# When uncertainty is calculated in non normalized eigenstates it doesn't give zero. Shouldn't it give zero?

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.

• Why do you think it should be zero if the eigenfunctions are not normalized? – garyp Jan 26 '16 at 19:45
• If a state is unnormalized then $\langle {\cal O}\rangle_\Psi = \frac{ \langle \Psi | {\cal O} | \Psi \rangle }{ \langle \Psi | \Psi \rangle }$ – Prahar Jan 26 '16 at 19:46
• Shoudn't the uncertainty be zero? @garyp – Jolie Jan 26 '16 at 19:46
• 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. – 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