# On wavefunctions

1. Nov 28, 2013

### Alpharup

In my electronics engineering program, we have topics on quantum mech, statistcal mech and so on(only intuitive treatment, not mathematical)....We follow'Introduction to modern physics' by Sir Arthur Beiser. I have a small doubt in the wavefunction of a system..
Let a system consist of two particles A and B and let their wave-fuctions by ψa and ψb respectively..Let ψc represent the wavefuction of the system..
In this book, it is given that
ψc is a product of ψa and ψb..
How is this possible?
We have learnt about the probability of existence of a particle by integrating the wave-function..Does this concept have anything to do with the wavefunction of the system..
Please explain me in an intuitive way without tough mathematics..

2. Nov 28, 2013

### jk22

The wavefunction of the system is not forcedly the product $$\Psi_a(x)\Psi_b(y)$$. It is the case only for separable systems, in general the wave-function of the system is a function of 2 coordinates set $$\Psi_c(x,y)$$.

3. Nov 28, 2013

### Jano L.

The reason is that in the probability theory, when we have no reason to believe that states of two systems A, B are correlated, the best choice for joint probability for the joint system is the product of the probabilities for the parts.

Then, the probability density for state where the particle $a$ is at $\mathbf r_a$ and the particle $b$ is at $\mathbf r_b$ is:

$$\rho_c(\mathbf r_a, \mathbf r_b) = \rho_a(\mathbf r_a) \rho_b(\mathbf r_b),$$

where $\rho_a$ describes probability density for the particle $a$ and so on.

Since in wave mechanics the probability densities for one-particle systems are given by the Born rule

$$\rho_a(\mathbf r_a) = |\psi_a(\mathbf r_a)|^2,$$
$$\rho_b(\mathbf r_b) = |\psi_b(\mathbf r_b)|^2,$$
$$\rho_c(\mathbf r_a, \mathbf r_b) = |\psi_c(\mathbf r_a, \mathbf r_b)|^2,$$

if we think the two systems are not correlated, we can derive

$$|\psi_c(\mathbf r_a, \mathbf r_b)| = |\psi_1(\mathbf r_a) \psi_2(\mathbf r_b)|.$$

Usually the absolute values are not required, either because we just need any two-particle function for construction of a basis and we just take the simplest one formally, or since it is assumed that the phase factor will not influence results. Then we arrive at function $\psi_c$ for joint system you have seen in Beiser.

To illustrate the difference between correlated and uncorrelated systems, let's use the symbol $\psi_1(\mathbf r_a)$ for psi function describing the system A and $\psi_2(\mathbf r_b)$ for the function describing the system B.

With these, we can construct psi function

$$\psi_c(\mathbf r_a, \mathbf r_b) = \frac{1}{\sqrt{2}} \left[ \psi_1(\mathbf r_a) \psi_2(\mathbf r_b) + \psi_1(\mathbf r_b) \psi_2(\mathbf r_a)|\right]$$

which we may use to describr the joint system. This function is not simple product of the two functions for particles, and if you calculate $\rho_a, \rho_b, \rho_c$ from the Born rule above, you will see that also the probability density for the joint system is not product of the corresponding one-particle probability densities. Such function $\psi_c$ describes correlated states, where the state of system A (position $\mathbf r_a$) is correlated with the state of the system B (position $\mathbf r_b$). Such psi functions come out as solutions to the Schoedinger equation

$$\hat H \psi_c = E\psi_c,$$

for the joint system, for example.