Lecture 12 - Decay into a squeezed vacuum field¶

Author: J. R. Johansson (robert@riken.jp), https://jrjohansson.github.io/

This lecture series was developed by J.R. Johannson. The original lecture notebooks are available here.

This is a slightly modified version of the lectures, to work with the current release of QuTiP. You can find these lectures as a part of the qutip-tutorials repository. This lecture and other tutorial notebooks are indexed at the QuTiP Tutorial webpage.

In [1]:

import matplotlib.pyplot as plt
import numpy as np
from qutip import (Bloch, about, basis, liouvillian, mesolve, n_thermal,
                   sigmam, sigmap, sigmax, sigmay, sigmaz, spost, spre)

%matplotlib inline

Introduction¶

We follow The theory of open quantum systems, by Breuer and Pretruccione, section 3.4.3 - 3.4.4, which gives the master equation for a two-level system that decays into an environment that is in a squeezed vacuum state:

$\frac{d}{dt}\rho = \gamma_0(N+1)\left(\sigma_-\rho(t)\sigma_+ - \frac{1}{2}\sigma_+\sigma_-\rho(t) - \frac{1}{2}\rho(t)\sigma_+\sigma_-\right)$

$ + \gamma_0 N \left(\sigma_+\rho(t)\sigma_- - \frac{1}{2}\sigma_-\sigma_+\rho(t) - \frac{1}{2}\rho(t)\sigma_-\sigma_+\right)$

$ -\gamma_0 M \sigma_+\rho(t)\sigma_+ -\gamma_0 M^* \sigma_-\rho(t)\sigma_-$

where the parameters $N$ and $M$ describes the temperature and squeezing of the environmental modes:

$\displaystyle N = N_{\rm th} ({\cosh}^2 r + {\sinh}^2 r) + \sinh^2 r$

$\displaystyle M = - \cosh r \sinh r e^{i\theta} (2 N_{\rm th} + 1)$

Alternatively, this master equation can be written in standard Lindblad form,

$\frac{d}{dt}\rho = \gamma_0\left(C\rho(t)C^\dagger - \frac{1}{2}C^\dagger C\rho(t) - \frac{1}{2}\rho(t)C^\dagger C\right)$

where $C = \sigma_-\cosh r + \sigma_+ \sinh r e^{i\theta}$.

Below we will solve these master equations numerically using QuTiP, and visualize at the resulting dynamics.

Problem parameters¶

In [2]:

w0 = 1.0 * 2 * np.pi
gamma0 = 0.05

In [3]:

# the temperature of the environment in frequency units
w_th = 0.0 * 2 * np.pi

In [4]:

# the number of average excitations in the
# environment mode w0 at temperture w_th
Nth = n_thermal(w0, w_th)

Nth

Out[4]:

0.0

Parameters that describes the squeezing of the bath¶

In [5]:

# squeezing parameter for the environment
r = 1.0
theta = 0.1 * np.pi

In [6]:

N = Nth * (np.cosh(r) ** 2 + np.sinh(r) ** 2) + np.sinh(r) ** 2

N

Out[6]:

1.3810978455418155

In [7]:

M = -np.cosh(r) * np.sinh(r) * np.exp(-1j * theta) * (2 * Nth + 1)

M

Out[7]:

(-1.7246746122879026+0.5603807511251908j)

In [8]:

# Check, should be zero according to Eq. 3.261 in Breuer and Petruccione
abs(M) ** 2 - (N * (N + 1) - Nth * (Nth + 1))

Out[8]:

0.0

Operators, Hamiltonian and initial state¶

In [9]:

sm = sigmam()
sp = sigmap()

In [10]:

H = (
    -0.5 * w0 * sigmaz()
)  # by adding the hamiltonian here, so we move back to the schrodinger picture

In [11]:

c_ops = [np.sqrt(gamma0 * (N + 1)) * sm, np.sqrt(gamma0 * N) * sp]

Let's first construct the standard part of the Liouvillian, corresponding the unitary contribution and the first two terms in the first master equation given above:

In [12]:

L0 = liouvillian(H, c_ops)

L0

Out[12]:

Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = (4, 4), type = super, isherm = False $ \\ \left(\begin{matrix}-0.119 & 0.0 & 0.0 & 0.069\\0.0 & (-0.094-6.283j) & 0.0 & 0.0\\0.0 & 0.0 & (-0.094+6.283j) & 0.0\\0.119 & 0.0 & 0.0 & -0.069\\\end{matrix}\right)$

Next we manually construct the Liouvillian for the effect of the squeeing in the environment, which is not on standard form we can therefore not use the liouvillian function in QuTiP

In [13]:

Lsq = -gamma0 * M * spre(sp) * spost(sp) - gamma0 * \
      M.conj() * spre(sm) * spost(sm)

Lsq

Out[13]:

Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = (4, 4), type = super, isherm = True $ \\ \left(\begin{matrix}0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & (0.086+0.028j) & 0.0\\0.0 & (0.086-0.028j) & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0\\\end{matrix}\right)$

The total Liouvillian for the master equation is now

In [14]:

L = L0 + Lsq

L

Out[14]:

Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = (4, 4), type = super, isherm = False $ \\ \left(\begin{matrix}-0.119 & 0.0 & 0.0 & 0.069\\0.0 & (-0.094-6.283j) & (0.086+0.028j) & 0.0\\0.0 & (0.086-0.028j) & (-0.094+6.283j) & 0.0\\0.119 & 0.0 & 0.0 & -0.069\\\end{matrix}\right)$

Evolution¶

We can now solve the master equation numerically using QuTiP's mesolve function:

In [15]:

tlist = np.linspace(0, 50, 1000)

In [16]:

# start in the qubit superposition state
psi0 = (2j * basis(2, 0) + 1 * basis(2, 1)).unit()

In [17]:

e_ops = [sigmax(), sigmay(), sigmaz()]

In [18]:

result1 = mesolve(L, psi0, tlist, [], e_ops)

In [19]:

fig, ax = plt.subplots(figsize=(12, 6))

ax.plot(result1.times, result1.expect[0], "r",
        label=r"$\langle\sigma_x\rangle$")
ax.plot(result1.times, result1.expect[1], "g",
        label=r"$\langle\sigma_y\rangle$")
ax.plot(result1.times, result1.expect[2], "b",
        label=r"$\langle\sigma_z\rangle$")

sz_ss_analytical = -1 / (2 * N + 1)
ax.plot(
    result1.times,
    sz_ss_analytical * np.ones(result1.times.shape),
    "k--",
    label=r"$\langle\sigma_z\rangle_s$ analytical",
)


ax.set_ylabel(r"$\langle\sigma_z\rangle$", fontsize=16)
ax.set_xlabel("time", fontsize=16)
ax.legend()
ax.set_ylim(-1, 1);

In [20]:

b = Bloch()
b.add_points(result1.expect, meth="l")
b.show()

Alternative master equation on Lindblad form¶

We can solve the alternative master equation, which is on the standard Lindblad form, directly using the QuTiP mesolve function:

In [21]:

c_ops = [np.sqrt(gamma0) *
         (sm * np.cosh(r) + sp * np.sinh(r) * np.exp(1j * theta))]

In [22]:

result2 = mesolve(H, psi0, tlist, c_ops, e_ops)

And we can verify that it indeed gives the same results:

In [23]:

fig, ax = plt.subplots(figsize=(12, 6))

ax.plot(result2.times, result2.expect[0], "r",
        label=r"$\langle\sigma_x\rangle$")
ax.plot(result2.times, result2.expect[1], "g",
        label=r"$\langle\sigma_y\rangle$")
ax.plot(result2.times, result2.expect[2], "b",
        label=r"$\langle\sigma_z\rangle$")

sz_ss_analytical = -1 / (2 * N + 1)
ax.plot(
    result2.times,
    sz_ss_analytical * np.ones(result2.times.shape),
    "k--",
    label=r"$\langle\sigma_z\rangle_s$ analytical",
)


ax.set_ylabel(r"$\langle\sigma_z\rangle$", fontsize=16)
ax.set_xlabel("time", fontsize=16)
ax.legend()
ax.set_ylim(-1, 1);