I have multiple different operators in matrix form and I need to find their common eigenstates. The challenge is that the common eigenstate is in a superposition of multiple states and isn't just a single eigenvector. Can someone give me an idea of how to do this? How would I figure out what a common eigenstate in the form of a linear combination of vectors is for multiple matrices? For context, I am using Python and NumPy.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Superposition is a linear combination of vectors which is a single vector. It may not be equal to some basis vector (from computational basis, for example), but this is not a big deal in general.
As for question, I suppose operators are normal (so we can apply the spectral theorem).
In the simplest case if operator $A$ has eigenvalue $\lambda$ with multiplicity 1, then the corresponding eigenvector $v$ will also be eigenvector for any commuting operator $B$.
Since $Av=\lambda v$ and $AB=BA$ then
$$
ABv = BAv= B\lambda v \implies A(Bv) = \lambda (Bv) \implies Bv = \mu v
$$
Here we obtained that $Bv$ is also eigenvector of $A$ with eigenvalue $\lambda$ and since it has multiplicity 1 it must be collinear to $v$.
So such eigenvector $v$ will automatically be common eigenvector for a set of commuting with $A$ operators.
In general case $A$ may not have eigenvalue with multiplicity 1. In this case $A$ has invariant subspaces (eigenspaces) $H_{\lambda_i}, \text{dim}H_{\lambda_i}>1$ such that $Av = \lambda_i v, \forall v \in H_{\lambda_i} $.They split space $H = H_{\lambda_1} \oplus ... \oplus H_{\lambda_k}$.
Similarly to the simple case, we can show that if $AB=BA$ then $H_{\lambda_i}$ are also invariant for $B: B(H_{\lambda_i}) \subset H_{\lambda_i}$. Now we can consider restrictions of operator $B$ on the subspaces $H_{\lambda_i}$. Every restriction has its own eigendecomposition. But every eigenspace of every such restriction of $B$ will be common eigenspace of $A$ and $B$ since $H_{\lambda_i}$ is eigenspace of $A$.
Actually, if we take the intersection of eigendecompositions $\{H_{\lambda_i}\}$ and $\{H_{\mu_i}\}$ correspondingly for $A$ and $B$ then we will obtain the decomposition of $H$ into common eigenspaces of $A$ and $B$.
Though it may not be that efficient to calculate intersection of all eigendecompositions for operators $A_i$.
I guess we can just pick some random numbers $r_i$ and find eigenvector of $\sum_i r_i A_i$. With high probability it will be the common eigenvector for commuting operators $A_i$.
