# Understanding Shankar's standard deviation: Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$?

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?

• Why do you think scalar multiplication isn’t one kind of operation on a vector? What is $V(\vec{r})$ in the Schrodinger equation? – G. Smith Jul 20 '19 at 2:24
• "Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$[...]" Yes. – Sean E. Lake Jul 20 '19 at 3:54
• 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. – G. Smith Jul 20 '19 at 3:59
• @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. – Sean E. Lake Jul 20 '19 at 4:00
• 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. – tparker Jul 20 '19 at 5:00

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