# Proof in heisenbergs uncertainty relation involving bra-ket

hey, can someone show me the step between these two lines of equations please:

$$(\Delta A)^2=<\psi|A^2|\psi>-<\psi|A|\psi>^2$$
$$=<\psi|(A-<A>)^2|\psi>$$

where A is an operator and $$\psi$$ is the wavefunction and $$<A>$$ is the expectation value of A

## Answers and Replies

Doc Al
Mentor
$$=<\psi|(A-<A>)^2|\psi>$$
Just expand this expression, realizing that <A> is just a number. ($$<A> = <\psi|A|\psi>$$)

Last edited:
thanks for quick reply! however im still not getting it. could you write it out expliitly please?

Doc Al
Mentor
thanks for quick reply! however im still not getting it. could you write it out expliitly please?
$$<\psi|(A-<A>)^2|\psi> = <\psi|A^2 -2A<A> + <A>^2 |\psi>$$

I'll let you do the rest.

Fredrik
Staff Emeritus
Science Advisor
Gold Member
Just a little LaTeX tip: Use \langle and \rangle instead of < and >. (Doc AI's answer is good, so I have nothing to add, except the complete solution, but you should try it yourself first. Note: "just a number" really means "just a number times the identity operator". OK, I guess I did have something to add ).

Doc Al
Mentor
I thought those bras and kets looked a bit off. :uhh: (Thanks, Fredrik!)

I thought those bras and kets looked a bit off. :uhh: (Thanks, Fredrik!)

Must... not... say... I... prefer... bras... off... must... not say... to staff...ARGGGH.. Too late. Be gentle!