# Wave function

1. Jul 15, 2015

### Molar

Ψ (x,0)= ΣcnΨn
Ψ is a linear combination of all Ψn and Ψn are the bases.

What does this statement actually mean..?? I mean, does it says that Ψ(x,0) contains c1 part of Ψ1 and c2 part of Ψ2 and so on...like we represent a vector :
R= 3x+4y+7z...??

2. Jul 15, 2015

### RUber

Pretty much. That is all it means. A linear combination of the bases implies that there is a unique set of c_n coefficients that define Psi(x,0). Just like any point in R^3 can be uniquely defined by ax + by + cz. Where a, b, and c are real coefficients and x, y, z represent the basis set.

3. Jul 15, 2015

### jfizzix

Indeed, the wavefunction can be thought of as a vector, since it is one in a mathematical sense.

The coefficients $c_{n}$ can be thought of as the components of $\Psi(x,0)$ in the basis given by the different values of $n$.

4. Jul 15, 2015

### Staff: Mentor

Yes, provided that by x, y, and z you mean the unit vectors along the coordinate axes, which usually have a caret or other distinguishing mark on them: $\vec R = 3 \hat x + 4 \hat y + 7 \hat z$

5. Jul 16, 2015

### alemsalem

Yes just like any point in space is given by (x,y,z) any point in "Function space" is given by (c1,c2,c3, ..., cn,..)

6. Jul 17, 2015

### Molar

I'm sorry,didn't understand this. Aren't those coefficients related to some kind of probability of respective states...???

7. Jul 17, 2015

### alemsalem

Yes, |cn|^2 is the probability to find the state Ψn.
mathematically they also represent the coordinates of the function Ψ (x,0) in the Ψn basis.

8. Jul 18, 2015

### Molar

Thanks for the information..didn't know that..

Here I'm having another confusion but don't know what it is...can you help me please ..??
The particle is in the state ψ(x,t) which is a combination of all Ψn , not a particular Ψn . So, how cn can be the probability of finding a particular Ψn state because we won't find the particle in Ψn, but ψ(x,t) , right..??

9. Jul 18, 2015

### blue_leaf77

This is one of the underlying postulates of QM, called the probabilistic interpretation, which is not so bad to understand how QM works. For example you can try calculating $\langle \psi(x,t) | \psi(x,t) \rangle$ for normalized $|\psi(x,t) \rangle$, you will find that the sum of all $|c_n|^2$ equals unity, that's how people agree that $|c_n|^2$ can be interpreted as a probability to find the corresponding state upon a measurement.
We should find the state after measurement in one of the $\psi_n$'s. By saying "we find", quantum mechanically means that we have conduct some measurement of whatever observable. $\psi(x,t)$ is the state right before the measurement. While $\psi_n$, one of the eigenstates of observable we are measuring, is the state right after the measurement.

10. Jul 22, 2015

### Staff: Mentor

Exactly.

It simply means the so called pure states form a vector space. Although its not pointed out in beginner, or even many intermediate texts, a state in QM is really a positive operator of unit trace, not an element of a vector space. But a certain subset, called pure, can be mapped onto a vector space and they are generally what most people mean when they talk about states. So really superpositions are a bit tautological. The principle of superposition only applies to pure states. What are pure states? Those that can be mapped to a vector space, which, by the definition of vector space, obeys the principle of superposition. Its rather strange, but still true, in mathematical physics things at a deep level often fall apart to some extent. Another interesting example in Noethers Theorem, and what energy and momentum is, and why its conserved - but that's another story for another time.

If you would like to see where states come from and what they are in a formal mathematical sense check out post 137:

The key is non-contextuality. Again its an example of how things sometimes fall apart when looked at deeply.

Thanks
Bill

11. Jul 22, 2015

### stevendaryl

Staff Emeritus
To say that a state is an operator is incomplete, if you don't say what it operates on, right? And normally, what it operates on are elements of the Hilbert space, which are the pure states. So it seems to me that you need the pure states in order to make sense of the more general mixed states.

Or maybe the whole thing can be done algebraically, where instead of talking about operators as functions on a Hilbert space, you can let "operators" be primitive terms with a certain multiplication?

12. Jul 22, 2015

### Staff: Mentor

Yes and no. You can interpret them as matrices.

But the theorem I proved in my link had them operating on a vector space without any interpretation on what the means. Its only after the proof of the Born rule you can interpret pure states that way.

Thanks
Bill