Exploring Quantum Physics - Week 1 Question 8

Question:

The variance $\sigma^2 = \langle(\hat{X} - \langle \hat{X} \rangle)^2 \rangle$ of an operator, $\hat{X}$, is a measure of how large a range its possible values are spread over (the standard deviation is given by $\sigma = \sqrt{\sigma^2}$). Suppose that $| X \rangle$ is an eigenstate of some operator $\hat{X}$, what is the variance of $\hat{X}$ in this state? You may assume that $|X\rangle$ is normalized ($\langle X | X\rangle = 1$).

Solution:

The real difficult part of this is notation and definitions. Let's start with "eigenstate". The statement $|X \rangle$ is an eigenstate of $\hat{X}$ means $\hat{X}| X \rangle = \lambda | X \rangle$ for some eigenvalue $\lambda$.

Next, we computes the innermost parenthesis, $\langle \hat{X} \rangle$. Recall that surrounding an operator using a pair of angle brackets means expectation, which means $\langle \hat{X} \rangle = \int\limits_{-\infty}^{\infty}{|X\rangle^* \hat{X} | X \rangle dx} = \int\limits_{-\infty}^{\infty}{|X\rangle^* \lambda | X \rangle dx} = \lambda$. The last equality follows because it is given that the eigenstate $| X \rangle$ is normalized.

Then we proceed to the variance. Again, the angle brackets means expectation. So the variance is given by

$\begin{eqnarray*} \sigma^2 &=& \langle(\hat{X} - \langle \hat{X} \rangle)^2 \rangle \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\hat{X} - \lambda)^2 | X \rangle dx} \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\hat{X}^2 - 2\lambda\hat{X} +\lambda^2) | X \rangle dx} \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\lambda^2| X \rangle - 2\lambda^2| X \rangle +\lambda^2| X \rangle) dx} \\ &=& 0 \end{eqnarray*}$

The last equality follows as the terms in the bracket cancel each other. I am surprised with this result. Does it mean if the state get into an eigenstate, a certain observable will not have any variances, it is hard to believe both questions leads to 0.