# Measurement and probability for quantum states

Suppose that the physical system is in generic state $|\psi\rangle$. Show that $\sum_{\lambda}p^2_{\lambda} = 1$ to an observable $O$, if and only if $\Delta O = 0$. ($\Delta O$ is a standard deviation).

Using the probability measure definition $p_{\lambda} = \langle\psi|P_{\lambda}|\psi\rangle$. Using hypothesis, $$\sum_{\lambda}\langle\psi|P_{\lambda}|\psi\rangle^2 = 1,$$ the fact that $\sum_{\lambda}\langle\psi|P_{\lambda}|\psi\rangle = 1$ and the definition $\Delta O = \sqrt{\langle O^2\rangle - \langle O\rangle^2 }$.

I tried $$\langle O^2\rangle - \langle O\rangle^2 = \langle\psi|O^2|\psi\rangle - (\sum_{\lambda}\lambda\langle\psi|P_{\lambda}|\psi\rangle)^2,$$ but dont know what else to do

• What does "Show that $\sum_{\lambda}p^2_{\lambda} = 1$ to an observable $O$" mean? – Michael Brown Apr 8 '13 at 14:58
• $p_{\lambda}$ is a probability of obtaining the eigenvalue $\lambda$ than result of measurement of the observable $O$ – juaninf Apr 8 '13 at 15:10

I assume all the eigenvalues are distinct, or there will be some issues.

The standard deviation is zero, if and only if the probability is peaked at one value, i.e $p_i=1$ for a single of $i$ and zero for the others.(try to prove this )

Now if the probability is peaked at one value, it clearly implies that, $\Sigma_\lambda p_\lambda^2 = 1$

Therefore $\Delta O = 0$ implies $\Sigma_\lambda p_\lambda^2 = 1$

Now the other way.

Consider $(\Sigma_\lambda p_\lambda)^2 - \Sigma_\lambda p_\lambda^2 = 0$. Since $p_\lambda \geq 0$, This equality implies only one of the $p_\lambda$'s are non zero. Therefore it is a peaked distributive, with $p_i = 1$ for a single $i$ and zero for the others. Therefore standard deviation is zero.

Therefore $\Sigma_\lambda p_\lambda^2 = 1$ implies that $\Delta O = 0$.

• Thank by your reply, I dont understand Why $(\Sigma_\lambda p_\lambda)^2 - \Sigma_\lambda p_\lambda^2 = 0$ implies only one of the $p_{\lambda}$ are non zero. Resolving $$(\Sigma_\lambda p_\lambda)^2 - \Sigma_\lambda p_\lambda^2 = 0 \iff \Sigma_\lambda p_\lambda^2 - \Sigma_\lambda p_\lambda^2 + 2(\lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \cdots)=0$$ And $2(\lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \cdots)$ never is zero – juaninf Apr 8 '13 at 18:46
• if 2 terms are non zero say $\lambda_1$ and $\lambda_2$, then the 2 $\lambda_1$ $\lambda_2$ term will be greater than zero, thus causing a contradiction. However if all but one term is zero, then the $\Sigma \lambda_i$ $\lambda_j$(i not equal to j ) will be identically zero – Prathyush Apr 8 '13 at 19:09
• I would like that you review this question physics.stackexchange.com/questions/60351/…, please. – juaninf Apr 8 '13 at 23:57