Expectation value of a SUM using Dirac notation

[rwooduk]

Homework Statement

Consider a one-dimensional particle subject to the Hamiltonian H with wavefunction \Psi(r,t) =\sum_{n=1}^{2} a_{n}\Psi _{n}(x)e^{\frac{-iE_{n}t}{\hbar}}
where H\Psi _{n}(x)=E_{n}\Psi _{n}(x) and where a_{1} = a_{2} = \frac{1}{\sqrt{2}}. Calculate the expectation value of the Hamiltonian with respect to \Psi (x,t)? Which energy eigenvalue is the most likely outcome when we measure the energy of particle once?

Homework Equations

Given in the question.

The Attempt at a Solution

\Psi(r,t) =\sum_{n=1}^{2} a_{n}\Psi _{n}(x)e^{\frac{-iE_{n}t}{\hbar}}

let \Psi_{1}(r,t) =a_{1}\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}} and \Psi_{2}(r,t) =a_{2}\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}

therefore \left \langle H \right \rangle = \left \langle \Psi _{1}+ \Psi _{2}|H |\Psi _{1}+ \Psi _{2}\right \rangle

which gives \left \langle H \right \rangle = (E_{1}+ E_{2}) \left \langle \Psi _{1}+ \Psi _{2}|\Psi _{1}+ \Psi _{2}\right \rangle

but not sure what to do now? is this the best way to do this? the trouble I am having is using bra ket notation to work with a sum of wavefunctions.

any advice on this would really be appreciated!

thanks in advance

 

[Vagn]

Are the wavefunctions orthonormal, if so what happens then?

 

[rwooduk]

Are the wavefunctions orthonormal, if so what happens then?

(E_{1}+ E_{2}) \left \langle \Psi _{1}+ \Psi _{2}|\Psi _{1}+ \Psi _{2}\right \rangle

Yes, hence the bra ket term above would equal 1. However I need to get the a_{1} and a_{2} out of there. we did a degenerate problem in bra and ket that gave:

a_{1}\left \langle \Psi _{1} |H| \Psi _{1} \right \rangle + a_{2}\left \langle \Psi _{2} |H| \Psi _{2} \right \rangle + a_{1}\left \langle \Psi _{1} |V| \Psi _{1} \right \rangle + a_{2}\left \langle \Psi _{2} |V| \Psi _{2} \right \rangle = Ea_{1}\left \langle \Psi _{1}|\Psi _{1} \right \rangle + Ea_{2}\left \langle \Psi _{1}|\Psi _{2} \right \rangle

and was much easier to simplify, for example the exponential terms in my question will not go to 1, because there are 2 different energy levels. how do I deal with these extra terms in bra ket notation?

many thanks for the reply!

 

[BvU]

let \Psi_{1}(r,t) =a_{1}\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}} and \Psi_{2}(r,t) =a_{2}\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}
therefore \left \langle H \right \rangle = \left \langle \Psi _{1}+ \Psi _{2}|H |\Psi _{1}+ \Psi _{2}\right \rangle
which gives $$\left \langle H \right \rangle = (E_{1}+ E_{2}) \left \langle \Psi _{1}+ \Psi _{2}|\Psi _{1}+ \Psi _{2}\right \rangle$$

By defining $\Psi_{1}(r,t)$ and then dropping the (r,t) you cause confusion between your $\Psi_{1}(r,t)$ and the eigenfunction of the Hamiltonian $\Psi_{1}(x)$.

The "which gives" that follows is not correct:
$$H \left |\; a_1\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}}+ a_2\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}\right \rangle = E_1 a_1 \left | \Psi _{1}(x) \right > e^{\frac{-iE_{1}t}{\hbar}} + E_2 a_2 \left | \Psi _{2}(x) \right > e^{\frac{-iE_{2}t}{\hbar}} $$

 

[rwooduk]

By defining $\Psi_{1}(r,t)$ and then dropping the (r,t) you cause confusion between your $\Psi_{1}(r,t)$ and the eigenfunction of the Hamiltonian $\Psi_{1}(x)$.

The "which gives" that follows is not correct:
$$H \left |\; a_1\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}}+ a_2\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}\right \rangle = E_1 a_1 \left | \Psi _{1}(x) \right > e^{\frac{-iE_{1}t}{\hbar}} + E_2 a_2 \left | \Psi _{2}(x) \right > e^{\frac{-iE_{2}t}{\hbar}} $$

Thanks, I didnt realize you could put everything in a ket like that, it's a little messy, but yes you answered my question. I will work through the problem and see where it goes.

Many thanks for all the help, really appreciated!

 

[BvU]

Don't forget to take complex conjugates when you take factors like $a_1\; e^{\frac{-iE_{1}t}{\hbar}}$ outside the bra state