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

 

[Doc Al]

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

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

 

[lavster]

thanks for quick reply! however I am still not getting it. could you write it out expliitly please?

 

[Doc Al]

thanks for quick reply! however I am 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]

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]

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

 

[Frame Dragger]

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

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