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On wavefunctions

  1. Nov 28, 2013 #1
    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. jcsd
  3. Nov 28, 2013 #2
    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].
     
  4. Nov 28, 2013 #3

    Jano L.

    User Avatar
    Gold Member


    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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