Author: J. R. Johansson (robert@riken.jp), 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.
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
from qutip import *
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.
w0 = 1.0 * 2 * pi
gamma0 = 0.05
# the temperature of the environment in frequency units
w_th = 0.0 * 2 * pi
# the number of average excitations in the environment mode w0 at temperture w_th
Nth = n_thermal(w0, w_th)
Nth
0.0
# squeezing parameter for the environment
r = 1.0
theta = 0.1 * pi
N = Nth * (cosh(r) ** 2 + sinh(r) ** 2) + sinh(r) ** 2
N
1.3810978455418155
M = - cosh(r) * sinh(r) * exp(-1j * theta) * (2 * Nth + 1)
M
(-1.7246746122879026+0.56038075112519081j)
# Check, should be zero according to Eq. 3.261 in Breuer and Petruccione
abs(M)**2 - (N * (N + 1) - Nth * (Nth + 1))
0.0
sm = sigmam()
sp = sigmap()
H = - 0.5 * w0 * sigmaz() # by adding the hamiltonian here, so we move back to the schrodinger picture
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:
L0 = liouvillian(H, c_ops)
L0
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
Lsq = - gamma0 * M * spre(sp) * spost(sp) - gamma0 * conjugate(M) * spre(sm) * spost(sm)
Lsq
The total Liouvillian for the master equation is now
L = L0 + Lsq
L
We can now solve the master equation numerically using QuTiP's mesolve
function:
tlist = np.linspace(0, 50, 1000)
# start in the qubit superposition state
psi0 = (2j * basis(2, 0) + 1 * basis(2, 1)).unit()
e_ops = [sigmax(), sigmay(), sigmaz()]
result1 = mesolve(L, psi0, tlist, [], e_ops)
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);
b = Bloch()
b.add_points(result1.expect, meth='l')
b.show()
We can solve the alternative master equation, which is on the standard Lindblad form, directly using the QuTiP mesolve
function:
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
result2 = mesolve(H, psi0, tlist, c_ops, e_ops)
And we can verify that it indeed gives the same results:
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);
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);
# for vacuum:
r = 0
theta = 0.0
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
result1 = mesolve(H, psi0, tlist, c_ops, e_ops)
# for squeezed vacuum:
r = 1.0
theta = 0.0
c_ops = [sqrt(gamma0) * (sm * cosh(r) + sp * sinh(r) * exp(1j*theta))]
result2 = mesolve(H, psi0, tlist, c_ops, e_ops)
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.
from qutip.ipynbtools import version_table; version_table()
Software | Version |
---|---|
matplotlib | 1.3.1 |
Cython | 0.20.1post0 |
Numpy | 1.8.1 |
SciPy | 0.13.3 |
Python | 3.4.1 (default, Jun 9 2014, 17:34:49) [GCC 4.8.3] |
IPython | 2.0.0 |
OS | posix [linux] |
QuTiP | 3.0.0.dev-5a88aa8 |
Thu Jun 26 14:09:19 2014 JST |