On wavefunctions

  • Thread starter Alpharup
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  • #1
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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..
 

Answers and Replies

  • #2
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The wavefunction of the system is not forcedly the product [tex]\Psi_a(x)\Psi_b(y)[/tex]. It is the case only for separable systems, in general the wave-function of the system is a function of 2 coordinates set [tex]\Psi_c(x,y)[/tex].
 
  • #3
Jano L.
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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?

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.
 

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