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I have read this:

http://en.wikipedia.org/wiki/EPR_paradox#Mathematical_formulation

and this: How does non-commutativity lead to uncertainty?

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,"

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Árpád Szendrei
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1 Answers1

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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.

G. Bergeron
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