While thinking about what it means for observables to be simultaneously measurable in quantum mechanics I came up with the following concepts, which I will call "linearly indexed" versions of standard linear algebra concepts.

  • A $W$-indexed linear map $L: V \to V$ is a linear map $W \otimes V \to W \otimes V$.
  • A $W$-indexed eigenspace for $L$ is a subspace $V'$ of $V$ such that $L = \lambda \otimes I$ on $W \otimes V'$.
  • $\lambda : W \to W$ is called the $W$-indexed eigenvalue of $L$ for the $W$-indexed eigenspace $V'$.

When the spaces are over $\mathbb{C}$, say, and $W = \mathbb{C}$, we just recover the normal definition of eigenspace and eigenvalue $\lambda \in \mathbb{C}$.

It seems to me that these concepts allow us to talk about multidimensional observables of a quantum mechanical system in a direct way. An observable on a space of states $V$ is a Hermitian $W$-indexed linear map $V \to V$, and the values it can take are its $W$-indexed eigenvalues, which should to be Hermitian operators $W \to W$.

Is this something that is known and studied, or is this a red herring?


The definition of $W$-indexing requires some further clarification. As stated right now, all three properties are satisfied by the classic Kronecker product.

To illustrate, suppose $V=\mathbb{C}^n$, $W=\mathbb{C}^m$. For argument's sake, let us assume that $L$ has simple eigenvalues, such that a diagonalization exists $L=X\Lambda X^{-1}$, and each eigenspace is dimension-1.

Then for any $w\in W$, $v\in V$, and any bijective linear map $D:W\to W$ which is also an invertible $m\times m$ matrix, we have

  • The $W$-indexed linear map $L$ is $\hat{L}=D\otimes L$. The domain is $w \otimes v \in W\times V$ and the range is $\hat{L}(w\otimes v)=(Dw)\otimes (Lv)\in W\times V$, so long as $L$ is nonsingular.
  • For any eigenpair $\lambda_i, x_i$ of $L$, we have $\hat{L}(w \otimes x_i)= (Dw) \otimes (Lx_i) = (Dw) \otimes (\lambda_i x_i) = (\lambda_i D\otimes I_n)(w \otimes x_i).$ Again, so long as $L$ is nonsingular, we have $\lambda_i\ne0$, so the range is $(\lambda_i D\otimes I_n)(w \otimes x_i) \in W \times x_i$
  • In the example above, the matrix $\lambda_i D$ is a $W\to W$ mapping that fits the definition of a $W$-indexed eigenvalue.

Of course if we set $D=1$, then we recover the usual eigenvalues and eigenvectors, as per stated.

Given any arbitrary $A$, $B$, the eigenvalues, eigenvectors, Jordan form, singular decomposition, etc, of $A\otimes B$ can all be expressed using properties of $A$ and $B$, and some basic linear algebra identities. See the wiki page.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.