# 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