Is the vacuum state a coherent state?

Question

I'm asking because I got introduced to the state $|0\rangle$ as a fock-state. Nevertheless: $$ \hat{a} |0\rangle = 0 |0 \rangle $$ It is an eigenstate of $\hat{a}$ with eigenvalue $0$, and it can be obtained the same way any other coherent states are obtained via the displacement operator with parameter 0: $$ \hat{D}(\alpha=0)|0 \rangle = e^{0 \hat{a}^\dagger - 0 \hat{a}}|0\rangle = |0\rangle $$

Would one consider the vacuum state a coherent state?

Answer 1

The coherent state $\vert \alpha\rangle$ is just a vacuum state $\vert 0\rangle$ translated in $x$ and $p$ space so $\alpha=x_0+ip_0$. Thus the vacuum state is a coherent state that has not been displaced, i.e. $x_0=p_0=0$.

In fact, a nice way to see this is in the Wigner function formalism. The vacuum state is just a Gaussian sitting at the centre of $(x,p)$ space whereas a coherent state is the same state displaced to another point. This is illustrated in the figures below, taken from this site: on the left is the Wigner function of the vacuum state, and on the right that of a coherent state.

Note also that the Wigner function for the coherent state is everywhere positive, and positivity of the Wigner function is sometimes taken as a marker of classicality so in this sense coherent states (and the vacuum state) are "classical states".

A short movie illustrating the time evolution of the Wigner function of a coherent state can be found on the coherent state wikipage; it shows the Wigner function does not deform and remains non-negative at all times Of course since the vacuum state is an eigenstate of the Hamiltonian and lies at the centre of $(x,p)$, its Wigner function would actually remain there at all times.

Answer 2

Coherent state is a superpostition of states with different particle number with a weight of Poisson distribution. In other words, $$|\alpha\rangle =e^{-{|\alpha|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle $$ where $|n\rangle$ is a state with $n$ number of particles.

Thus when you say that my system is in coherent state, you mean that your system does not have a definite particle number. This is why if you remove a particle from that state with $$a|\alpha\rangle=\alpha|\alpha\rangle$$ you dont change the state. That's why coherent state is an eigenstate of annihilation operator.

But the reason that vacuum is an eigenstate of annihilation op is; it does not have any particle so there is nothing to annihilate. that's why it does not change. So in this sense, I would not say vacuum is a coherent state because it has a definite particle number $0$. and the reason of this similarity under the act of annihilation operator is due to different reasons.

Answer 3

The ordinary coherent states may be generated in several ways: (1) as minimum uncertainty states, (2) as eigenstates of the annihilation operator, and (3) as states displaced from the ground state via the operator $D (\alpha ) =e^{\alpha a^{\dagger}-\alpha^{*} a}$, where $\alpha$ is a complex number. The states generated by these three methods are equivalent. I will show how they are equivalent, and by their equivalence, it is enough to show that the ground state $|0\rangle$ is a minimum uncertainty state.

Proof: \begin{align} x^{2} &=\frac{\hbar}{2 m \omega}\left(a+a^{\dagger}\right)^{2} \\ p^{2} &=-\frac{m \omega \hbar}{2}\left(a-a^{\dagger}\right)^{2} \end{align}

Using the second definition of Coherent state (as eigenstates of the annihilation operator), \begin{align} a|\alpha\rangle=\alpha|\alpha\rangle \end{align} which implies $\left\langle\alpha\left|a^{\dagger} a\right| \alpha\right\rangle=|\alpha|^{2}$. It is trivial to check that this indeed defines a minimal wave-packet

\begin{align} \begin{array}{r} \left\langle\alpha\left|\left(a+a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha+\alpha^{\star}\right) \\ \left\langle\alpha\left|\left(a-a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha-\alpha^{\star}\right) \\ \left\langle\alpha\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha+\alpha^{\star}\right)^{2}+1 \\ \left\langle\alpha\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha-\alpha^{\star}\right)^{2}-1 \end{array} \end{align}

from which follows

\begin{align} \left\langle(\Delta x)^{2}\right\rangle_{\alpha} &=\left\langle x^{2}\right\rangle_{\alpha}-\langle x\rangle_{\alpha}^{2}=\frac{\hbar}{2 m \omega} \\ \left\langle(\Delta p)^{2}\right\rangle_{\alpha} &=\left\langle p^{2}\right\rangle_{\alpha}-\langle p\rangle_{\alpha}^{2}=\frac{\hbar m \omega}{2} \end{align}

and accordingly

\begin{align} \left\langle(\Delta x)^{2}\right\rangle_{\alpha}\left\langle(\Delta p)^{2}\right\rangle_{\alpha}=\frac{\hbar}{4} \end{align}

So the states $|\alpha\rangle$ satisfy the minimum uncertainty relation.

In the $|n\rangle$ base the coherent state looks like: \begin{align} |\alpha\rangle=\sum_{n} c_{n}|n\rangle=\sum_{n}|n\rangle\langle n \mid \alpha\rangle \end{align} Since \begin{align} |n\rangle=\frac{\left(a^{\dagger}\right)^{n}}{\sqrt{n !}}|0\rangle \end{align} we have \begin{align} \langle n \mid \alpha\rangle=\frac{\alpha^{n}}{\sqrt{n !}}\langle 0 \mid \alpha\rangle \end{align} and thus \begin{align} |\alpha\rangle=\langle 0 \mid \alpha\rangle \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle \end{align} The constant $\langle 0 \mid \alpha\rangle$ is determined by normalization as follows: \begin{align} 1=\sum_{n}\langle\alpha|n\rangle \langle n| \alpha\rangle=|\langle 0 \mid \alpha\rangle|^{2} \sum_{m=0}^{\infty} \frac{|\alpha|^{2 m}}{m !}=|\langle 0 \mid \alpha\rangle|^{2} e^{|\alpha|^{2}} \end{align} solving for $\langle 0 \mid \alpha\rangle$ we get: \begin{align} \langle 0 \mid \alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} \end{align} up to a phase factor. Substituting, we obtain the final form: \begin{align} |\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle \end{align}

Now,

\begin{align} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle=\sum_{n=0}^{\infty} \frac{\alpha^{n}}{n !}\left(a^{\dagger}\right)^{n}|0\rangle=e^{\alpha a^{\dagger}}|0\rangle \end{align}

which implies \begin{align} |\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} e^{\alpha a^{\dagger}}|0\rangle . \end{align}

Due to $a|0\rangle=0$, we have $e^{-\alpha^{*}a}|0\rangle=|0\rangle$. Thus, the above equation can be written as

\begin{align} |\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} e^{\alpha a^{\dagger}} e^{-\alpha^{*} a}|0\rangle . \end{align}

Using the Baker-Campbell-Hausdorff $(\mathrm{BCH})$ formula for any two operators $A$ and $B$ which commute with the commutator of $A$ and $B$ $$ e^{-\frac{1}{2}[A, B]} e^{A} e^{B}=e^{A+B} $$ and the relation \begin{align} \left[\alpha a^{\dagger}, \alpha^{*} a\right]=|\alpha|^{2}, \end{align}

we arrive at, \begin{align} |\alpha\rangle = e^{\alpha a^{\dagger}-\alpha^{*} a}|0\rangle = D(\alpha)|0\rangle \end{align}

and $|\alpha \rangle $ is a “displaced vacuum state”. $D(\alpha)$ is the displacement operator.

So, we showed $(2) \equiv (1) $ and $(2) \equiv (3)$.

Now to show that the ground state $|0\rangle$ is a minimum uncertainty state:

Since

\begin{align} \left\langle 0\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| 0\right\rangle &=\left\langle 0\left|a a^{\dagger}\right| 0\right\rangle=1 \\ \left\langle 0\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| 0\right\rangle &=-\left\langle 0\left|a a^{\dagger}\right| 0\right\rangle=-1 \end{align}

it follows that

\begin{align} \left\langle x^{2}\right\rangle_{0}\left\langle p^{2}\right\rangle_{0}=-\frac{\hbar^{2}}{4} 1(-1)=\frac{\hbar^{2}}{4} \end{align}

and finally, since $\langle x\rangle_{0}=\langle p\rangle_{0}=0$, it follows that

\begin{align} \left\langle(\Delta x)^{2}\right\rangle_{0}\left\langle(\Delta p)^{2}\right\rangle_{0}=\frac{\hbar^{2}}{4} \end{align}

But note that $|n\rangle$ is not a minimum uncertainty state.

\begin{align} \left\langle n\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| n\right\rangle=\left\langle n\left|a a^{\dagger}+a^{\dagger} a\right| n\right\rangle=\left\langle n\left|2 a^{\dagger} a+\left[a, a^{\dagger}\right]\right| n\right\rangle=2 n+1 \end{align}

and similarly

\begin{align} \left\langle n\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| n\right\rangle=-(2 n+1) \end{align}

which implies

\begin{align} \left\langle(\Delta x)^{2}\right\rangle_{n}\left\langle(\Delta p)^{2}\right\rangle_{n}=\frac{\hbar^{2}}{4}(2 n+1)^{2} \end{align}

so $|n\rangle$ is not minimal!