The definition of the expectation value for a continuous domain f(x) is given by

$$<f(x)>=\int{f(x)p(x)dx}$$ where p(x) is the probability density function corresponding to {x}.

In quantum mechanics the expectation value of an operator (for 1-d space) $\hat{Q}$ is given by

$$<\hat{Q}> = \int{\psi^{*}{(x)}\hat{Q}\psi{(x)} dx}$$

My question, and confusion, is twofold.

1) f(x) is a domain of values. $\hat{Q}$ is an operator that can only be considered a domain of values when it operates on $\psi$. Thus it strikes me that one wouldn't be able to evaluate $<\hat{Q}>$ but rather $<\hat{Q}\psi{(x)}>$. This however wouldn't work because this would be evaluated as $$ <\hat{Q}\psi{(x)}>=\int{\hat{Q}\psi{(x)}p(x)dx}=\int{\hat{Q}\psi{(x)}|\psi{(x)}|^2dx}=\int{\hat{Q}\psi{(x)}\psi{(x)}^*\psi{(x)} dx}$$

This is obviously does not fit the definition of the expectation value of an operator used in QM

2) Say I'm missing something in the definition of an expectation value or this is just some notational convenience I'm unaware of. Let us then assume I can take the expectation value of an operator alone.

$$ <\hat{Q}>=\int{\hat{Q}p(x)dx}=\int{\hat{Q}|\psi{(x)}|^2dx}=\int{\hat{Q}\psi{(x)}^*\psi{(x)} dx}$$

But this still does not make sense because $\int{\hat{Q}\psi{(x)}^*\psi{(x)} dx}$ does not necessarily equal $\int{\psi^{*}\hat{Q}\psi dx}$

I know I'm doing something wrong, but what?

  • $\begingroup$ I don't understand how you're using the term "domain". What does it mean that "$f(x)$ is a domain of values"? $\endgroup$
    – joriki
    Jun 5, 2016 at 20:44
  • $\begingroup$ @joriki I was thinking of the expectation value as a transformation with a domain; this domain is the image of f(x) where f(x) has a domain {x}. $\endgroup$ Jun 5, 2016 at 21:10
  • $\begingroup$ I'm not sure if this is answering the question but if $\phi_k(x)$ is a basis of eigenstates of $\hat{Q}$ with eigenvalues $q_k$ then one could write $\psi(x) = \sum_k a_k \phi_k(x)$. Then subbing in this formula gives $$\int \overline{\psi(x)} \hat{Q}\psi(x) dx = \sum_k |a_k|^2 q_k$$ Since $|a_k|^2$ is the probability of being in the kth eigenstate the two definitions of expectation value that you give agree. $\endgroup$ Aug 15, 2020 at 17:13

1 Answer 1


To answer the end of 2) $Q$ is Hermitian ($Q = {Q}^*$ in this notation [I know its poor notation]), so $\langle Q \rangle \equiv \langle \psi , Q \psi \rangle = \langle Q \psi , \psi \rangle$ which shows that $\int \psi^* Q \psi \; dx = \int {Q}^* \psi^* \psi \; dx = \int {Q} \psi^* \psi \; dx.$

1) The $Q$ versus $Q$ acting on some domain issue is easily fixed since we let that the symbol $\langle Q \rangle$ implicitly mean "the expectation of $Q$ acting on this particular state function" - it doesn't mean "the expectation of $Q$ forever and always in all situations." So we are not finding the expectation value of the operator, but the expectation of the operator acting on this system.

One could define $\langle Q \rangle \equiv \langle \psi | Q | \psi \rangle \equiv \int \psi^* Q \psi \; dx$ to be the expectation of $Q$ acting on $\psi$, that fixes the problem but is not satisfying for the question you posed. Perhaps using what you did in problem 2 in combination with the fact that $Q$ is Hermitian solves your problem.

  • $\begingroup$ It seems to me that the first paragraph obscures more than it clarifies. When you write $Q\psi^*\psi$ without parentheses, there's a risk of equivocation between $(Q\psi^*)\psi$ and $Q(\psi^*\psi)$. The latter is the sense in which the OP had used this expression, deriving it from $Q|\psi|^2$ in analogy with $Qf(x)$, with $|\psi|^2$ playing the role of a classical probability density. $\endgroup$
    – joriki
    Jun 5, 2016 at 20:47
  • $\begingroup$ Thanks, Merkh! I see where my problem is now with (2). The reason I didn't do what you did with the inner product and the Hermitian property of Q was that I got some algebra mixed up and treated $\psi$ as if it was an operator, giving $(Q\psi)^*=\psi^* Q^*$, which is obviously wrong since $\psi$ isn't an operator. And thanks for the clarification with (1). That's what I thought but didn't want to assume. $\endgroup$ Jun 5, 2016 at 21:06
  • $\begingroup$ Ah wait, the error mentioned in the above comment of mine did not fix it all the way, only partly. At any rate, thanks for the help. $\endgroup$ Jun 5, 2016 at 21:40

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