Mean value of energy <E> for a QM state?

[helpmeprepls]

Homework Statement

If the system is in a state

|ψ> = 1/sqrt(6) |v1> + 1/sqrt(3) |v2> - i/sqrt(2) |v3>

with Hamiltonian satisfying H|v_j> = (2-j)a|v_j>

Find the mean value of energy <E> and the root mean square deviation √(<E²> - <E>² ) that would result from making a number of measurements of the energy of the system in state |ψ>

Homework Equations

<E> = <ψ|E|ψ>

for a free particle E = p² / 2m

The Attempt at a Solution

[/B]
To find the mean value of Energy <E> is it just eigenvalues (a, 0 -a) multiplied by the probability of it being in the corresponding state P = (1/6, 1/3, 1/2)?

= a/6 - a/2 = -2a/6.
For E² then do you just have the same, but the eigenvalues squared multiplied by the probabilities?

 

[helpmeprepls]

except divide the energy by two? as its a mean, so it would be -a/6 ?

 

[BvU]

Why divide by 2 ? What would be the mean value for the identity operator 1 ?

 

[helpmeprepls]

Why divide by 2 ? What would be the mean value for the identity operator 1 ?

1, so do you just find the mean of the eigenvalues? So 0...

 

[BvU]

For the identity operator you get $\ {1\over 6} + {1\over 3} + {(-i^*)(-i)\over 2} = 1 \$. No dividing by 2. The state is properly normalized.

For <H> you get $\ {1\over 6} (a) + {1\over 3} (0) + {(-i^*)(-i)\over 2} (-a) = {\displaystyle -a\over 3} \$ as you did. No dividing by 2 either.

So for <H²> ...