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I would like to refresh my mind on eigenvalues/eigenvector of tensor product of operator (for Quantum Mechanics purposes).

Consider two hermitic matrices $A$ and $B$ living in Hilbert spaces of finite dimensions $N_A$ and $N_B$ respectively.

I would like to know the most general properties about the eigenvalues of the operator $U=A \otimes B$.

Let's assume I know the spectrum of $A$ and $B$:

$$A |a^i\rangle = a |a^i\rangle$$ $$B |b^j\rangle = b |b^j\rangle$$

Where the indices $i$ and $j$ are here to take in account possible degeneracy in the eigenvalues.


Proposition 1

A possible eigenbasis of $U$ is the set of all $|a^i\rangle \otimes |b^j\rangle$ with eigenvalues $ab$.

Proof:

$|a^i\rangle \otimes |b^j\rangle$ is an eigenvector of $U$ with eigenvalue $ab$.

And as I have $N_A$ orthogonal vectors $|a^i\rangle$ and $N_B$ orthogonal vectors $|b^j\rangle$, I have found $N_A*N_B$ orthogonal vectors that are eigenstates of $U$.


However, there might have quantum state of $U$ that are not of the form of a tensor product.

Proposition 2

$U=A \otimes B$ admits one degenerated eigenvalue if and only if there exist eigenvectors of $U$ that are not a tensor product.

Proof:

Necessary condition:

As $\{ |a^i\rangle \otimes |b^j\rangle \}$ are basis of eigenstate of $U$, and there is a degenerated eigenvalue of $U$, then there exist $(a,b) \neq (a',b')$ such that $|a\rangle \otimes |b\rangle$ and $|a'\rangle \otimes |b'\rangle$ are eigenstates of $U$ with the same eigenvalue. Thus $|a\rangle \otimes |b\rangle + |a'\rangle \otimes |b'\rangle$ also. Then I found an eigenstate of $U$ that is not a product state.

Sufficient condition:

If there is an eigenvector of $U$ that is not a tensor product, then it must be a linear combination of different $|a^i\rangle \otimes |b^j \rangle$ as they diagonalise $U$. And if a linear combination of eigenvector is an eigenvector, then the two initial eigenvector must have the same eigenvalue.

Then, $U$ has degenerated eigenvalues.

Do you agree ?

Especially with my second proposition (I am quite confident about the first one).

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I use my notations.

i) Let $A\in M_n,B\in M_m$ be diagonalizable matrices over $\mathbb{C}$. Let $(u_i)_{i\leq n},(v_j)_{j\leq m})$ be bases of eigenvectors associated to the eigenvalues $spectrum(A)=(\lambda_i),spectrum(B)=(\mu_j)$. Then $(u_i\otimes v_j)_{i,j}$ is a basis of eigenvector of $A\otimes B$ because $(A\otimes B)(u_i\otimes v_j)=A(u_i)\otimes B(v_j)=\lambda_i\mu_j u_i\otimes v_j$.

ii) Assume that $\tau=\lambda_i\mu_j=\lambda_k\mu_l$ where $(i,j)\not= (k,l)$.

Then $(A\otimes B)(u_i\otimes v_j+u_k\otimes v_l)=\tau(u_i\otimes v_j+u_k\otimes v_l)$. Assume that $u_i\otimes v_j+u_k\otimes v_l$ is a tensor product $a\otimes b$; then, $ab^T$, the associated $n\times m$ matrix has rank $1$ and is the sum of two matrices of rank $1$: $u_i{v_j}^T+u_k{v_l}^T$ (one has the same result for the transposes). This can only happen if the images are the same, that is $span(u_i)=span(u_k)$ or, in the same way, if $span(v_j)=span(v_l)$; in other words, $u_i\otimes v_j+u_k\otimes v_l$ is a tensor product iff $i=k,j\not=l$ or $j=l,i\not= k$ iff $\lambda_i$ is a multiple eigenvalue of $A$ or $\mu_j$ is a multiple eigenvalue of $B$.

iii) Let $z$ be an eigenvector of $A\otimes B$ that is not a tensor product. Then $z=\sum_{(i,j)\in K}z_{i,j}u_i\otimes v_j$ where $z_{i,j}\not= 0$. There is $\tau$ s.t.

$(A\otimes B)(z)=\sum_{i,j} z_{i,j}\lambda_i\mu_j u_i\otimes v_j=\sum_{i,j}\tau z_{i,j}u_i\otimes v_j$, that implies

for every $(i,j)\in K$, $\tau=\lambda_i\mu_j$, and $A\otimes B$ has a multiple eigenvalue.

$\textbf{Conclusion.}$. Finally, your equivalence is valid when $A,B$ have simple eigenvalues.

For example, for $A=diag(1,2),B=I_2$, then all the eigenvectors of $A\otimes B$ are tensor products.

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  • $\begingroup$ To the OP. Do you read the answers ? $\endgroup$
    – user91684
    Sep 4 '19 at 19:16
  • $\begingroup$ Sorry, I always read them but often with some delay. I answer you friday at the latest. Thank you very much. $\endgroup$
    – StarBucK
    Sep 4 '19 at 22:44
  • $\begingroup$ Oh ok I think I see my mistake for the necessary condition given your proof. If $(a,b) \neq (a',b')$ my vector $|a\rangle | b \rangle$ and $|a'\rangle | b' \rangle$ will indeed be different, but their sum might still be a tensor product (for example if $a=a'$). Do you agree it is the exact place where my proof failed for the necessary condition ? $\endgroup$
    – StarBucK
    Sep 6 '19 at 19:45

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