Suppose we have an operator $Q$ with eigenvalue $q$. Expectation value is $\langle Q \rangle$ and dispersion $D(Q) = \sqrt{\langle \left( Q - \langle Q \rangle \right)^2 \rangle} $. I want to find values for the expectation value and dispersion.


Expectation value is simply the average value of the observable associated with $Q$ right? Dispersion is the variance of the observable associated with $Q$ right?

$$ \langle Q \rangle = \langle \psi | Q |\psi \rangle = q \langle \psi|\psi \rangle = q $$

$$ D(Q) = \sqrt{\langle \left( Q - \langle Q \rangle \right)^2 \rangle} = \sqrt{ \langle Q^2 -2Q \langle Q\rangle + \langle Q \rangle^2 \rangle } = \sqrt{ \langle Q^2 - 2Qq + q^2 \rangle } $$

Now $\langle Q^2 \rangle = q^2 $ by applying $Q$ twice.

Does this mean that $$D(Q) = 0 $$?


Yes. There's an even easier notation for dispersion, or standard deviation which is

$$D(Q) = \sqrt{ \langle Q^2\rangle - \langle Q\rangle^2 }$$

Both those terms are the same. So $D(Q)$ is zero.

  • $\begingroup$ No idea how to use the math notation on here yet. $\endgroup$ – jhobbie Jun 19 '14 at 16:57
  • $\begingroup$ Then what's the dispersion $D(x)$ of this wavefunction? $\left( \frac{2}{\pi a^2} \right)^{\frac{1}{4}} exp\left(-\frac{x^2}{a^2} + ikx \right)$ ? Surely it can't be zero. $\endgroup$ – user44840 Jun 19 '14 at 17:06
  • $\begingroup$ Is the above equation an eigenstate of x? $\endgroup$ – jhobbie Jun 19 '14 at 17:10
  • $\begingroup$ yes, it's an eigenstate of x $\endgroup$ – user44840 Jun 19 '14 at 17:10
  • $\begingroup$ If you do it out (I used integral notation in the position basis) it's a lot easier. In your question above, you made it seem as if Q did not depend on q. $\endgroup$ – jhobbie Jun 19 '14 at 17:16

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