# Homework Help: Even parity => symmetric space wave function?

1. Mar 5, 2007

### mrandersdk

If I have af wavefunction that is a product of many particle wavefunctions

$$\Psi = \psi_1(r_1)\psi_2(r_2) ... \psi_n(r_n)$$

If I then know that the parity of $$\Psi$$ is even. Can I then show that the wavefunction i symmetric under switching any two particles with each other. That is

$$\psi_1(r_1)\psi_2(r_2) ...\psi_i(r_i) ... \psi_j(r_j) ... \psi_n(r_n) = =\psi_1(r_1)\psi_2(r_2) ...\psi_j(r_j) ... \psi_i(r_i) ... \psi_n(r_n)$$

for any i and j between 1 and n?

It may be used that the parity operator commutes with the hamilton of the system if nessesary, and that the interaction between the particles only depends on the distance between any two particles.

It is clear that if the system only consist of two particles, and we use one of the particles as the 0-point of our coordinatesystem, the parity operator does the same as changing the particles, and then even parity means even space function, but when n is greater than 2, I can't see it.

Hope someone understand what i'm asking, because the result is used frequently in my course of particle physics.

2. Mar 5, 2007

### Meir Achuz

For two particles, parity changes r_12 into -r_12, which has the same effect as interchanging r_1 and r_2. For your many body wave function this isn't true, so parity is unrelated to particle interchange. Parity changes all r_i to -r_i. You have to look at the detailed wave function to see what this does. It depends on how your WF relates all the coordinates.

3. Mar 5, 2007

### mrandersdk

ok that was my thought too. Let me give an example how it is used, because I would like to know how th author solves the excercise. It's from nuclear and particle physics by B.R. Martin.

Consider a scenario where overall hadronic wavefunction $$\Psi$$ consist of just spin and space part, i.e. $$\Psi = \psi_{space} \psi_{spin}$$. What would be the resulting multplet structure of the lowest-lying baryon states composed of u,d and s quarks?

The autors own solution:
'Low lying' implies that the internal orbital momentum between the quarks is zero. Hence the parity is P = +1 and space is symmetric. Since the Pauli principle requires the overall wavefunction to be antisymmetric under the interchange of any pair of like quarks, it follows that $$\psi_{spin}$$ is antisymmetric...

The rest of the excercise makes sense, but how does he conclude that the space part i symmetric, to me it seems like he use parity symmetric and symmetric under switching of particles.

4. Mar 5, 2007

### Meir Achuz

That is a three body case. If the three body WF is written using Jacobi coordinates, the state with Daliz angular momenta L and l both equal zero is spatially symmetric. This is most easily seen for a harmonic oscillator wave function. Martin has made a common (historic) mistake of assuming more about parity than is needed. There are positive parity three body states that are not spatially symmetric.

5. Mar 5, 2007

### mrandersdk

ok i never heard of Jacobi coordinates or Daliz angular momenta. I'm not sure what you are saying, is it correct of him to say that positive parity gives a symmetric WF under switching any pair of quarks ?

6. Mar 5, 2007

### Meir Achuz

No. There are many positive parity quark wave functions that are not spatially symmetric. The Jacobi coordinates are r=r1-r2, and rho=r3-(r1-r2)/2, but normalized with some square roots to make them more symmetric.
For three bodies there are only two independent angular momenta in the cm system. Dalitz introduced them to analyze three body systems. One is related to r1-r2, and the other to the rho coordinate.

7. Mar 5, 2007

### mrandersdk

thanks for your help. There are many faults in this book so sometimes it's hard to figure out what to believe, so you got to be critical.

I was wandering if the parity of a system is independent of what coordinatsystem you use? Or do you got to have a coordinatsystem before you find the parity?

8. Mar 5, 2007

### Meir Achuz

The parity is independent of the coord system, but it is easier to see in some systems. It is easiest if all independent angular momenta are known.