# I Calculating the probability density of a superposition

1. Mar 26, 2017

### Archeon

This ought to be some simple gap in my knowledge, but it bugs me nonetheless. Let me present the argument as I see it, I'm fairly certain that there is just some tiny part that I didn't learn correctly.
Let us assume a wavefunction $$\Psi$$ is defined as a superposition of two wavefunctions:

\Psi(x) = \alpha \psi_1 + \beta \psi_2

with both subwavefunctions being normalized.

Now (and correct me if I'm wrong here) even if the wavefunctions arent orthogonal, the norm squared and thus the probability density can be expressed like so:

|\Psi(x) |^2 = |\alpha \psi_1 + \beta \psi_2|^2 = |\alpha|^2 + |\beta|^2 + \lambda = 1

with $$\lambda$$ being some real number as a result of the multiplication.
And here is the problem I face. Does this mean that $$|\alpha|^2$$ and $$|\beta|^2$$ only represent probabilities if the wavefunctions are orthogonal? Or am I missing something else here?

Thanks

2. Mar 26, 2017

### Staff: Mentor

Where does $\lambda$ come from? And why must it be real?

3. Mar 27, 2017

### Archeon

It would be the double product of ψ1 with ψ2: $$\lambda = \alpha^* \psi_1^* \beta \psi_2 + \beta^* \psi_2^* \alpha \psi_1 = 2 Re(\alpha^* \psi_1^* \beta \psi_2)$$
which can be seen as some complex number plus its conjugate, thus twice its real part and a real number.

4. Mar 27, 2017

### blue_leaf77

You need to take the integral over all $x$ instead of only using $|\psi(x)|^2$ in order to come to the probabilistic interpretation.

5. Mar 27, 2017

### Staff: Mentor

Probabilities of what? If they are not orthogonal, they cannot be two distinct measurement results of the same type of measurement. There is no measurement that would give either the first or the second state, therefore there is no reason why the squared prefactors should sum to 1.

6. Mar 27, 2017

### Archeon

Ah, I think I see the error in my thinking. Also, what I meant to write was that the integral of the Real part would have to add up with the factors squared to be equal to one.
I see now, of course the two wavefunctions that make up the superposition would have to be eigenstates of some observable that gets measured, but can I assume that such eigenstates are always orthogonal for every measurable quantity?

7. Mar 27, 2017

### Staff: Mentor

That is given by linear algebra.

8. Mar 27, 2017

### Jilang

Enough already...!The normalisation factor of the superposition will be different. You cannot add two normalised wavefunctions together and say the result is normalised!