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Meir Achuz

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The probabiliity of the particle being anywhere must be one.

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I know about it, but I have seen using normalised wavefunction in calculating the expectation only so?

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SpectraCat

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The measurement postulate defines the expectation value of an arbitrary Hermitian operator A for an arbitrary wavefunction [tex]\Psi[/tex] as:

[tex]<A>=\frac{<\Psi|\hat{A}|\Psi>}{<\Psi|\Psi>}[/tex]

this is valid for all wavefunctions, including unnormalized ones. If the wavefunction is normalized, the integral in the denominator is just 1, so you only need the numerator.

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jtbell

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Of course, the denominator is just the normalization integral. So one way or another, you end up normalizing the wave function. Either you or somebody else does it beforehand, or you do it as part of finding the expectation value.

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To take the expectation value of an operator you first act on the function (quantum state) with the operator, which gives you a new function. Then you dot the new function with the original function to see how much of the new function is linear with the original function. If you think of the functions as vectors this is how much the new vector lies along the original vector.

However the dot product could have any amplitude if the function is not normalized, which is not of much interest. If we then normalize the result or start with a normalized function then a result of 1 means the new vector is linear with the original vector and 0 means it is orthogonal. And everything in between 0 and 1 means there is a percentage that is linear and orthogonal and we know exactly what that is.

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