# Quantum Mechanics question

## Homework Statement

Assume that a particle in a one-dimensional box is in its first excited state. Calculate the expectation values [x], [p], and [E], and the uncertainties
delta(x), delta(p), and delta(E). Verify that delta(x)*delta(p)>=h_bar/2.

## Homework Equations

Psi=sqrt(2/a) cos(pi*x/a) e^(-i*E*t/h_bar)

[x]=Int(Psi_star x Psi, -a/2, a/2)

[p]=Int(Psi_star (-i*h_bar*d/dx) Psi, -a/2, a/2)

[E]=Int(Psi_star (i*h_bar*d/dt) Psi, -a/2, a/2)

## The Attempt at a Solution

After evaluating the above integrals, I get:

[x]=0

[p]=0

[E]=h_bar*pi^2*n^2 / 2*m*a^2

I am trying to calculate the quantities delta(x), delta(p), and delta(E) but I am having trouble doing that. Can you please suggest some hints on how to proceed. Thank you.

Related Advanced Physics Homework Help News on Phys.org
What are the definitions of delta(x) and delta(p) in terms of expectation values?

This is a trivial question, but would I be able to approximate it using the relations:

delta(x)=h_bar/sqrt(2m(V0 - E))

This is a trivial question, but would I be able to approximate it using the relations:

delta(x)=h_bar/sqrt(2m(V0 - E))
That's a new relation to me. Just look at the definition of the variance and follow the prescription.

That's a new relation to me. Just look at the definition of the variance and follow the prescription.
Yep. You need to calculate six expectation values in order to calculate deltax, deltap and deltaE, and only three of them are <x>, <p> and <E>.

dextercioby
Do you know a reason why $\Delta E$ is zero for this problem ? Besides the actual computation of it, which can be avoided by knowing this reason.