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