Suppose $\Psi$ is an eigenstate of observable $\text H$ with eigenvalue $E_1$. Then uncertainty in the value of $\text H$,
$(\Delta E)^2=\langle E^2\rangle-\langle E\rangle^2$ which gives,
$(\Delta E)^2=E_1^2\bigg(\langle\Psi_1|\Psi_1\rangle-\big|\langle\Psi_1|\Psi_1\rangle\big|^2\bigg)$
If the eigenstate is not normalized then the right hand side is not zero. But the measurement of $\text H$ on $\Psi$ must yield $E_1$ with no uncertainty.

  • $\begingroup$ Why do you think it should be zero if the eigenfunctions are not normalized? $\endgroup$ – garyp Jan 26 '16 at 19:45
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    $\begingroup$ If a state is unnormalized then $\langle {\cal O}\rangle_\Psi = \frac{ \langle \Psi | {\cal O} | \Psi \rangle }{ \langle \Psi | \Psi \rangle }$ $\endgroup$ – Prahar Jan 26 '16 at 19:46
  • $\begingroup$ Shoudn't the uncertainty be zero? @garyp $\endgroup$ – Jolie Jan 26 '16 at 19:46
  • $\begingroup$ Normalizing a wave function is (effectively) multiplying the wave function by a scalar term: $\psi_{N}=A\psi_{UN}$. Not sure how/why its absence would lead to a zero eigenstate. $\endgroup$ – Kyle Kanos Jan 27 '16 at 11:29

It is part of the postulates of quantum mechanics that the expectation value of the observable corresponding to the hermitian operator $A$ in the normalized state $|\psi\rangle$ is given by $\langle A\rangle_\psi =\langle\psi|A|\psi\rangle$. Alternatively, you can postulate that the expectation value is given by $\langle A \rangle_\psi = \frac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}$. See, for example, the Dirac--von Neumann axioms

Either way the normalization is necessary, because otherwise you have to give physical meaning to the normalization.

Edit: So the reason your calculation is giving a strange result is you are calculating expectation values wrong.


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