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.
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.
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