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

• helpmeprepls
You have to square the Hamiltonian operator so you get ##\ H^2|v_j> = (2-j)^2a^2|v_j> \ ##. Then you have to find the probability of being in each state, and for each state you have to multiply the probability by the corresponding eigenvalue squared <v_j|v_j> = 1 so you don't have to do anything for that state. You get##\ <H^2> = {\displaystyle 1\over 6} a^2 + {\displaystyle 1\over 3} 0 + {\displaystyle 1\over 2} (-a)^2 = \ ####\ = {\displaystyle a

## 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|vj> = (2-j)a|vj>

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

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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?

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

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

BvU said:
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...

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> ...