conserved quantities before you even start getting into the math. That would remove so much confusion about QM...

[@ProfessorMdoesScience]

Thanks for the feedback! This specific video is aimed at explaining the mathematical details behind the projection operator. We use it in a number of other videos, and hope to do so in even more videos, but we are not quite there yet!

[@schmetterling4477]

@@ProfessorMdoesScience I understand that. The problem is that we have plenty of such mathematical videos and almost none that connect the formalism to actual physics. In classical mechanics everybody understands that if an object changes its motion there has to be a force that leads to that change. Moreover, we teach that all such forces must also come with a reaction force on the system that exerts that force. That is the entire point of teaching classical mechanics: to motivate that the world is a balanced play of energy, momentum and angular momentum conservation. Non-relativistic quantum mechanics violates all of that because the theory can not describe the back-reaction of "the particle" on the system that created the potential in our Hamiltonian. Neither do we explain where the momentum/angular momentum change from the measurement process comes from. These all seem to be "magical" quantum effects to the novice, even though they don't exist in nature, at all. These are all defects of non-relativistic quantization procedures that we do not explain correctly. I have gone through a very similar style of presentation of the theory in my QM 101 class and I can assure you that I was utterly confused for a very long time. That confusion wasn't lifted until I understood the relativistic theory somewhat. I don't think it is necessary to leave the student hanging in this limbo. It is easy enough to draw a system diagram and to say what really happens and why we can't describe the entire process in the non-relativistic theory. And with that Copenhagen becomes a perfectly natural framework rather than a "shut up and calculate" formalism. Now, don't get me wrong, you didn't invent this style of QM 101 teaching and I wasn't blaming you for it. You are basically just repeating what you were taught and your teachers have been repeating the same material at least for two or three generations since the late 1920s or early1930s. It is nevertheless incomplete, even if it has been established as the standard way of teaching the material. End of rant.

[@kaushikgupta1410]

Why are we taking eigenvalue as zero?

[@ProfessorMdoesScience]

We show that the only possible eigenvalues of the projection operator are 0 or 1. Is your question about the proof of this?

[@kaushikgupta1410]

@@ProfessorMdoesScience yes sir. is it possible to have zero as an eigenvalue?

[@ProfessorMdoesScience]

Yes, it is possible, and in the case of the projection operator, any state that is orthogonal to the state onto which we are projecting is an eigenstate with zero eigenvalue. I hope this helps!

[@kaushikgupta1410]

@@ProfessorMdoesScience yes, crystal clear. Thx for being so amazing :)

[@danielblumowski34]

@@ProfessorMdoesScience why do they have to be orthogonal? couldn't they be in any "orientation" if lambda is equal to 0?

[@amaljeevk3950]

❤

[@Godakuri]

If the cross of a vector with itself is zero, how does the projection operator even exist

[@ProfessorMdoesScience]

Not sure I follow your argument, could you please expand?

[@Godakuri]

@@ProfessorMdoesScience Oh let me edit it. I used the wrong word. I've always been taught that the outer product of a vector with itself is zero. I don't understand how the projection operator even exist

[@ProfessorMdoesScience]

@@Godakuri The outer product of a vector with itself is an operator, and in general it is not zero. We first introduce outer products as operators in this video: kzbin.info/www/bejne/pn-pn5Rtr7-Vnac I hope this helps!

[@Godakuri]

@@ProfessorMdoesScience Alright I'll check it out