# I'm soooo lost in quantum!

1. Apr 6, 2009

### DummBlonde

1. The problem statement, all variables and given/known data
Okay, here it is: The wavefunction is psi=1*[1/sqrt(6)]*psi_1 + 2*psi_2 and I have to find the total probability (yes, there is a "weight" of 2 in front of the second part of the wavefunction).

2. Relevant equations
I know that psi^2 gives the probability, but I don't understand how to square this. Do I foil or something?

3. The attempt at a solution
psi_squared: (1/6)psi_1+ (2/3)psi_2 ????

2. Apr 6, 2009

### Pengwuino

Are you sure isn't a $$\frac{2}{{\sqrt 3 }}$$ instead of a 2 for the 2nd wave function? Although even with that, your probabilities don't add up to 1 if the stationary states are normalized as they should...

You do foil it though, yes. Also remember to consider orthogonality between different psi's.

3. Apr 7, 2009

### DummBlonde

I *think* it's just a 2, not 2/sqrt(3)...

By the way, I don't see how the probabilities add up to 1 if it's 2/sqrt(3)...

4. Apr 7, 2009

### Pengwuino

It won't. I should have asked this first but is Psi and psi_1, psi_2 normalized in the problem you're having?

5. Apr 7, 2009

### DummBlonde

One more thing: since the weight of the second part of the wavefunction is a 2, which is twice as great as the first part, it would seem to me that the second wavefunction is twice as likely to occur. That is, psi_1 would be 33% and psi_2 would be 67%. However, I don't understand how the 1/sqrt(6) fits in...

6. Apr 7, 2009

### Pengwuino

Well you have to multiply the 1 and the 1/sqrt(6). Your wavefunction then looks like:

$$\Psi = \frac{1}{{\sqrt 6 }}\psi _1 + 2\psi _2$$

Thus the probability of finding $$\psi_2$$ is $$2\sqrt 6$$ times as likely as finding $$\psi_1$$. Though something is seemingly missing in this problem because the coefficients of each $$\psi$$ squared must add up to 1 since the particle has to exist...

7. Apr 7, 2009

### millitiz

I am soooo lost too...
obviously, the total probability should be one, because one must find something at some where...
And yes, the wave function is not properly normalized, that only makes things even worse.

Yes, the probability of a wavefunction is l$$\Psi$$l^2, as long as the function is NORMALIZED.
And how to find it? Well, pretty much you just simply do it. And you'll see you can get some terns.
And using the orthogonality of wavefunctions could let you discard some of the terms (although technically, from just your question, I don't really know whether they are orthogonal or not). And I hope at this point you'll see the problem of your question. What is the probability after your calculation? And what should the total probability be?