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Why does the system has lower energy if its wave function is symmetric?

  1. Jan 27, 2009 #1
    Hi all:
    I am confused that in general case, if [H,p]=0 (where H is Hamiltonian of system and P is parity operator), system wave function is either symmetric or antisymmetric. How do we know that system is in lower energy state if its wave function is symmetric by comparing that system is described by antisymmetric wave function? What I said is true or not? If true, what is physical significance behind?


  2. jcsd
  3. Jan 27, 2009 #2
    It is not generally true that symmetric (wrt parity) wave-functions always have lower energy than anti-symmetric ones. However, there is a "node-counting theorem" (at least for bound states) which basically states the energies of the states are ordered according to the number of nodes in the wave-function. I.e. a state with a wave-function with n nodes has a lower energy than a state with n+1 nodes and so on. Since any anti-symmetric wave-function must have at least one node there usually (always?) exists a symmetric wave-function with zero nodes which has the lowest energy.

    At least this is what I can remember, please correct me if I am wrong (I'm too lazy to look it up)
    Last edited: Jan 27, 2009
  4. Jan 28, 2009 #3

    thanks for your answer. By the way, where can I find node-counting theorem? which textbook talk about this theorem?
  5. Jan 28, 2009 #4
    Unfortunately I don't know a good reference discussing this. It seems to be very overlooked in QM books. Probably it is more likely to find this in some book on differential equations. Some googling led me to this http://www.emis.de/journals/AM/98-1/kusano.ps. Not sure if it helps you.

    Many people refer to this theorem without giving any references. I remember having a very hard time finding a proof of this theorem and now I don't remember where I found it. Good luck!
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