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

  • $\begingroup$ What does "Show that $\sum_{\lambda}p^2_{\lambda} = 1$ to an observable $O$" mean? $\endgroup$ – Michael Brown Apr 8 '13 at 14:58
  • $\begingroup$ $p_{\lambda}$ is a probability of obtaining the eigenvalue $\lambda$ than result of measurement of the observable $O$ $\endgroup$ – 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 $\Delta O = 0$.

  • $\begingroup$ 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 $\endgroup$ – juaninf Apr 8 '13 at 18:46
  • $\begingroup$ 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 $\endgroup$ – Prathyush Apr 8 '13 at 19:09
  • $\begingroup$ I would like that you review this question physics.stackexchange.com/questions/60351/…, please. $\endgroup$ – juaninf Apr 8 '13 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.