# Energy Probability of Electron in 1d box

1. Feb 15, 2015

### Lamebert

1. The problem statement, all variables and given/known data
We're given an unnormalized state function ψ(x) of an electron in a 1 dimensional box of length pi. The state function is a polynomial. We're asked to find the probability that a measurement of its energy would find it in the lowest possible energy state.

2. Relevant equations
H = -h2 / 2m d2/dx2
E = h2n2π2/ 2mL2

3. The attempt at a solution
I took several approaches to finding the lowest possible energy state probability (n=1). I attempted to do it in the same way that you would do a position probability, except I also included the Hamiltonian. So I tried something along the lines of <E> = ∫(0 to pi) ψ(x)*⋅ H⋅ ψ(x) dx which I thought might give me a number that I could then divide into the second equation above (using n=1 to find energy). This returned a number that is clearly not a probability. I'm not going to explain every approach I took because I don't want to take away from my main questions. But another approach I took is finding the expectation value and dividing the first energy level value by this expectation value, which returned a very small value for n=1 (I assume n=1 should be somewhat probable, with decreasing values for n=2 and so forth).

I do not understand the energy dependence on position. How does energy only depend on position? Also, is it necessary to normalize my energy equations (since it has to do with probability). I'm not sure I'm even on the right track with expectation values or my integration of ψ(x)*⋅ H⋅ ψ(x), there's some kind of connection between the quantum mechanical ideas that I'm just not grasping. Any help is appreciated.

2. Feb 16, 2015

### Orodruin

Staff Emeritus
What you are trying to do is to find the expectation value of the energy. You cannot find the probability of being in a particular energy eigenstate by doing this. In fact, a system can have an expected energy equal to one of the eigenstate energies with probability zero of actually being found in that state.

Instead, you need to find the coefficient of the lowest eigenstate wavefunction when you expand the wavefunction in the eigenstates.

And yes, you should normalise your wave function or you will have to take into account at every alternate step that you are working with a non-normalised wave function. Better just do it once and for all.

3. Feb 16, 2015

### vela

Staff Emeritus
Look up probability amplitude in your textbook.

4. Feb 16, 2015

### BruceW

yeah. expectation values are not the right track to be on.
"I do not understand the energy dependence on position. How does energy only depend on position?" In this case, the Hamiltonian is just the kinetic energy of the particle. And the problem is implicitly assuming that the particle has zero chance of appearing outside the box. Hence the Hamiltonian does not include any potential energy terms.
In this problem, the main thing is that you are given the wavefunction ψ(x), and you also need to know the wavefunction of the ground state. From there, how do you calculate the probability of collapsing from ψ(x) into the ground state? (this is what vela was hinting at).

5. Feb 17, 2015

### Lamebert

I ended up using a Fourier expansion to solve for each wavefunction coefficient. Is this what you were hinting at?

6. Feb 17, 2015

### Orodruin

Staff Emeritus
Yes. The coefficient of the ground state is the amplitude of being in that state.

7. Mar 3, 2015

### Lamebert

Looking back on this two weeks later, what you were saying makes sense. I guess I just hadn't made connections between the math and concepts at that point. Thanks for your help.