# Can an eigenvalue be a function?

When we say that $$\hat{E}(\psi(x))=\alpha\psi(x),$$ where $$\hat{E}$$ is an operator and $$\alpha$$ is the eigenvalue.

Is $$\alpha$$ a fixed constant(like a number) or can it's value keep on varying?

For example, $$\hat{X}(\psi(x))=x\psi(x)$$ where $$\hat{X}$$ is the position operator and $$x$$ is the position, can $$x$$ be considered as an eigenvalue?

• I'm not quite sure what you have in mind when you ask if the eigenvalue can be a function. In particular, a function of what? When you further ask "can its value keep on varying", I get the impression that you're thinking of a function of time. I wonder if you're using the word "function" when you might be thinking of "parameter" or even "label". For example, for the position eigenket $|x\rangle$, the $x$ is a label that essentially invites one to replace $x$ with any real number for the coordinate $x$ so that, e.g., $\hat X|2\rangle = 2|2\rangle$. Is this what you mean by 'function'? – Alfred Centauri Sep 25 '19 at 22:18

It has to be a constant, because, by definition, an eigenvalue belongs to the field of scalars $$\mathbb{K}$$ underlying the vector space which is the domain and codomain of the linear operator in question, i.e. the one whose eigenvector and corresponding eigenvalue we are talking about.

In quantum mechanics of the kind you are describing, the vector space is the space $$\mathbf{L}^2(\mathbb{R})$$ of equivalence classes of square integrable functions of one real variable over the field of scalars $$\mathbb{C}$$. So we know that the eigenvalue must be a member of this field. Indeed, in quantum mechanics, we can go further, since we are talking about self adjoint operators, so the eigenvalue must be real, but this is further to what you need to know here.

It has to be constant. Otherwise essentially every function would be an eigenfunction of the x operator (except perhaps where $$x=0$$ or $$\psi=0$$).

You are really asking a question about conventions of the bra-ket notation, $$\psi(x)=\langle x| \psi\rangle$$.

$$\hat{E}(\psi(x))=\alpha\psi(x)$$ is a synonym for $$\langle x| \hat{E} |\psi\rangle= \langle x| \alpha |\psi\rangle = \alpha \langle x| \psi\rangle ,$$ so that, likewise, $$\langle x| \hat{X} |\psi\rangle= \langle x| x |\psi\rangle = x \langle x| \psi\rangle .$$

In both cases $$\alpha$$ and x are constant eigenvalues of eigenvectors $$|\psi\rangle$$ and $$|x\rangle$$, respectively.

As these states alter, the eigenvalues follow suit (co-vary with the alteration). But for each eigenvector they are constants.