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This question already has an answer here:

The coherent state is defined such that $a|\alpha\rangle =\alpha|\alpha\rangle $.

We can calculate the uncertainty using

$$\sqrt{\langle x^2\rangle-\langle x\rangle ^2}\sqrt{\langle p^2\rangle-\langle p\rangle ^2}$$

Substituting $$x=\frac{1}{\sqrt{2}}(a+a^\dagger)$$ and $$p=i\frac{1}{\sqrt{2}}(a-a^\dagger)$$ with $\omega=\hbar=m=1$ we find: $$x^2=\frac 1 2 (a+a^\dagger)(a+a^\dagger)=\frac 1 2 (aa+aa^\dagger+a^\dagger a+a^\dagger a^\dagger)$$

$$p^2=-\frac 1 2 (a-a^\dagger)(a-a^\dagger)=\frac 1 2 (-aa+aa^\dagger+a^\dagger a-a^\dagger a^\dagger)$$

$$\langle \alpha| x| \alpha\rangle =\frac 1 {\sqrt 2}(\alpha+\alpha^*)=\sqrt 2R(\alpha)$$ $$\langle \alpha| x| \alpha\rangle^2 =2R^2(\alpha)$$ $$\langle \alpha| x^2| \alpha\rangle =(R^2(\alpha)-I^2(\alpha)+|\alpha|^2)$$

$$\langle \alpha| p| \alpha\rangle =\frac i {\sqrt 2}(\alpha-\alpha^*)=-\sqrt 2 I(\alpha)$$ $$\langle \alpha| p| \alpha\rangle^2 = 2 I^2(\alpha)$$ $$\langle \alpha| p^2|\alpha\rangle =(I^2(\alpha)-R^2(\alpha)+|\alpha|^2)$$

$$\sqrt{\langle x^2\rangle-\langle x\rangle^2}\sqrt{\langle p^2\rangle-\langle p\rangle^2}=0\ne\frac 1 2$$ Heisenberg's uncertainty principle states that this should be more than a half.

Where is the problem with this derivation?

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marked as duplicate by ZeroTheHero, Jon Custer, Kyle Kanos, Cosmas Zachos, Bill N Jan 31 at 20:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I think the mistake is in your calculation of $\langle \alpha| x^2|\alpha\rangle$ and $\langle \alpha| p^2|\alpha\rangle$.

$$ \begin{align} \langle \alpha| x^2 |\alpha\rangle =& \langle \alpha| \frac 1 2 \left(aa+aa^\dagger+a^\dagger a+a^\dagger a^\dagger\right)|\alpha\rangle \\ =& \frac 1 2 \left(\langle \alpha|aa|\alpha\rangle + \langle \alpha|aa^\dagger|\alpha\rangle + \langle \alpha|a^\dagger a|\alpha\rangle + \langle \alpha|a^\dagger a^\dagger|\alpha\rangle \right) \end{align} $$

Here, with the help of $a|\alpha\rangle = \alpha|\alpha\rangle$, $\langle\alpha |a^\dagger = \alpha^* \langle\alpha |$, and $\langle\alpha | \alpha\rangle = 1$, you could simplify the first, third and forth term.

$$ \begin{align} \langle \alpha|aa|\alpha\rangle &= \alpha^2 \\ \langle \alpha|a^\dagger a|\alpha\rangle &= \alpha^* \alpha \\ \langle \alpha|a^\dagger a^\dagger|\alpha\rangle &= \alpha^{*2} \end{align} $$

However, it is not possible to simplify the second term $\langle \alpha|aa^\dagger|\alpha\rangle$ in a similar way, because $|\alpha\rangle$ is not an eigenvalue of $a^\dagger$.

The same reasoning applies for calculating $\langle \alpha| p^2|\alpha\rangle$.

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As Thomas pointed out, we can't operate $a^\dagger$ on $|\alpha\rangle $ so we need to keep the second term of $\langle x^2\rangle $. The full derivation leads us to: $$x^2=\frac 1 2 (a+a^\dagger)(a+a^\dagger)=\frac 1 2 (aa+aa^\dagger+a^\dagger a+a^\dagger a^\dagger)$$

$$p^2=-\frac 1 2 (a-a^\dagger)(a-a^\dagger)=\frac 1 2 (-aa+aa^\dagger+a^\dagger a-a^\dagger a^\dagger)$$

$$\langle \alpha| x| \alpha\rangle =\frac 1 {\sqrt 2}(\alpha+\alpha^*)=\sqrt 2R(\alpha)$$ $$\langle \alpha| x| \alpha\rangle^2 =2R^2(\alpha)$$ $$\langle \alpha| x^2| \alpha\rangle =(R^2(\alpha)-I^2(\alpha)+\frac 1 2 |\alpha|^2+\frac 1 2 \langle \alpha|aa^\dagger|\alpha\rangle )$$

$$\langle \alpha| p| \alpha\rangle =\frac i {\sqrt 2}(\alpha-\alpha^*)=-\sqrt 2 I(\alpha)$$ $$\langle \alpha| p| \alpha\rangle^2 = 2 I^2(\alpha)$$ $$\langle \alpha| p^2|\alpha\rangle =(I^2(\alpha)-R^2(\alpha)+\frac 1 2 |\alpha|^2+\frac 1 2 \langle \alpha|aa^\dagger|\alpha\rangle ))$$

$$\sigma_x\sigma_y=\sqrt{\langle x^2\rangle-\langle x\rangle^2}\sqrt{\langle p^2\rangle-\langle p\rangle^2}=\frac 1 2 \langle \alpha|aa^\dagger|\alpha\rangle-\frac 1 2 |\alpha|^2=\frac 1 2 \langle \alpha|aa^\dagger|\alpha\rangle-\frac 1 2\langle \alpha|a^\dagger a|\alpha\rangle$$

Since we have $[a,a^\dagger]=1$, which can be easily verified from the canonical commutation relation, we find the uncertainty to be

$$\sigma_x\sigma_y=\frac 1 2 \langle \alpha|\alpha\rangle [a,a^\dagger]=\frac 1 2$$

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All coherent states are obtained when the ground states of the harmonic oscillator is operated with an unitary operator called displacement operator and they lie in the middle of the minimum uncertainty curve where the uncertainty product is 1/2.

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