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Expected value of operators

  1. Apr 19, 2012 #1
    Often in quantum mechanics, there appears statements of the type :

    Expected value of operator = a value

    I am told that operators are instructions and I do not understand how an instruction can have a value, expected or otherwise. Even in the case where the operator is of the form "muliply the argument by the value of variable x", this is not the same as the value of x per se. In other cases the incompatability is starker. Can someone please resolve this problem, preferably in words, because the explanations I have seen to date seem to make the problem disappear by making a symbol have diffeent meanings in different locations.
  2. jcsd
  3. Apr 19, 2012 #2


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    If you already have some background in QM, I'd recommend to study the postulates of QM. There, the connection between the mathematical concepts and the physical meaning are made.

    In QM, an operator A is an object which acts on a state vector |ψ>. The result of A|ψ> is again a state vector, |ψ'>. The expectation value of A in the state |ψ> is the inner product <ψ|ψ'>, so this is a number. If you have trouble with the notation, replace |ψ> by your favourite vector notation, maybe v. If you are not familiar with vectors and inner products, you should probably learn some linear algebra first.
  4. Apr 19, 2012 #3


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    I suspect that when you were told "operators are instructions", they meant that operators are instructions to do some specific thing to a wave function so that you get a value. The "expected value" of such an operator is the average of all possible values.
  5. Apr 20, 2012 #4
    Thanks,KITH and PF MENTOR. Of course I agree that when an operator acts on something the result may have a value. But what disturbs me is the way in which a symbol is used sometimes to represent a property and at other times to represent the corresponding operator,in the same problem.
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