A coherent state of the quantum harmonic oscillator is defined as an eigenvector $|\alpha\rangle$ of the annihilation operator $\hat a$ with eigenvalue $\alpha$ or as spatial translations of the ground state of the QHO:$$T_{x_0}|0\rangle = \exp(-\frac i \hbar\hat px_0)|0\rangle:=|\bar x_0\rangle$$the definitions are equivalent when $\alpha$ is real. Coherent states exhibit certain semi-classical properties, such as the following:$$\langle \bar x_0|\hat x_H(t)|\bar x_0\rangle= x_0\cos\omega t,$$where $\omega$ is the angular frequency of the harmonic oscillator, and the $H$ subscript represents a Heisenberg operator. We have also $$\langle \bar x_0|\hat p_H(t)|\bar x_0\rangle = -m\omega x_0 \sin \omega t.$$ So that both the expectation value of position and momentum oscillate with time, in contrast to the energy eigenstates of the harmonic oscillator, which have vanishing expectation values for these operators. So, my question is: why do the coherent states actually behave like oscillators, when the harmonic oscillator energy eigenstates do not? What is a physical intuition for why the expectation values vanish for the QHO energy eigenstates?
Why do coherent states behave semi-classically, but harmonic oscillator states do not
coherent-statesharmonic-oscillatorquantum mechanics
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There are many ways to go around this. You can start from the coherent states and apply the unitary $\hat{O}$ directly on them. That will not be that simple because you will get a term $\hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}$. Now, the typical approach would be to exchange the order of the operators to get something like $e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}\hat{O}$ (up to some extra term due to the commutator). This not really a simple task but once you are done, you can Taylor expand the operator $\hat{O}$ and keep only the zeroth order (all other terms contain annihilation operators acting on vacuum). I am not going to dig into the calculation in more detail, there are many ways to do it and none of them is really pleasant.
But there is a better way. You can define a displacement operator by the action $\hat{D}(\alpha)|0\rangle = |\alpha\rangle$ and then you have $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha) = \hat{a}+\alpha$. You can combine this with the formulas for $\hat{O}\hat{a}\hat{O}^\dagger$, $\hat{O}\hat{b}\hat{O}^\dagger$ to see how the annihilation operators are transformed. What you should get is a beam-splitting of the two coherent states, i.e., $$|\alpha,\beta\rangle\to|t\alpha+r\beta,t\beta-r\alpha\rangle$$, where $t = \cos\theta$, $r = \sin\theta$.
Let $\alpha \in {\Bbb C}$, and let $\vert{n}\rangle $ be the harmonic oscillator state with energy $(n+\textstyle\frac{1}{2})\hbar\omega$. At $t=0$, the coherent state $\vert {\alpha(0)}\rangle $ is defined by $$ \vert{\alpha(0)}\rangle= e^{-\vert \alpha \vert^2/2}\,\left( \sum_{n=0}^{\infty} \displaystyle{\alpha^n\over \sqrt{n!}}\,\vert{n}\rangle\right) \tag{1} $$
What is $\vert{\alpha(t)}\rangle$, the coherent state at time $t$? Start with (1). Since $\left\vert n\right\rangle $ is an eigenstate of the harmonic oscillator hamiltonian $\hat{H}=\left( \hat a^{\dagger }\hat a+\frac{1}{2}\right) \hbar \omega $ with eigenvalue $\left( n+\frac{1}{2}\right) \hbar \omega ,$ the time evolution of $\left\vert n\right\rangle $ is simply $\left\vert n(t)\right\rangle =e^{-i(n+\frac{1}{2})\omega t}\left\vert n\right\rangle $ and thus \begin{equation} \left\vert \alpha (t)\right\rangle =e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{% 1}{2})\omega t}\left\vert n\right\rangle \right) . \end{equation} It is easy to show that $\left\vert \alpha (t)\right\rangle $ is normalized.
Now we first need to show that $a\vert{\alpha(t)}\rangle=\alpha e^{i\hbar \omega t}\vert{\alpha(t)}\rangle$. Recall that $\hat{a}\left\vert n\right\rangle =\sqrt{n}\left\vert n-1\right\rangle .$ \ Then, since $\hat{a}$ is linear, \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}% e^{-i(n+\frac{1}{2})\omega t}\hat{a}\left\vert n\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{1}{2})\omega t}\sqrt{n}\left\vert n-1\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{\left( n-1\right) !}}e^{-i(n+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) , \\ &=&\alpha e^{-i\omega t}e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n-1}}{\sqrt{\left( n-1\right) !}}% e^{-i(n-1+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) .\quad \end{eqnarray} The sum properly starts at $n=1$ since the $n=0$ term does not exist. Thus, setting $m=n-1,$ we can rewrite this sum in terms of $m,$ with $m$ starting at $m=0.$ Hence \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&\alpha e^{-i\omega t}\left[ e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{m=0}^{\infty }\frac{% \alpha ^{m}}{\sqrt{m!}}e^{-i(m+\frac{1}{2})\omega t}\left\vert m\right\rangle \right) \right] \\ &=&\alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle . \end{eqnarray} A useful secondary result, which follows immediately from above, is \begin{eqnarray} \left[ \hat{a}\left\vert \alpha (t)\right\rangle \right] ^{\dagger } &=&\left\langle \alpha (t)\right\vert \hat{a}^{\dagger } \\ &=&\left[ \alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle \right] ^{\dagger }=\alpha ^{\ast }e^{i\omega t}\left\langle \alpha (t)\right\vert \end{eqnarray}
Now $\langle \hat p(t) \rangle$ and $\langle \hat x(t)\rangle$ for $\vert{\alpha(t)}\rangle$. Starting from the definitions $$ \hat{a} =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}+\frac{i}{% m\omega }\hat{p}\right) , \qquad \hat{a}^{\dagger } =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}-% \frac{i}{m\omega }\hat{p}\right) , $$ we have $$ \hat{x} =\sqrt{\frac{\hbar }{2m\omega }}\left( \hat{a}^{\dagger }+% \hat{a}\right) , \qquad \hat{p} =i\sqrt{\frac{m\omega \hbar }{2}}\left( \hat{a}^{\dagger }-% \hat{a}\right) , $$ and thus \begin{eqnarray} \left\langle x(t)\right\rangle &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle +\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right]\, , \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] , \end{eqnarray} which is real, as expected. We can clean this up by writing $\alpha =\left\vert \alpha \right\vert e^{i\theta }$ to obtain% \begin{equation} \left\langle x(t)\right\rangle =\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left( \omega t-\theta \right) \tag{2} \end{equation}
Likewise, \begin{eqnarray} \left\langle p(t)\right\rangle &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle -\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right] \\ &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&-\sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left( \omega t-\theta \right) \tag{3} \end{eqnarray} which is again real.
In your specific case you are starting with a coherent state for which, at $t=0$, we have $$ \langle x(0)\rangle= b\sqrt{2}x_0\, ,\qquad \langle p(0)\rangle=0 $$ so this implies from (2) and (3) evaluated at $t=0$ that $$ b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left(\theta \right)\, , \qquad 0= \sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left(\theta \right) $$ Comparing with your initial conditions gives $\theta=0$ and $b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }} \alpha $ with $\alpha$ real.
Finally, $\hat{x}^{2}$ and $\hat{p}^{2}.$ From $\hat{x}$ and $\hat{p},$ we find \begin{eqnarray} \hat{x}^{2} &=&\frac{\hbar }{2m\omega }\left( \hat{a}^{\dagger }+ \hat{a}\right) ^{2}=\frac{\hbar }{2m\omega }\left( \left( \hat{a} ^{\dagger }\right) ^{2}+\hat{a}^{\dagger }\hat{a}+\hat{a}\hat{a} ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&\frac{\hbar }{2m\omega }\left( \left( \hat{a}^{\dagger }\right) ^{2}+2 \hat{a}^{\dagger }\hat{a}+1+\left( \hat{a}\right) ^{2}\right) , \\ \hat{p}^{2} &=&-\frac{m\omega \hbar }{2}\left( \hat{a}-\hat{a} ^{\dagger }\right) ^{2}=-\frac{m\omega \hbar }{2}\left( \left( \hat{a} ^{\dagger }\right) ^{2}-\hat{a}^{\dagger }\hat{a}-\hat{a}\hat{a}% ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&-\frac{m\omega \hbar }{2}\left( \left( \hat{a}^{\dagger }\right) ^{2}-2% \hat{a}^{\dagger }\hat{a}-1+\left( \hat{a}\right) ^{2}\right) , \end{eqnarray} where \begin{equation} \hat{a}\hat{a}^{\dagger }=\hat{a}\hat{a}^{\dagger }-\hat{a}% ^{\dagger }\hat{a}+\hat{a}^{\dagger }\hat{a}=\left[ \hat{a},% \hat{a}^{\dagger }\right] +\hat{a}^{\dagger }\hat{a}=1+\hat{a}% ^{\dagger }\hat{a} \end{equation} has been used. Thus, \begin{eqnarray} \left\langle x^{2}(t)\right\rangle &=&\frac{\hbar }{2m\omega }\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle +2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad +1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}+2\alpha ^{\ast }\alpha +1+\left( \alpha e^{-i\omega t}\right) ^{2}% \right]\, ,\\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right) ^{2}+1\right] , \\ &=&\frac{\hbar }{2m\omega }\left[ 4\left\vert \alpha \right\vert ^{2}\cos ^{2}\left( \omega t-\theta \right) +1\right] . \\ \left\langle p^{2}(t)\right\rangle &=&-\frac{m\omega \hbar }{2}\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle -2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad -1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}-2\alpha ^{\ast }\alpha -1+\left( \alpha e^{-i\omega t}\right) ^{2}\right]\, ,\\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right) ^{2}-1\right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ -4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) -1\right]\, ,\\ &=&\frac{m\omega \hbar }{2}\left[ 4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) +1% \right] . \end{eqnarray}
Best Answer
The discovery and study of coherent states represents one aspect of one of the biggest problems physicists have faced with the birth and the subsequent development, supported by excellent experimental results, of the quantum mechanics: the search for a correspondence between the new theory, conceived for the analysis of microscopic systems, and classical physics, still fully valid for the description of the macroscopic world.
The history of coherent states begins immediately after the advent of mechanics quantum: their introduction on a conceptual level dates back to a article published in 1926, in which Schrödinger reports the existence of a class of states of the harmonic oscillator that show, in a certain sense, behavior analogous to that of a classic oscillator: for these states it is verified that the energy mean corresponds to the classical value and the position and momentum averages have oscillatory forms in constant phase relation.
Returning to Schrödinger's article, the "almost classical" states from him identified present, in addition to the characteristics already mentioned, an important aspect: being represented by Gaussian wave packets that do not change shape in the time, guarantee the minimization of the product among the uncertainties about position and on the impulse, that is the condition closest to the possibility of measuring simultaneously the aforesaid quantities with arbitrary precision, allowed from classical physics.
So, starting from the following relations: \begin{equation} a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle \quad a|n\rangle=\sqrt{n}|n-1\rangle \end{equation} it is noted that, by virtue of the orthonormality of the states stationary, the diagonal matrix elements of the position and momentum operators are null in the representation of energy, which means that the expectation values of position and momentum on any stationary state are zero instant by instant.
The stationary states just analyzed are characterized by distributions of constant probabilities with respect to the position over time; the wait values of the position e of the impulse are null at all times: this aspect is a fundamental one difference with the states of the classic oscillator, for which, once the energy is defined (as long as different from zero), the observables position and momentum evolve over time according to sinusoidal functions and are always in phase quadrature with each other. Also, if yes calculate the uncertainties on position and momentum for a steady state with n photons, yes gets the uncertainty relation $\Delta x \Delta p=(n+1/2)\hbar$.
it is therefore possible to obtain the minimization of the product of the uncertainties on impulse and position, which represents the maximum similarity with classical mechanics.
A state that is as similar as possible to the classical case must therefore have the following characteristics:
1): The evolution over time of the position and momentum expectation values must be of a simple periodic type, with a constant phase ratio between position and impulse.
2): The wave functions must be as narrow as possible around the value average of the position, so that the probability distribution with respect to the position may tend, by varying appropriate parameters, to a delta function of Dirac;
3): The product of the uncertainties on the position and on the impulse must be minimal.
So you can see this "classical" behavior, that admits these particular states, as an intrinsic property of QHO.
Furthermore I have say that: also the harmonic oscillator energy eigenstates actually behave like oscillators.
Maybe this answer is a bit too long but I hope it can help you.