Classical limit of coherent state in Jaynes–Cummings model

Question

Im dealing with an exercise on the Jaynes Cummings model in a resonat single mode approximation. The interaction Hamiltonian in rotating wave approximation is

$$H_{int}=g\, \sigma_+\,a\,+g^*\,\sigma_-\,a^\dagger \tag{I}$$

where $a$ and $a^\dagger$ are the annihilation and creation operators for the bosonic state, which is assumed to be in a coherent state $|\alpha\rangle$. $\sigma_+$ and $\sigma_-$ are the raising and lowering operators for the atom in the cavity.

Now im required to do some calculations and for that I get the hint that $H_{int}$ can be replaced by the expectation value $$H_{int}\to H_c=\langle\alpha|H_{int}|\alpha\rangle=g\, \sigma_+\,\alpha\,+g^*\,\sigma_-\,\alpha^* \tag{II}$$

in the classical limit $|\alpha|>>1 $.

I want to know why this approximation of the Hamiltonian is justified (Why we can take (II) instead of (I) for our Hamiltonian), in this limit.

My thoughts: Since $a|\alpha\rangle=\alpha|\alpha\rangle $, our exact interaction Hamiltonian differs from the approximation only by the fact that $a^\dagger $ has been replaced by $a^*$. This seems not too far fetched since this is just the complex conjugate of the eigenvalue that we get by applying the annihilation operator on $|\alpha\rangle$. Then I think we can somehow argue that since $|\alpha|>>1 $, the creation operator will not change the coherent state much. But I cannot really cook up a solid argument and the mathematics.

EDIT

I was thinking that maybe in the classical limit, the standard deviation of $\alpha^\dagger$ is neglectable vs. the mean value. But if I havnt done a mistake, we have

$$SD_\alpha(a^\dagger)=\sqrt{\langle\alpha|a^\dagger a^\dagger|\alpha\rangle-\langle\alpha|a^\dagger|\alpha\rangle^2}=\sqrt{a^*a^*-(a^*)^2}=0, \tag{III}$$

independent of the value of $|\alpha|$. Which confuses me even more.

Answer 1

Since $a\vert \alpha \rangle=\alpha\vert \alpha \rangle$ for the coherent state, it follows that \begin{align} \langle \alpha \vert a\vert\alpha\rangle=\alpha \tag{1} \end{align} by normalization and that (as you suspect) by taking the complex conjugate: $$ \alpha^*=\langle \alpha \vert a\vert \alpha\rangle^*=\langle \alpha \vert a^\dagger\vert\alpha\rangle \tag{2} $$ which is true for any $\alpha$. The condition $\vert\alpha\vert\gg 1$ must enter elsewhere since (1) and (2) are exact, irrespective of $\vert \alpha\vert$.

Answer 2

Let's see what happens when we unitarily evolve some initial state $|\psi\rangle\otimes|\alpha\rangle$ over time, where the first mode is for the atomic system and the second for the bosonic mode. The evolution is to $$|\Psi(t)\rangle=U(t)|\psi\rangle\otimes |\alpha\rangle=e^{-it (g\sigma_+ a+g^*\sigma_-a^\dagger)/\hbar}|\psi\rangle\otimes |\alpha\rangle.$$ We will use a series expansion for the exponential and some commutation relations to attain the desired result.

First, let's show that the probability the field ends in the same state $|\alpha\rangle$ is approximately unity when $|\alpha|$ is sufficiently large. To do so, we compute $$P(\alpha)=|\langle \alpha|\Psi(t)\rangle|^2=|\langle g|\otimes \langle \alpha| U(t)|\psi\rangle\otimes |\alpha\rangle|^2+|\langle e|\otimes \langle\alpha| U(t)|\psi\rangle\otimes |\alpha\rangle|^2,$$ equivalent to tracing out the atomic system using the $\{|g\rangle,|e\rangle\}$-basis and projecting the resultant state onto $|\alpha\rangle\langle \alpha|$. Expanding $U(t)=\sum_{n=0}^\infty \frac{1}{n!}(-\frac{i t}{\hbar}H_{int})^n$, there are lots of terms to compute: \begin{align} \langle i|\otimes \langle \alpha| \frac{1}{0!}(-\frac{i t}{\hbar}H_{int})^0|\psi\rangle\otimes |\alpha\rangle=&\langle i|\psi\rangle=\langle i|\frac{1}{0!}(-\frac{i t}{\hbar}H_{c})^0|\psi\rangle,\\ \langle i|\otimes \langle \alpha| \frac{1}{1!}(-\frac{i t}{\hbar}H_{int})^1|\psi\rangle\otimes |\alpha\rangle=&\langle i|(-\frac{i t}{\hbar}H_c)^1|\psi\rangle,\\ \langle i|\otimes \langle \alpha| \frac{1}{2!}(-\frac{i t}{\hbar}H_{int})^2|\psi\rangle\otimes |\alpha\rangle=&\frac{1}{2!}(-\frac{i t}{\hbar})^2\langle i|\otimes \langle \alpha|(g^2\sigma_+^2a^2+g^{*2}\sigma_-^2a^{\dagger 2}+gg^* \sigma_+\sigma_- a a^\dagger+g^*g \sigma_-\sigma_+ a^\dagger a) |\psi\rangle\otimes |\alpha\rangle\\ =&\frac{1}{2!}(-\frac{i t}{\hbar})^2\langle i|\otimes \langle \alpha|(g^2\sigma_+^2a^2+g^{*2}\sigma_-^2a^{\dagger 2}+gg^* \sigma_+\sigma_- (a^\dagger a+1)+g^*g \sigma_-\sigma_+ a^\dagger a) |\psi\rangle\otimes |\alpha\rangle\\ =&\frac{1}{2!}(-\frac{i t}{\hbar})^2\langle i|(g^2\sigma_+^2\alpha^2+g^{*2}\sigma_-^2\alpha^{*2}+gg^* \sigma_+\sigma_- (|\alpha|^2+1)+g^*g \sigma_-\sigma_+ |\alpha|^2) |\psi\rangle,\\ \approx &\frac{1}{2!}(-\frac{i t}{\hbar})^2\langle i|(g^2\sigma_+^2\alpha^2+g^{*2}\sigma_-^2\alpha^{*2}+gg^* \sigma_+\sigma_- (|\alpha|^2)+g^*g \sigma_-\sigma_+ |\alpha|^2) |\psi\rangle\\ =&\langle i|\frac{1}{2!}(-\frac{i t}{\hbar}H_c)^2|\psi\rangle\\ \vdots&. \end{align} Other than expanding terms, liberally using the eigenvalue equations $a|\alpha\rangle=\alpha|\alpha\rangle$ and $\langle\alpha|a^\dagger=\alpha^*\langle\alpha|$, using $\langle \alpha|\alpha\rangle=1$, and recollecting terms in $H_c$, the extra manipulation used in the $n=2$ case is that $a a^\dagger=a^\dagger a+1$ and the approximation $|\alpha|^2+1\approx |\alpha|^2$. So far, it looks like the action on $|\psi\rangle$ is the same as that of the "classical" unitary $\exp(-i H_c t/\hbar)$, at least for the first few terms.

Now we need to formalize this approximation for arbitrary orders $n$. When expanding $H_{int}^n$, each term in the polynomial will be a product of $n$ factors that either take the form $g\sigma_+a$ or $g^*\sigma_-a^\dagger$. Every time $aa^\dagger$ appears in this product, it can be replaced by $a^\dagger a+1$ to move all of the $a^\dagger$ operators to the far left and all of the $a$ operators to the far right. These movements can happen a maximum of $(n/2)^2$ times (if all $a$ start on the left of all $a^\dagger$ and there are an equal number of each). After the movements, the operators all see $\langle \alpha|$ on the left and $|\alpha\rangle$ on the right, so we can use the eigenvalue equations to replace the operators by scalars. When it moves to the left or the right, each $a$ or $a^\dagger$ remains the same, the total number of each operator remains the same, so the coefficient of any term like $(g\sigma_+)(g^*\sigma_-)(g\sigma_+)(g\sigma_+)\cdots(g^*\sigma_-)$ will be $\alpha \alpha^*\alpha\alpha\cdots\alpha^*$, for example, if we ignore all of the $+1$ factors that come from the commutation relation. This directly yields the same results as $H_c$.

In general, how can we guarantee that ignoring the results of the commutation relations will be valid? Since each term with a $+1$ can also be put into the normal order, it will also eventually be replaced by a product $\alpha^k\alpha^{*l}$ for some nonnegative integers $k$ and $l$, but the degree of this monomial will always be 1 less in each of $\alpha$ and $\alpha^*$ than the $a^\dagger a$ term it came with. As such, if the term that retains all of the operators is $\alpha^k\alpha^{*n-k}$, the next leading order term will be $\alpha^{k-1}\alpha^{*n-k-1}$. There can be at most $(n/2)^2$ such correction terms, so the magnitude of the classical term is $|\alpha^k\alpha^{*n-k}|=|\alpha|^n$ and the magnitude of the leading correction term is at most $(n/2)^2|\alpha|^{n-2}$. This pattern continues, the next order correction term has magnitude at most $|\alpha|^{n-4}$ and can occur at most $((n-1)/2)^2$ times. Even if we round up and say each of the correction terms always occur $(n/2)^2$ times, all of the corrections will at most sum to $\sum_{k=1}^{n/2}(n/2)^2|\alpha|^{n-2k}=(n/2)^2\frac{|\alpha|^n-1}{|\alpha|^2-1}$. For any $n$, there is a value of $|\alpha|$ sufficiently large that these corrections are negligible, with $$\frac{|\alpha|^n}{ (n/2)^2\frac{|\alpha|^n-1}{|\alpha|^2-1}}\approx \frac{|\alpha|^2}{ (n/2)^2}\gg 1.$$ We can thus conclude that, for sufficiently large $|\alpha|$, $$\langle i|\otimes \langle \alpha| \frac{1}{n!}(-\frac{i t}{\hbar}H_{int})^n|\psi\rangle\otimes |\alpha\rangle\approx \langle i|\frac{1}{n!}(-\frac{i t}{\hbar}H_{c})^n|\psi\rangle.$$ This allows us to re-exponentiate to $$\langle i|\otimes \langle \alpha| U(t)|\psi\rangle\otimes |\alpha\rangle\approx\langle i|\exp(-\frac{i t}{\hbar}H_{c})|\psi\rangle$$ and, since unitary operations preserve the norm of a state, we find $$P(\alpha)\approx |\langle g|\exp(-\frac{i t}{\hbar}H_{c})|\psi\rangle|^2+|\langle e|\exp(-\frac{i t}{\hbar}H_{c})|\psi\rangle|^2=\langle \psi|e^{\frac{i t}{\hbar}H_{c}}(|g\rangle\langle g|+|e\rangle\langle e|)e^{\frac{-i t}{\hbar}H_{c}}|\psi\rangle=1.$$

After establishing the above, the effective Hamiltonian is almost trivial. We know that the field stays in approximately the same state $|\alpha\rangle$ as the state in which it began, so tracing over the field mode is almost equivalent to projecting it onto $|\alpha\rangle$. Specifically, if we trace out the bosonic mode in an orthonormal basis that includes $|\alpha\rangle$ and states orthogonal to it (this is not the usual identity formed by overcomplete coherent states, this is partial trace in some specific basis), the probability of obtaining any of the orthogonal states will be approximately zero because the probability of obtaining $|\alpha\rangle$ is approximately unity, so $$\mathrm{Tr}_{field}(|\Psi(t)\rangle\langle \Psi(t)|)=\langle \alpha |\Psi(t)\rangle\langle \Psi(t)|\alpha\rangle+\sum_{n}\langle \lambda_n|\Psi(t)\rangle\langle \Psi(t)|\lambda_n\rangle\approx \langle \alpha |\Psi(t)\rangle\langle \Psi(t)|\alpha\rangle$$ for some set of orthonormal states $\{|\lambda_n\rangle\}$ that are all orthogonal to $|\alpha\rangle$ and span the kernel of $|\alpha\rangle\langle\alpha|$. But then this is a quantity that we have already computed, because we know that $$\langle \alpha |\Psi(t)\rangle\langle \Psi(t)|\alpha\rangle\approx e^{\frac{-i t}{\hbar}H_{c}}|\psi\rangle\langle \psi|e^{\frac{i t}{\hbar}H_{c}}.$$ So the evolution of the atomic state $\mathrm{Tr}_{field}(|\Psi(t)\rangle\langle \Psi(t)|)$ alone indeed looks like a unitary evolution under the effective Hamiltonian $H_c$.

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There are extra questions of convergence that are more subtle. If $|\alpha|^n$ is really that large, do the different terms in expanding the exponential conflict with each other? At short times or small interaction strengths they won't be, if we take $|tg\alpha/\hbar|\ll 1$ even though $|\alpha|$ is large. At sufficiently long times, deviations from this approximate (semiclassical) model will be seen, such as the famous collapses and revivals in the Jaynes-Cummings model. The correction terms will be checked by the $1/n!$ that will always eventually cause $|\alpha|^n/n!$ to get small for a sufficiently large $n$, so we don't actually have to worry about an infinite number of terms here for any finite time or finite yet large coherent state strength. And the correction terms from operator ordering won't all be as large as the maxima I used, so their effects are even smaller than the rough bounds used here.

Answer 3

Perhaps a simpler way to understand this semi-classical approximation is to consider the system on phase space in terms of Wigner functions. The Wigner function of a coherent state is a Gaussian function $$ \text{Wigner}\{|\alpha\rangle\langle\alpha|\} = 2\exp(-2|\alpha-\alpha_0|^2) . $$ Here $\alpha$ is a complex phase space variable and $\alpha_0$ is a complex parameter that identifies the coherent states.

For a strong coherent states $|\alpha_0|\gg 1$, which means that the Gaussian function sits very vary away from the origin. The width of the Wigner function of the coherent state (let's denote it with $w$) is fixed at the smallest value that an isotropic Wigner function can have. Therefore, with a strong coherent state the width is much smaller than the distance of the coherent state from the origin $w\ll |\alpha_0|$.

We can now do a change of variables $\alpha\rightarrow\alpha_0+\epsilon$, where $\epsilon$ is a new complex phase space variable relative to the location of the coherent state. Due to the comparative small value of $w$, the values of $\epsilon$ for which the coherent state is non-zero is also small. Therefore, we can expand the relevant expression in terms of $\epsilon$. (One can call this approach strong field perturbation theory.) The leading order terms represent the semi-classical approximation. All sub-leading order terms are severely suppressed relative to the leading order terms due to the size of $|\alpha_0|$. In effect the coupling constant in the leading order terms become $g\rightarrow g|\alpha_0|$, which is much stronger than $g$. Therefore, we can safely discard all sub-leading orders terms and only retain the leading order semiclassical approximation.

As for the dynamics, the evolution of a state is usually represented in terms of the Schroedinger equation or the von Neumann equation (which contains the Hamiltonian). On phase space it becomes a Fokker-Planck equation for the Wigner function of the state. In such an equation, the operators in the Hamiltonian are converted to phase space variables. In the semi-classical limit, as described above, these variables are also replaced by the coherent state parameter.