Degenerate parametric amplifier: quadratures

[CharlieCW]

Homework Statement

The degenerate parametric amplifier is described by the Hamiltonian:

$$H=\hbar \omega a^\dagger a-i\hbar \chi /2 (e^{(2i\omega t)}a^2-e^{(-2i\omega t)}(a^\dagger)^2)$$

Where $a$ and $a^\dagger$ as just the operators of creation and anhiquilation and $\chi$ is just a real constant.

(a) Find the equations of motion in the interaction (Dirac) picture for $a$ and $a^\dagger$ and solve them.

(b) If we define the quadratures as:

$$X_1=a+a^\dagger \ \ \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ \ \ X_2=a-a^\dagger$$

Show that the quadratic fluctuations (uncertainties) of these quadratures satisfy the equations:

$$(\Delta X_i)^2(t)=e^{2\chi t}(\Delta X_i)^2(0)$$

Homework Equations

$$\frac{A_I (t)}{dt}=\frac{i}{h}[A_I,H_0]$$

$$\frac{d}{dt}|\Psi_I (t)\rangle=H_1|\Psi_I (t)|$$

The Attempt at a Solution

(a) This one is pretty straight forward, as we can easily separate:

$$H=\underbrace{\hbar \omega a^\dagger a}_{H_0}+\underbrace{i\hbar \chi /2 (e^{(2i\omega t)}a^2-e^{(-2i\omega t)}(a^\dagger)^2)}_{H_1}$$

Where $H_0$ is just the hamiltonian for the harmonic oscillator (with known solution) and $H_1$ is just a perturbation. This allows to find the equations of motion for $a$ and $a^\dagger$ in the interaction picture:

$$\frac{a_I (t)}{dt}=\frac{i}{\hbar}[a_I,H_0]=i\omega a_I \ \ \rightarrow \ \ \ \ a_I(t)=a_I(0)e^{i\omega t}$$

$$\frac{a^{\dagger}_I (t)}{dt}=\frac{i}{\hbar}[a^{\dagger}_I,H_0]=-i\omega a^{\dagger}_I \ \ \ \rightarrow \ \ \ \ a^{\dagger}_I(t)=a_I^{\dagger} (0)e^{-i\omega t}$$

b) I'm stuck here as I'm not sure how to get the form of $(\Delta X_i)^2 (t)$ shown above. I tried getting the equations for the expectation values of $a^2$ and $(a^\dagger)^2$:

$$(a^2)(t)=(a^2)(0)e^{2i\omega t}$$
$$((a^\dagger)^2)(t)=((a^\dagger)^2)(0)e^{-2i\omega t}$$

But they don't give me the form specified above. Moreover, I don't know how to get the exponential factor, as it only appears in the perturbed hamiltonian $H_1$.

PD: I haven't studied time-dependent perturbation theory so I'm not sure if it's necessary to solve this problem.

 

[mark726]

To solve part (b), we can use the Heisenberg uncertainty principle, which states that for any two operators $A$ and $B$, the uncertainty in their measurements is given by:

$$(\Delta A)^2 (\Delta B)^2 \geq \frac{1}{4} \langle [A,B]^2 \rangle$$

In this case, we can use the operators $X_1=a+a^\dagger$ and $X_2=a-a^\dagger$, and apply the uncertainty principle to them. Note that these operators are Hermitian, so their expectation values are real.

Using the commutation relations, we can calculate:

$$[X_1,X_2]=2[a,a^\dagger]=2$$

Therefore, the uncertainty principle gives:

$$(\Delta X_1)^2 (\Delta X_2)^2 \geq \frac{1}{4} \langle [X_1,X_2]^2 \rangle=\frac{1}{4} \langle 4 \rangle = 1$$

Now, we can use the fact that the operators $a$ and $a^\dagger$ are related to the quadratures through:

$$X_1=a+a^\dagger$$
$$X_2=i(a-a^\dagger)$$

So, we can write:

$$(\Delta X_1)^2 = (\Delta (a+a^\dagger))^2 = (\Delta a)^2 + (\Delta a^\dagger)^2 + 2 \langle a^\dagger a \rangle$$

$$(\Delta X_2)^2 = (\Delta (a-a^\dagger))^2 = (\Delta a)^2 + (\Delta a^\dagger)^2 - 2 \langle a^\dagger a \rangle$$

Substituting these expressions into the uncertainty principle, we get:

$$(\Delta a)^2 (\Delta a^\dagger)^2 \geq 1 - \frac{1}{2} \langle a^\dagger a \rangle$$

Now, we can use the equations of motion for $a$ and $a^\dagger$ that we found in part (a):

$$a_I (t)=a_I (0)e^{i \omega t}$$
$$a_I^\dagger (t)=a_I^\dagger (0)e^{-i \omega t}