# Electron in two Potential Wells

1. May 31, 2014

### kq6up

1. The problem statement, all variables and given/known data

An electron can be in one of two potential wells that are so close that it can “tunnel” from one to the other (see §5.2 for a description of quantum- mechanical tunnelling). Its state vector can be written
|ψ⟩ = a|A⟩ + b|B⟩, (1.45)
where |A⟩ is the state of being in the first well and |B⟩ is the state of being in the second well and all kets are correctly normalised. What is the probability of finding the particle in the first well given that: (a) a = i/2; (b) b = e^(i*pi); (c) b = 1/3 + i/√2?

2. Relevant equations

a*a is the probability of finding a particle in state A

3. The attempt at a solution

The question is confusing me. I don't know what the second b is for. Also, these are supposed to be normalized according to the question, but b*b (for the first b) would be 1 all by itself. Is this question ok, and I am just missing something?

Thanks,
Chris Maness
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 31, 2014

### hilbert2

Actually, the probability to be in state $A$ is $|a|^{2}$ (square modulus, different from simple square for complex numbers). If $|b|$ has value $1$, it just means that $a$ must be zero if the state is normalized. If one of the constants $a,b$ is given, you can deduce the absolute value of the another from the normalization condition.

3. May 31, 2014

### kq6up

Yes, I show that the probability for finding the system in state a is a*a. Where a* is the complex conjugate of a. This is the mod squared for a complex number. The only problem I am having is that he states that it is already normalized. I expect |a|^2 +|b|^2=1, but it is not, and then there is another b too. This confuses me more.

Chris

4. May 31, 2014

### hilbert2

Sorry, I confused the star "*" with a multiplication sign. The parts (a), (b) and (c) are separate problems and have different states of form $a\left|A \right> + b\left|B\right>$ as answers. You are not given two values of $b$ for solving the same problem.

5. May 31, 2014

### kq6up

Perfect, thanks. I have been doing too many math problems that have the phrase "show that" in them :D

Chris KQ6UP