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A doubt on the meaning of the bra ket product

  1. Mar 15, 2015 #1
    Hello everyone, I have thi doubt:
    If I have a state, say psi1, associated with the energy eigenvalue E1, the integral over a certain region gives me the probability of finding the particle in that region with the specified energy E1. Now if I put an operator between the states I obtain its mean value of the observable. If now I have two different states psi1 and psi2, associated with the two energy E1 and E2, what does their product mean? And what's the meaning if I put inside an operator?
     
  2. jcsd
  3. Mar 15, 2015 #2

    bhobba

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    What do you mean by product? Do you mean different states in different systems or a superposition in the same system?

    Thanks
    Bill
     
  4. Mar 15, 2015 #3
    I mean the hermitian product of two states. For example consider the hamiltonian eigenstates of the harmonic oscillator, what does it mean that the product of two different states is zero?
     
  5. Mar 15, 2015 #4

    bhobba

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    Obviously it means they are orthogonal - but I think you want a bit more than that.

    Ok we need to go to the Born Rule. Given a state |u> the expected value of observing it with an observable O is E(O) = <u|O|u>. Now suppose we have two orthogonal states |b1> and |b2>. Consider the following observable O = |b1><b1| which is the observable that gives 1 if the system gives the state |b1> after observation (the 1 is interpreted as a yes, otherwise a no, and its easy to see this is the probability it will be in state |b1>) and apply the Born Rule to each state. E(O) = <b1|b1><b1|b1> = 1. So when it is observed with O it will always be in state |b1> after observation and always give a yes. In fact the state is not altered by the observation. Now apply it to |b2> and you get <b2|b1><b1|b2> = 0. Thus the state will always give no. The state of the system is also unchanged after. Similarly if you use the observable |b2><b2|.

    What orthogonal states imply is we can find an observable that will give the probability of the system being in one of those states after observation and a yes. If its in that state it will always give a yes and not change the state. If its in the other it will never give a yes and not change the state.

    Thanks
    Bill
     
  6. Mar 15, 2015 #5
    And what happen if the system is in the superposition of b1 and b2, say: |u> = a|b1> + c|b2>
     
  7. Mar 15, 2015 #6

    bhobba

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    Then go back to the Born Rule. Probability of it being in |b1> after the observation is (a*<b1| + c*<b2|)|b1><b1|(a|b1> + c|b2>) = |a|^2. You can work out |b2>.

    Thanks
    Bill
     
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