# What does probability amplitude mean

1. Apr 22, 2012

### latnoa

I'm not exactly sure what it means when my book says that the wave function ψ(x,t) is a probability amplitude of knowing where the particle is in position x at time t.

2. Apr 22, 2012

### Staff: Mentor

It's simply a definition of the term "probability amplitude." In order to get the actual probability density, you have to multiply the probability amplitude by its complex conjugate:

$$P(x,t)= \Psi^*(x,t) \Psi(x,t)$$

"Probability amplitude" is more general than $\Psi$ because it also applies to other functions. For example, there is a momentum-space probability amplitude $\phi(x,t)$, for which $\phi^*(x,t) \phi(x,t)$ gives you the probability density for the particle's momentum.

3. Apr 23, 2012

### zhangyang

You can treat it as a field.
Or you can multiply it with electronic charge of electron,than the result is the distribution of charge in the space.

4. Apr 23, 2012

### Steely Dan

It means that if you attempt to locate the particle at position $x$ at time $t$, the value of the wave function tells you the probability that you will actually find it there.

More specifically, the wave function multiplied by its complex conjugate gives you this probability directly. So if the value of the wave function is 1/2 at some location at some time, then the interpretation is that if one out of every four times you repeat the experiment and look at that location, you should find the particle there.

5. Apr 23, 2012

### Naty1

latnoa: welcome to the club....beyond the mathematical descriptions lots of people have
had that problem....Post #2 has the math definition if that's what you were after.

If you are looking for interpretations of that matematics:

Try here for some good discussion and animations....
http://en.wikipedia.org/wiki/Wavefunction

but how a 'real' particle is represented by 'complex' numbers is where it gets really interesting.

Another view is this:

https://www.physicsforums.com/showthread.php?t=551554&page=2

6. Apr 23, 2012

### QuantumTaco

I could be wrong, but is your question why use a "probability amplitude", when it's not the actual probability? In other words, what is it actually good for?

7. Apr 23, 2012

### Staff: Mentor

The OP needs to come back and explain what he means by the word "means".

8. Apr 23, 2012

### latnoa

Thank you everyone. I've read more in detail this section but I will tell you my own way of explaining this and tell me if I'm wrong.

The wave function depends both on its position (x) and time (t). We can split the wave function into two parts, one that represents the time and the other one that represents the position.

ψ(x,t)= $\psi$(x) * f(t)

Furthermore if we split the schrodinger equation into it's time independent part we see that a function of time is equation to a function of position which also means that these functions equal a constant. If we set the function equal to that constant we get a differential equation and once we solve the differential equation we get

f(t) =e-iEt/h

Where E is the the constant. Furthermore if we can replace f(t) into the wave function.

ψ(x,t)= $\psi$(x) * e-iEt/h

We know that if we try to find where the particle is in a certain time we can do so by finding the integral of the wave function times its complex conjugate. Then we can say that the probability of finding the particle from negative infinity to infinity has to be 1

$\int$ψ(x,t)*ψ(x,t)= 1

Where we integrate from negative infinity to infinity. Then by integrating the f(t) cancels out( I don't know how) so we are just left with

$\int$|$\psi$(x)|2= 1

Which means that the probability of the position for the particle is time independent.

Edit: Sorry I didn't understand what complex conjugates were or how we can use them to find the probability but I posted this to make sure I actually understand it now.

9. Apr 23, 2012

### Staff: Mentor

"Complex conjugate" means replace i with -i everywhere.

The f(t) cancels out even before you take the integral:

$$\Psi(x,t) = \psi(x)e^{-iEt/\hbar}$$
$$\Psi^*(x,t) = \psi^*(x)e^{+iEt/\hbar}$$
$$\Psi^*\Psi = \psi^*\psi e^{+iEt/\hbar}e^{-iEt/\hbar} = \psi^*\psi e^{+iEt/\hbar-iEt/\hbar} = \psi^*\psi e^0 = \psi^*\psi$$

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