# QM explanation on entanglement for spin and the original EPR experiment

But it does not give me a specific explanation on the math description how to prove these specifically for spin in the original EPR paradox:

And finally I think I understand logically where it says : "It remains only to show that Sx and Sz cannot simultaneously possess definite values in quantum mechanics. One may show in a straightforward manner that no possible vector can be an eigenvector of both matrices. More generally, one may use the fact that the operators do not commute,"

But I have not found anything on the mathematical formulation on how these can be shown. My questions are about the math formulation how to prove these.

Question:

1. Can somebody please help me how to show in QM math that "It remains only to show that Sx and Sz cannot simultaneously possess definite values in quantum mechanics."

2. And can somebody please help me with showing in QM math this specifically "One may show in a straightforward manner that no possible vector can be an eigenvector of both (Sx and Sz) matrices."

3. And can somebody please help me show in math how to "More generally, one may use the fact that the operators do not commute,"

• Possible duplicate: physics.stackexchange.com/q/10362/50583 – ACuriousMind Dec 5 '16 at 18:50
• OK I edited. That is a nice question on this site but what I did not find in that either is I am looking for the specific QM math for spins in the original EPR experiment (electron-positron, Alice, Bob etc.). I know if somebody is pro in QM then he can figure it out somehow there, but I am not really, just trying to learn, and if I could get a specific QM math explanation on the exact original EPR experiment in the wikipedia page, and the questions I raised, then I would understand it. – Árpád Szendrei Dec 5 '16 at 18:55
• The spin commutation relations are always the same no matter the circumstance. Nothing of what you're asking about is in any way specific to the EPR experiment. – ACuriousMind Dec 5 '16 at 18:56
• I know. It is specific to my amateurness. I am asking please for a specific math formulation on this specific (Sx Sz matrices cannot have a common eigenvektor) case because then I would understand. – Árpád Szendrei Dec 5 '16 at 19:01
• The "original EPR paradox" in the famous 1935 Physical Review paper of Einstein, Podolski, and Rosen was not formulated for the spin of two particles but for the location and momentum. – freecharly Dec 5 '16 at 20:00

A state in QM is given by a vector of an Hilbert space, in this case, of finite dimension. Observables are represented by Hermitian operators on that Hilbert space. The possible values that can be obtained after a measurement by such an observable operator $\mathcal{O}$ are given by its eigenvalues. Now, for a state to have a definite value, we want a state (vector of the Hilbert space $|x\rangle$) that is associated to a definite eigenvalue. Those vectors are called eigenvectors and satisfy

$\mathcal{O} |x\rangle = x |x\rangle$

This then becomes a problem in linear algebra. Your question really is about linear algebra. Given $S_x$ and $S_z$ as per the article, one can show that they cannot share an eigenvector by supposing the converse and arriving at a contradiction. We have (omitting the constants):

$S_x = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)\quad$ and $\quad S_z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$

Now suppose that $(x,y)$ is an eigenvector of $S_z$ and $S_x$. This implies that both

$(x,-y)=\alpha (x,y)\quad$ and $\quad (y,x)=\beta (x,y)$

So we have that

$x=\alpha x, \quad -y = \alpha y, \quad y = \beta x, \quad x=\beta y \implies x=0$ or $y=0$ which contradicts the last two equations unless $x=y=0$. In both cases we have the zero vector which cannot be a physical state (not normalizable). In general, finding eigenvectors of a matrix is really basic linear algebra and I suggest that you read up on the topic: https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors, as there are much more efficient technique than I what I used in this simple case. What you basically want to do for a matrix $A$ is to calculate the determinant of $A-\lambda Id$ and find the zeroes in the variable $\lambda$ of the resulting polynomial equation. This works because in the diagonal form, substracting $\lambda Id$ will leave one row with only zeroes leading to a null determinant. Now once the eigenvalues are found, you use them to obtain linear equations on the components of the eigenvector associated.

Finally, it is a result of linear algebra that non-commuting matrices cannot be simultaneously diagonalized, which means they do not share a common eigenbasis. This really means in this context that since: $[S_x,S_z]=S_x S_z - S_z S_x \neq 0,$ there will be some states that do not have definite value for both operators. As ACuriousMind noted, to go further requires the fact this pair of operators do not commute on any subspace of the Hilbert space they act on.

• thank you. Now this is an answer to 1. right? I have done linear algebra at the university, I graduated, but maybe I don't remember well any more since I am working as a comp programmer. Still can't believe that simply lin algebra answers it? so Sx and Sz cannot have an eigenvector. Can you please help me with how to write the converse and come to a conrtadiction in this specific case? – Árpád Szendrei Dec 5 '16 at 18:59
• Ohh OK so now I read "Non-commuting observables, in general, can have common eigenvectors, just not a common eigenbasis. The fact that the spin operators do not share a single eigenvector is specific to their commutation relation." so now I do not understand. I thought I got it to be as simple as Sx and Sz cannot have a common eigenvektor (I think I understand what an eigenvektor and eigenvalue is in 2D matrices visually, that is in the case of image transformation). But it would really help me if someone could help with the math for Sx and Sz. Because I suppose this is for 3D spins. – Árpád Szendrei Dec 5 '16 at 19:04
• I added a simple proof of the non-existence of simultaneous eigenvectors. – G. Bergeron Dec 5 '16 at 19:26
• OK I think now I understand this far: "Finally, it is a result of linear algebra that non-commuting matrices cannot be simultaneously diagonalized, which means they do not share a common eigenbasis. This really means in this context that since: [Sx,Sz]=SxSz−SzSx≠0," – Árpád Szendrei Dec 5 '16 at 19:29
• If no common eigenbasis exists, then some vectors will have non-zero components for the basis vectors where the eigenbasis are not equal. If I pick an eigenvector of the first operator with non-zero components where the eigenbasis are not equal, it will not have a definite eigenvalue for the second operator. – G. Bergeron Dec 5 '16 at 19:35