# Uncertainty for delta E

1. Oct 31, 2011

### Uniquebum

1. The problem statement, all variables and given/known data
Determine the energy uncertainty $\Delta E = \sqrt{<E^2> - <E>^2}$ for a particle described by a wave function
$\Psi (x) = c_1 \psi (x)_1 + c_2 \psi (x)_2$
where $\psi_1$ and $\psi_2$ are different (orthonormal) energy eigenstates with eigenvalues $E_1$ and $E_2$.

2. Relevant equations
I'd presume you need to know
$\hat E = i \hbar \frac{\partial}{\partial t}$
$\hat E_{kin} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$

3. The attempt at a solution
First, i'm not sure whether i should throw in the kinetic or total energy operator. If i put in the total energy operator, i'll have to derivate the function in respect to time which in this case would result in 0. If i put in the kinetic energy operator it just might work but i'm not sure how i work with those expectation values when they're of the form <E^2> or <E>^2.

Assuming i'd use the total energy operator, should it look like
$<E^2> = \int_{-\infty}^{\infty} \psi^* (x) i^2 \hbar^2 \frac{\partial^2}{\partial t^2} \psi (x)$
$<E>^2 = \int_{-\infty}^{\infty} \psi^* (x)^2 i^2 \hbar^4 \frac{\partial^2}{\partial t^2} \psi (x)^2$
?

Any help would be appreciated.

2. Oct 31, 2011

### CompuChip

Note that
$$\langle E \rangle_\Psi \equiv \langle \Psi \mid E \mid \Psi \rangle$$

Since you are given that the wave function is a linear combination of orthonormal energy
eigenstates you can work out what <E> is in terms of c1,2 and E1,2. Squaring that will give you <E>2.

You can do a similar trick for

$$\langle E^2 \rangle_\Psi \equiv \langle \Psi \mid E^2 \mid \Psi \rangle = \left( \langle \Psi \mid E \right) \left( E \mid \Psi \rangle \right)$$

3. Nov 1, 2011

### Uniquebum

Hmm... so in this case $\hat E_{12} = E_{12}$? Meaning i don't have to mess around with the operators but i can just put in the constants E_1 and E_2?

4. Nov 1, 2011

### CompuChip

Sorry, what do you mean by $\hat E_{12}$?

Maybe I wasn't entirely clear with:
What I meant is:
But yes, you can work it out in terms of eigenvalues rather than using the explicit form of the operator - but of course you will still need to work it out to some extent (you can't just call the energy E, you do need to get it in terms of the given constants)

Just to be completely clear on what I mean: can you tell me (from the given information) what $\hat E { |\Psi\rangle}$ is?

5. Nov 2, 2011

### Uniquebum

$\int_{-\infty}^{\infty} \Psi^* \hat E \Psi dx$, where $\hat E = i \hbar \frac{\partial}{\partial t}$

But i figured i can just ignore the operator and use E instead. Which would lead into

$<E> = \int_{-\infty}^{\infty} \Psi^* \hat E \Psi dx = \int_{-\infty}^{\infty} (c_1^* \psi (x)_1^* + c_2^* \psi (x)_2^* )\cdot(c_1 E_1 \psi (x)_1 + c_2 E_2 \psi (x)_2)dx$

6. Nov 2, 2011

### vela

Staff Emeritus
You can't just ignore stuff. That's a sign that you're making a mistake. In this case, you're using the wrong operator to find the energy. You should be using the Hamiltonian $\hat{H}$ because your wave function is a solution to the time-independent Schrodinger equation $$\hat{H}\psi(x) = E\psi(x)$$ where E is a constant, not an operator. The operator $$\hat{E} = i\hbar\frac{\partial}{\partial t}$$ works on solutions $\Psi(x,t)$ to the time-dependent Schrodinger equation $$\hat{H}\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t).$$
The expected value of the energy E is then
$$\langle E \rangle = \int_{-\infty}^\infty \psi^* \hat H \psi \, dx = \int_{-\infty}^\infty [c_1^* \psi_1^*(x) + c_2^* \psi_2^*(x)][c_1 E_1 \psi_1(x) + c_2 E_2 \psi_2(x)]\,dx$$

7. Nov 2, 2011

### Uniquebum

Ok, so the way i thought you'd do it led to the right equation but it was pure luck since i just guessed the way it'd go. This definitely helped even more than just considering the solution to the main question.
Anyway, the rest should be fairly simple since i just have to take into account that the wave functions are orthonormal. Which mainly means that the wave functions will disappear since they'll be multiplying each others and thus be 1 or 0. I'll give it a shot. Thanks!