On page 128 of Shankar's book, he defines the standard deviation as follows: $$\Delta\Omega=[\langle\psi|(\Omega-\langle\psi|\Omega|\psi\rangle)^2|\psi\rangle]^{\frac{1}{2}} $$ Now, this equation is undefined as is, because $\langle\psi|\Omega|\psi\rangle$ is a scalar and $\Omega$ is an operator. Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$, or is it something different?

  • $\begingroup$ Why do you think scalar multiplication isn’t one kind of operation on a vector? What is $V(\vec{r})$ in the Schrodinger equation? $\endgroup$
    – G. Smith
    Jul 20, 2019 at 2:24
  • 4
    $\begingroup$ "Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$[...]" Yes. $\endgroup$ Jul 20, 2019 at 3:54
  • $\begingroup$ I say no. To multiply a scalar times a vector, there is no need to insert a pointless identity operator. One of the axioms of a vector space is that scalar multiplication is allowed. $\endgroup$
    – G. Smith
    Jul 20, 2019 at 3:59
  • 3
    $\begingroup$ @G.Smith That's right, but $\langle\psi|\Omega|\psi\rangle$ is a scalar and $\Omega$ is an operator, so OP wants to know how those two are being subtracted. $\endgroup$ Jul 20, 2019 at 4:00
  • 4
    $\begingroup$ This argument is entirely about notation. Everyone here is agreeing that a scalar is not an operator, but scalar multiplication is an operator. The only disagreement is whether the notation in the OP makes it sufficiently obvious that the appearance of the scalar actually denotes the scalar multiplication operator. I think that the very existence of this question demonstrates that (to at least one person) that fact is not obvious, so in a first course in QM it may be worth inserting the identity operator explicitly, just to show that the scalar actually denotes scalar multiplication. $\endgroup$
    – tparker
    Jul 20, 2019 at 5:00

1 Answer 1


So if you want to explicitly combine $(\Omega-\langle\psi|\Omega|\psi\rangle)^2$ into a single operator without actually using it on a state, then you would need to put the identity operator in for the math to make sense.

However, as pointed out in the comments, multiplying a state by a scalar is a perfectly valid operation. Therefore, you could easily apply the distributive law for the operator/scalar terms and then be perfectly fine with not even thinking about the identity operator. i.e.

$$\langle(\Omega-\langle\Omega\rangle)^2\rangle=\langle \Omega^2\rangle+\langle\langle\Omega\rangle^2\rangle-2\langle\langle\Omega\rangle\Omega\rangle=\langle \Omega^2\rangle-\langle\Omega\rangle^2$$ where I have condensed the notation to avoid clutter of $|\psi\rangle$ terms.

In my opinion, there is nothing wrong with putting in the identity operator, so if that is what makes the most sense to you then go ahead and think of it like that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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