Solving Liboff 5.33: Showing <E> \geq E_1 and Condition on \psi(x)

[quasar987]

There's this question ngyah (Liboff 5.33) that says

"We consider a particle in an infinite 1-dimensional well. The particle is described by an arbitrary wave function \psi(x). (a) For this particle, show that &lt;E&gt; \geq E_1. (b) What is the condition on \psi(x) such that &lt;E&gt; = E_1?"

My solution: First of all, I'm assuming that by <E> he means <H>. But <H> = E. And E is the energy associated with the superposition of the sine waves that make up \psi(x). So

E = \sum_{i=1}^{\infty}c_i E_i

where the c_i are either 1 or 0 depending on wheter or not \psi_i(x) = sin(k_ix) is present in the Fourier expansion of \psi(x). This said, if \psi_1(x) is in the expansion of \psi(x), then E = E_1 +... \geq E_1. If not, then E = 0\cdot E_1 + ... + 1\cdot E_j +... \geq E_1.

And for (b), the condition on psi is that psi be exactly psi_1.

That looks pretty neat to me. The only point I'm unsure of is wheter or not, it is true that the energy of a psi made up of a linear combination of other psi_i is the sum of the energy of the psi_i. Actually, this seems FALSE to me. For suppose psi_1 and psi_2 are solution of the time ind. SE with respective energy eigenvalues E_1 and E_2. Then let's see if psi = psi_1 + psi_2 is a solution with eigenvalue E_1 + E_2.

(E_1+E_2)\psi = (E_1+E_2)(\psi_1 + \psi_2) = E_1\psi_1 + E_2\psi_2 + E_1\psi_2 + E_2\psi_1 = H\psi_1 + H\psi_2 + E_1\psi_2 + E_2\psi_1 = H\psi + E_1\psi_2 + E_2\psi_1 \neq H\psi

So the energy of psi_1 + psi_2 is not E_1 +E_2. Correct? Hopefully I'm missing something.

 

[quasar987]

Arg, it's assuredly wrong, for any function that requires an infinity of sine waves to be constructed will have an infinite energy.

 

[CarlB]

Arg, it's assuredly wrong, for any function that requires an infinity of sine waves to be constructed will have an infinite energy.

Yes. If you have a wave function that is not an eigenstate of the Hamiltonian, it's energy is not a precise number. What the instructor was asking for was <H>, which means the average of the energy.

If c_n are the (complex) amount of \psi_n that you have to use to make up your (normalized) wave function \psi = \sum \psi_n, then


\langle E \rangle = \langle \psi |E| \psi \rangle
= \sum_n \sum_m c_n^* c_m \langle E_n| H |E_m\rangle
= \sum_n \sum_m c_n^* c_m E_m \langle E_n | E_m\rangle
= \sum_n \sum_m c_n^* c_m E_m \delta_n^m
= \sum_n c_n^* c_n E_n
= \sum_n |c_n|^2 E_n &lt; \infty

Since \sum_n |c_n|^2 = 1, you can see that the maximum that <E> could be would be the highest of the E_n, which is less than infinity.

Hope this helps.

Carl

 

[quasar987]

I'm not used to the notation you used. What is <psi|E|psi> ?

 

[CarlB]

I'm not used to the notation you used. What is <psi|E|psi> ?

I don't know what book you're being taught out of, but here goes:

\psi(x) = \langle x|\psi\rangle

That is, |\psi\rangle is just the ket associated with the state \psi.

It could be that in your book / class, they haven't covered the bra-ket formalism yet. In that case,

\langle \psi | M| \psi \rangle = \int_{-\infty}^\infty \psi^*(x) M \psi(x) dx

for wave functions in one dimension. In the above, M is an operator.

Did this help?

Carl

 

[quasar987]

It could be that in your book / class, they haven't covered the bra-ket formalism yet. In that case,

\langle \psi | M| \psi \rangle = \int_{-\infty}^\infty \psi^*(x) M \psi(x) dx

Now we're talking the same language!

 

[quasar987]

Your way seems kinda complicated CarlB with double sums and stuff. Here's mine..

&lt;E&gt; = \sum_{n=1}^{\infty}E_n|c_n|^2
&lt;E&gt; = E_1|c_1|^2 + E_2|c_2|^2+... \geq E_1|c_1|^2+E_1|c_2|^2+... = E_1 \sum_{n=1}^{\infty}|c_n|^2 = E_1