# Wave equation indicates the probability density

1. Dec 15, 2015

### Small bugs

First, we know for every wave function
$$p(x)=\psi(x)^*\psi(x)$$ indicates the probability density of a particle appearing at the point x.
So if we calculate $$P=\int _M p \text{d}x$$ this gives the probability of the particle appearing in the range M.
On the other side, I was thinking about superposition:
$$\Psi=\frac{1}{2}(\psi_A+\sqrt{3}\psi_B)$$
So it means that as long as we measure the system, at 25% the wave will collapses to state A, 75% to B.
And e.g. in state A, the particle has probability PA to appear in range M; while in B, has PB.
So there would be $$\frac{P_A+3P_B}{4}$$ to appear in range M. Right??
However, if we calculate in this way:
$$P(x)= \int _M \Psi^*\Psi\text{d}x=\frac{1}{4}\int _M (\psi_A^*+3\psi_B^*)(\psi_A+3\psi_B)\text{d}x$$
$$=\frac{1}{4}\int _M \psi_A^*\psi_A+3\psi_B^*\psi_B+\psi_A^*\psi_B+\psi_A\psi_B^*\text{d}x$$
First two terms can be eliminated to probability,
$$=\frac{1}{4}\left[P_A+3P_B+\int _M \psi_A^*\psi_B+\psi_A\psi_B^*\text{d}x\right]$$
We know: if M is the whole space, then the last two terms are 0, but they can NOT vanish for every function and every interval. So the calculation would be the same!!! Anything wrong???
Look forward to your help, thanks!

Last edited: Dec 15, 2015
2. Dec 16, 2015

### Orodruin

Staff Emeritus
You are assuming that the measurement constraining the system to state A or B commutes with the measurement of position. This is generally not the case.

3. Dec 16, 2015

### PietKuip

Eigen states are orthogonal; the whole-space integral is zero for basis functions.

4. Dec 16, 2015

### Orodruin

Staff Emeritus
But he has already realised this and it is not his question. The problem is that he is implicitly taking the measurements of position and energy to be commutative, which they are not unless the hamiltonian commutes with the position operator.

5. Dec 16, 2015

### PietKuip

That was part of the problem (OP: "can not vanish for every interval").
I just added that the whole-space density is still normalized because the wavefunctions A and B are not just any functions.

6. Dec 16, 2015

### Orodruin

Staff Emeritus
But the OP already said this:

7. Dec 16, 2015

### Demystifier

Wrong!
The probabilities are to be added only when the events are mutually exclusive. On the other hand, in general, being in A and being in B are not mutually exclusive events.

Let me take an example from geography (rather than quantum physics). If A=Europe and B=America, then A and B are mutually exclusive. You cannot be in both Europe and America at the same time. But if A=Europe and B=Turkey, then A and B are not mutually exclusive. There is a north part of Turkey which is in Europe, so you can be in both Europe and Turkey at the same time.

Now let us get back to physics. An important quantity is $f(x)=p_A(x)p_B(x)$. Is $f(x)=0$ for all $x$?
1) If it is, then A and B are mutually exclusive, so your addition of probabilities above is OK, but then
$\psi_A^*(x) \psi_B(x)=\psi_A(x) \psi^*_B(x)=0$ for all $x$. So everything is OK.
2) If it is not, then A and B are not mutually exclusive, so your addition of probabilities above is wrong.

In your case you didn't specify what are A and B. So either the case 1) or case 2) must apply (I cannot say which one), and for each case the problem disappears, in one way or another.

Last edited: Dec 16, 2015
8. Dec 16, 2015

### PeroK

To add to what has been said, you have to be careful with the "addition of probabilities" argument when $|\Psi|^2$ is your pdf and not $\Psi$. Just looking at probability theory:

If $p(x) = \alpha a(x) + \beta b(x)$ then $<x> = \alpha <x>_a + \beta <x>_b$

as $\int xp(x)dx = \alpha \int xa(x)dx + \beta \int xb(x)dx$

(where $p, a, b$ are normalised pdf's and $\alpha + \beta = 1$

But if $p(x) = \alpha a(x) + \beta b(x)$ where $p^2, a^2, b^2$ are normalised pdf's, $a, b$ are othogonal and $\alpha^2 + \beta^2 = 1$, then

$<x> = \int xp^2(x)dx = \alpha^2 \int xa^2(x)dx + \beta^2 \int xb(x)dx + 2 \alpha \beta \int xa(x)b(x)dx$

$= \alpha^2 <x>_a + \beta^2 <x>_b + 2 \alpha \beta \int xa(x)b(x)dx$

Moreover, if you add the time-dependent part of the wave-function you will find that, for example:

$<x> = \alpha^2 <x>_a + \beta^2 <x>_b + 2 \alpha \beta cos(\frac{E_a - E_b}{\hbar}t) \int x\psi_a(x)\psi_b(x)dx$

(assuming we have real-valued $\alpha, \beta, \psi$)

9. Dec 16, 2015

### Small bugs

Thanks for your answer. I understand it, but my mean problem is: why the this two events are not mutually exclusive?
In my case, I wanted to calculate the event that electron appears in the interval e.g. (0,1). But the system is in superposition state which is superposed by to eigenstate A and B. So I thought like this: in state A, the probability of the event that the electron appears there was PA. The same for in the state B, PB. So, now that, it is superposed situation if collapses to A, then wouldn't to B. In this case, they are exclusive to each other, aren't they?

10. Dec 16, 2015

### Orodruin

Staff Emeritus
If you want to use the probability distributions for A and B you must first measure which of these states the system is in and then make the position measurement.

This is not the same thing as directly measuring the position. The measurement affects the state.

11. Dec 16, 2015

In any case, even integrating over the whole domain, the addition you are doing only works for observables of which the states are eigenfunctions - normally energy. With an observable like position you'll get the extra cross term. See post $8. That's a mistake in addition to one that Orodruin is pointing out. 12. Dec 16, 2015 ### PeroK For example. You could wrongly argue that: The particle is in the state$\psi_A$25% of the time with$<x>_A$and in state$\Psi_B$75% of the time with$<x>_B$so for the superposition you would have$<x> = \frac{<x>_A}{4} + \frac{3<x>_B}{4}$But that fails to work. Mathematically it's because$|\Psi^2|$is the pdf for$x$and not$\Psi$. When calculating energy, the cross term disappears as$\psi_A$and$\psi_B$are eigenfunctions of the operator. So you do have:$<E> = \frac{E_A}{4} + \frac{3E_B}{4}##

13. Dec 16, 2015

### Small bugs

Thanks, I understand!! And I also feeling "bad" about measurement. How can my "eyeball" influences the system?
And if the wave function collapses after measurement, does the system stay the same state afterwards which means I measure it again, will I still get the same eigenstate 100%?

14. Dec 16, 2015

### Orodruin

Staff Emeritus
You are making a common mistake about what constitutes a measurement. It is not the fact that you see something which is a measurement, it is the interaction with the environment which leads to decoherence.

15. Dec 16, 2015

### Staff: Mentor

It doesn't. It only gets involved when and if light that has interacted with the system reaches your eyeball, and it's that interaction that makes the difference.

(It is somewhat unfortunate that for historical reasons the word "observation" is used, when "measurement" or "interaction" would be less confusing).

16. Dec 16, 2015

### Small bugs

So if the wave function collapses after measurement, if I measure it again, will I still get the same eigenstate 100%?
If we solve the time-dependent Schrodinger equation. After our measurement, will it be still time-dependent? But the "collapse" seems to make it into stationary state

17. Dec 16, 2015

### Heinera

If the time interval until you measure it again is extremely short, then you will get the same eiegenstate with close to 100% probability, yes.
It will still be time dependent, but the starting point will now be the eigenstate that you measured. And it will evolve from there like usual, so it will not be a stationary state. As time passes, you again increasingly introduce uncertainty about the eigenstate you will see upon the next measurement.

18. Dec 16, 2015

### Orodruin

Staff Emeritus
Well, this depends. If the measured state is also an eigenstate of the Hamiltonian, there will be no uncertainty introduced as time evolves.