# Proof in heisenbergs uncertainty relation involving bra-ket

lavster
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

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

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

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.

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