3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ What does "Show that $\sum_{\lambda}p^2_{\lambda} = 1$ to an observable $O$" mean? $\endgroup$
    – Michael
    Apr 8, 2013 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, 2013 at 15:10

1 Answer 1

1
$\begingroup$

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 $\Delta O = 0$.

$\endgroup$
3
  • $\begingroup$ 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, 2013 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, 2013 at 19:09
  • $\begingroup$ I would like that you review this question physics.stackexchange.com/questions/60351/…, please. $\endgroup$
    – juaninf
    Apr 8, 2013 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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