Prove equivalence of the definition for coherent states

Question

Quote:

The coherent states $|z\rangle$ is defined as $$|z\rangle =e^{-|z^2|/2}\sum_{n=0}^\infty\frac{z^n}{\sqrt{n!}}|n\rangle =e^{-|z^2|/2} e^{a^\dagger z}|n\rangle,$$

which was, very understandable and mathematically easy.

However, I encountered an article that using translation operator to define the coherent states, See reference here page 28. Quote:

$$|\tilde{x}_0\rangle \equiv T_{x_0}|0\rangle=e^{-\frac{i}{\hbar}\hat{p}x_0}|0\rangle.\tag{4.14}$$

But this has suddenly be somewhat confusing, the former stated that coherent states is a superposition of infinite states of probability amplitude distribution of Gaussian wave pack. The later sates that coherent states is a spacial translation of $|0\rangle$ eigenstates.

Further, the first definition involve only $\hat{a}^\dagger$ while the latter involved $\hat{p}$, a linear combination of $\hat{a}$ and $\hat{a}^\dagger$.

Could you explain to me what's going on? Especially, what's the inituition between define coherent states from $T_{x_0}$? How to prove they are same or not the same?

Answer 1

Let's recall a simple but very powerful trick: $e^{k\hat{a}}|0\rangle=\big(1+k\hat{a}+...\big)|0\rangle=|0\rangle$ where $k$ is a scalar, $\hat{a}$ is the annihilation operator, and $|0\rangle$ is the ground state. You see, when we open the bracket, only the first term survives because the rest of them just act on the ground state and annihilate it. So, we have the identity

$e^{k\hat{a}}|0\rangle=|0\rangle$

We will use this in a moment. So, we can say, using the definition of the ladder operators, that $$e^{i\hat{p}x_0}|0\rangle=e^{iCx_0(\hat{a}^\dagger-\hat{a})}|0\rangle$$where $C$ is the usual scalar constant coming out of the definition of the ladder operators, a known expression of $m,\omega$. Now, using the BCH formula and the fact that $[a, a^\dagger]=1$, we can write $$e^{iCx_0\hat{a}^\dagger}e^{-iCx_0\hat{a}}=e^{iCx_0(\hat{a}^\dagger-\hat{a})-\frac{C^2x_0^2}{2}}=e^{iCx_0(\hat{a}^\dagger-\hat{a})}e^{-\frac{C^2x_0^2}{2}}$$Thus, we finally write $$e^{i\hat{p}x_0}|0\rangle=e^{iCx_0(\hat{a}^\dagger-\hat{a})}|0\rangle=e^{\frac{C^2x_0^2}{2}}e^{iCx_0\hat{a}^\dagger}e^{-iCx_0\hat{a}}|0\rangle=e^{\frac{C^2x_0^2}{2}}e^{iCx_0\hat{a}^\dagger}|0\rangle$$where we used the identity $e^{k\hat{a}}|0\rangle=|0\rangle$ in the last step.

Now, if you plug in the exact expression for $C$ then you should get the expected equality.

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Notice that you have misquoted one result in your question. If you can verify, you will see that $$|z\rangle =e^{-|z^2|/2}\sum_{n=0}^\infty\frac{z^n}{\sqrt{n!}}|n\rangle =e^{-|z^2|/2} e^{a^\dagger z}|n\rangle$$ is incorrect, the correct result is $$|z\rangle =e^{-|z|^2/2}\sum_{n=0}^\infty\frac{z^n}{\sqrt{n!}}|n\rangle =e^{-|z|^2/2} e^{a^\dagger z}|0\rangle$$ and this is what you'd find equivalent to the other definition according to the method I described. But I raised this point rather than simply editing your question because it is interesting to point out that in order to replace the whole series with a sum over $n$ with just the zeroth term in the last step in the quoted result, they have used the same identity $e^{k\hat{a}}|0\rangle=|0\rangle$. So, it's a pretty useful identity.

Answer 2

Note that later in the text on p. 34 the formula (4.54) explains exactly how Zwiebach's coherent states $|\tilde{x}_0\rangle$ are related to the standard coherent states.

The intuition is that actual position-eigenstates does not belong to the Hilbert/Fock space, so instead we roughly speaking consider Gaussian wavepackets, centered around $x_0$.