# What is the eigenvalue of an arbitrary state?

I found this questions here assumes that eigenvalue of $$|0001\rangle$$ is $$-1$$. Can someone please explain how does this work for this example? How does it work in general for any state?

An eigenvalue is measured relative to a particular operator/matrix. For example, in the linked question, it was considering the operator $$Z\otimes Z\otimes Z\otimes Z$$.

Recall that for any operator/matrix $$M$$, a state $$|\psi\rangle$$ is an eigenvector with eigenvalue $$\lambda$$ if and only if it can be written as $$M|\psi\rangle=\lambda|\psi\rangle.$$ So, in the present case, you simply need to evaluate $$Z\otimes Z\otimes Z\otimes Z|0001\rangle=(Z|0\rangle)\otimes (Z|0\rangle)\otimes (Z|0\rangle)\otimes (Z|1\rangle).$$ Using the definition of $$Z$$ that $$Z|0\rangle=|0\rangle$$ and $$Z|1\rangle=-|1\rangle$$, you'll find $$Z\otimes Z\otimes Z\otimes Z|0001\rangle=-|0001\rangle,$$ and hence it is an eigenvector with eigenvalue -1.

Note that without a matrix, you cannot say anything. Even with a matrix, most of the time, a given state will not be an eigenvector. It is a very specific property.

Eigenvalues and eigenstates are properties of operators.

For an operator $$A$$, its eigenstates are the states $$x$$ that imply that $$Ax=\lambda x$$

So all $$x$$ that imply this for a given $$x$$ is the eigenstates of $$A$$, and its eigenvalue is the scalar $$\lambda$$ that is coming out of $$Ax$$.

In your case there was some operator, the $$x$$ was $$|0001\rangle$$, and $$\lambda$$ the eigenvalue was $$-1.$$