Lecture 12: Decay into a squeezed vacuum field

Author: J. R. Johansson ([email protected]), http://dml.riken.jp/~rob/

The latest version of this IPython notebook lecture is available at http://github.com/jrjohansson/qutip-lectures.

The other notebooks in this lecture series are indexed at http://jrjohansson.github.com.

In [1]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
In [2]:
from qutip import *

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 [3]:
w0 = 1.0 * 2 * pi
gamma0 = 0.05
In [4]:
# the temperature of the environment in frequency units
w_th = 0.0 * 2 * pi
In [5]:
# the number of average excitations in the environment mode w0 at temperture w_th
Nth = n_thermal(w0, w_th)

Nth
Out[5]:
0.0

Parameters that describes the squeezing of the bath

In [6]:
# squeezing parameter for the environment
r = 1.0
theta = 0.1 * pi
In [7]:
N = Nth * (cosh(r) ** 2 + sinh(r) ** 2) + sinh(r) ** 2

N
Out[7]:
1.3810978455418155
In [8]:
M = - cosh(r) * sinh(r) * exp(-1j * theta) * (2 * Nth + 1)

M
Out[8]:
(-1.7246746122879026+0.56038075112519081j)
In [9]:
# Check, should be zero according to Eq. 3.261 in Breuer and Petruccione
abs(M)**2 - (N * (N + 1) - Nth * (Nth + 1))
Out[9]:
0.0

Operators, Hamiltonian and initial state

In [10]:
sm = sigmam()
sp = sigmap()
In [11]:
H = - 0.5 * w0 * sigmaz()  # by adding the hamiltonian here, so we move back to the schrodinger picture
In [12]:
c_ops = [sqrt(gamma0 * (N + 1)) * sm, 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 [13]:
L0 = liouvillian(H, c_ops)

L0
Out[13]:
Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = [4, 4], type = super, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}-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{array}\right)\end{equation*}

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 [14]:
Lsq = - gamma0 * M * spre(sp) * spost(sp) - gamma0 * conjugate(M) * spre(sm) * spost(sm)

Lsq
Out[14]:
Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = [4, 4], type = super, isherm = True\begin{equation*}\left(\begin{array}{*{11}c}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{array}\right)\end{equation*}

The total Liouvillian for the master equation is now

In [15]:
L = L0 + Lsq

L
Out[15]:
Quantum object: dims = [[[2], [2]], [[2], [2]]], shape = [4, 4], type = super, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}-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{array}\right)\end{equation*}

Evolution

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

In [17]:
tlist = np.linspace(0, 50, 1000)
In [18]:
# start in the qubit superposition state
psi0 = (2j * basis(2, 0) + 1 * basis(2, 1)).unit()
In [19]:
e_ops = [sigmax(), sigmay(), sigmaz()]
In [20]:
result1 = mesolve(L, psi0, tlist, [], e_ops)
In [23]:
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(shape(result1.times)), '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 [24]:
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 [25]:
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
In [26]:
result2 = mesolve(H, psi0, tlist, c_ops, e_ops)

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

In [27]:
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(shape(result2.times)), '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);

Compare the two forms of master equations

In [28]:
fig, axes = plt.subplots(3, 1, sharex=True, figsize=(12, 9))

axes[0].plot(result1.times, result1.expect[0], 'r', label=r'$\langle\sigma_x\rangle$ - me')
axes[0].plot(result2.times, result2.expect[0], 'b--', label=r'$\langle\sigma_x\rangle$ - me lindblad')
axes[0].legend()
axes[0].set_ylim(-1, 1);

axes[1].plot(result1.times, result1.expect[1], 'r', label=r'$\langle\sigma_y\rangle$ - me')
axes[1].plot(result2.times, result2.expect[1], 'b--', label=r'$\langle\sigma_y\rangle$ - me lindblad')
axes[1].legend()
axes[1].set_ylim(-1, 1);

axes[2].plot(result1.times, result1.expect[2], 'r', label=r'$\langle\sigma_y\rangle$ - me')
axes[2].plot(result2.times, result2.expect[2], 'b--', label=r'$\langle\sigma_y\rangle$ - me lindblad')
axes[2].legend()
axes[2].set_ylim(-1, 1);
axes[2].set_xlabel("time", fontsize=16);

Compare dissipation into vacuum and squeezed vacuum

In [29]:
# for vacuum: 
r = 0
theta = 0.0
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
In [30]:
result1 = mesolve(H, psi0, tlist, c_ops, e_ops)
In [31]:
# for squeezed vacuum: 
r = 1.0
theta = 0.0
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
In [32]:
result2 = mesolve(H, psi0, tlist, c_ops, e_ops)
In [33]:
fig, axes = plt.subplots(3, 1, sharex=True, figsize=(12, 9))

axes[0].plot(result1.times, result1.expect[0], 'r', label=r'$\langle\sigma_x\rangle$ - vacuum')
axes[0].plot(result2.times, result2.expect[0], 'b', label=r'$\langle\sigma_x\rangle$ - squeezed vacuum')
axes[0].legend()
axes[0].set_ylim(-1, 1);

axes[1].plot(result1.times, result1.expect[1], 'r', label=r'$\langle\sigma_y\rangle$ - vacuum')
axes[1].plot(result2.times, result2.expect[1], 'b', label=r'$\langle\sigma_y\rangle$ - squeezed vacuum')
axes[1].legend()
axes[1].set_ylim(-1, 1);

axes[2].plot(result1.times, result1.expect[2], 'r', label=r'$\langle\sigma_y\rangle$ - vacuum')
axes[2].plot(result2.times, result2.expect[2], 'b', label=r'$\langle\sigma_y\rangle$ - squeezed vacuum')
axes[2].legend()
axes[2].set_ylim(-1, 1);
axes[2].set_xlabel("time", fontsize=16);

From this comparison it's clear that dissipation into a squeezed vacuum is faster than dissipation into vacuum.

Software versions

In [34]:
from qutip.ipynbtools import version_table; version_table()
Out[34]:
SoftwareVersion
matplotlib1.3.1
Cython0.20.1post0
Numpy1.8.1
SciPy0.13.3
Python3.4.1 (default, Jun 9 2014, 17:34:49) [GCC 4.8.3]
IPython2.0.0
OSposix [linux]
QuTiP3.0.0.dev-5a88aa8
Thu Jun 26 14:09:19 2014 JST