Development notebook: Tests for QuTiP's stochastic master equation solver

Copyright (C) 2011 and later, Paul D. Nation & Robert J. Johansson

In this notebook we test the qutip stochastic master equation solver (smesolve) with a few textbook examples taken from the book Quantum Optics, by Walls and Milburn, section 6.7.


In [1]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np

In [2]:
from qutip import *

Photo-count detection

Theory

Stochastic master equation in Milburn's formulation

$\displaystyle d\rho(t) = dN(t) \mathcal{G}[a] \rho(t) - dt \gamma \mathcal{H}[\frac{1}{2}a^\dagger a] \rho(t)$

where

$\displaystyle \mathcal{G}[A] \rho = \frac{A\rho A^\dagger}{\mathrm{Tr}[A\rho A^\dagger]} - \rho$

$\displaystyle \mathcal{H}[A] \rho = \frac{1}{2}(A\rho + \rho A^\dagger - \mathrm{Tr}[A\rho + \rho A^\dagger] \rho) $

and $dN(t)$ is a Poisson distributed increment with $E[dN(t)] = \gamma \langle a^\dagger a\rangle (t)dt$.

Formulation in QuTiP

In QuTiP we write the stochastic master equation on the form (in the interaction picture, with no deterministic dissipation):

$\displaystyle d\rho(t) = D_{1}[A]\rho(t) dt + D_{2}[A]\rho(t) dW$

where $A = \sqrt{\gamma} a$, so we can identify

$\displaystyle D_{1}[A]\rho(t) = - \frac{1}{2}\gamma \mathcal{H}[a^\dagger a] \rho(t) = -\gamma \frac{1}{2}\left( a^\dagger a\rho + \rho a^\dagger a - \mathrm{Tr}[a^\dagger a\rho + \rho a^\dagger a] \rho \right) = -\frac{1}{2}\left( A^\dagger A\rho + \rho A^\dagger A - \mathrm{Tr}[A^\dagger A\rho + \rho A^\dagger A] \rho \right)$

$\displaystyle D_{2}[A]\rho(t) = \mathcal{G}[a] \rho = \frac{A\rho A^\dagger}{\mathrm{Tr}[A\rho A^\dagger]} - \rho$

and

$dW = dN(t)$

and $A = \sqrt{\gamma} a$ is the collapse operator including the rate of the process as a coefficient in the operator.

Reference solution: deterministic master equation


In [3]:
N = 10
w0 = 0.5 * 2 * np.pi
times = np.linspace(0, 15, 150)
dt = times[1] - times[0]
gamma = 0.25
A = 2.5
ntraj = 50
nsubsteps = 50

In [4]:
a = destroy(N)
x = a + a.dag()

In [5]:
H = w0 * a.dag() * a

In [6]:
#rho0 = coherent(N, 5)
rho0 = fock(N, 5)

In [7]:
c_ops = [np.sqrt(gamma) * a]

In [8]:
e_ops = [a.dag() * a, x]

In [9]:
result_ref = mesolve(H, rho0, times, c_ops, e_ops)

In [10]:
plot_expectation_values(result_ref);


Solve using stochastic master equation

$\displaystyle D_{1}[a, \rho] = -\gamma \frac{1}{2}\left( a^\dagger a\rho + \rho a^\dagger a - \mathrm{Tr}[a^\dagger a\rho + \rho a^\dagger a] \right) \rightarrow - \frac{1}{2}(\{A^\dagger A\}_L + \{A^\dagger A\}_R)\rho_v + \mathrm{E}[(\{A^\dagger A\}_L + \{A^\dagger A\}_R)\rho_v]$

$\displaystyle D_{2}[A, \rho(t)] = \frac{A\rho A^\dagger}{\mathrm{Tr}[A\rho A^\dagger]} - \rho \rightarrow \frac{A_LA^\dagger_R \rho_v}{\mathrm{E}[A_LA^\dagger_R \rho_v]} - \rho_v$

Using QuTiP built-in photo-current detection functions for $D_1$ and $D_2$


In [11]:
result = photocurrent_mesolve(H, rho0, times, c_ops=[], sc_ops=c_ops, e_ops=e_ops, 
                  ntraj=ntraj, nsubsteps=nsubsteps,
                  store_measurement=True, noise=1234)


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80.0%. Run time:   4.64s. Est. time left: 00:00:00:01
90.0%. Run time:   5.27s. Est. time left: 00:00:00:00
Total run time:   5.85s

In [12]:
plot_expectation_values([result, result_ref]);



In [13]:
for m in result.measurement:
    plt.step(times, dt * m.real)


Solve problem again, with the same noise as the previous run

photocurrentmesolve does not take custom noise, but you can set the seed.


In [14]:
result = photocurrent_mesolve(H, rho0, times, c_ops=[], sc_ops=c_ops, e_ops=e_ops, 
                  ntraj=ntraj, nsubsteps=nsubsteps, store_measurement=True, noise=1234)


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90.0%. Run time:   5.57s. Est. time left: 00:00:00:00
Total run time:   6.07s

In [15]:
plot_expectation_values([result, result_ref]);



In [16]:
for m in result.measurement:
    plt.step(times, dt * m.real)


Homodyne detection


In [17]:
H = w0 * a.dag() * a + A * (a + a.dag())

In [18]:
result_ref = mesolve(H, rho0, times, c_ops, e_ops)

Theory

Stochastic master equation for homodyne in Milburn's formulation

$\displaystyle d\rho(t) = -i[H, \rho(t)]dt + \gamma\mathcal{D}[a]\rho(t) dt + dW(t) \sqrt{\gamma} \mathcal{H}[a] \rho(t)$

where $\mathcal{D}$ is the standard Lindblad dissipator superoperator, and $\mathcal{H}$ is defined as above, and $dW(t)$ is a normal distributed increment with $E[dW(t)] = \sqrt{dt}$.

In QuTiP format we have:

$\displaystyle d\rho(t) = -i[H, \rho(t)]dt + D_{1}[A]\rho(t) dt + D_{2}[A]\rho(t) dW$

where $A = \sqrt{\gamma} a$, so we can identify

$\displaystyle D_{1}[A]\rho(t) = \gamma \mathcal{D}[a]\rho(t) = \mathcal{D}[A]\rho(t)$


In [19]:
L = liouvillian(H, c_ops=c_ops).data
def d1_rho_func(t, rho_vec):
    return cy.spmv(L, rho_vec)

$\displaystyle D_{2}[A]\rho(t) = \sqrt{\gamma} \mathcal{H}[a]\rho(t) = A\rho + \rho A^\dagger - \mathrm{Tr}[A\rho + \rho A^\dagger] \rho \rightarrow (A_L + A_R^\dagger)\rho_v - \mathrm{Tr}[(A_L + A_R^\dagger)\rho_v] \rho_v$


In [20]:
n_sum = spre(c_ops[0]) + spost(c_ops[0].dag())
n_sum_data = n_sum.data

def d2_rho_func(t, rho_vec):
    e1 = cy.cy_expect_rho_vec(n_sum_data, rho_vec, False)
    out = np.zeros((1,len(rho_vec)),dtype=complex)
    out += cy.spmv(n_sum_data, rho_vec) - e1 * rho_vec
    return out

In [21]:
result = general_stochastic(ket2dm(rho0), times, d1=d1_rho_func, d2=d2_rho_func, 
                            e_ops=[spre(op) for op in e_ops], ntraj=ntraj, solver="platen",
                            m_ops=[spre(a + a.dag())], dW_factors=[1/np.sqrt(gamma)],
                            nsubsteps=nsubsteps, store_measurement=True, map_func=parallel_map)


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100.0%. Run time:  79.44s. Est. time left: 00:00:00:00
Total run time:  79.46s

In [22]:
plot_expectation_values([result, result_ref]);



In [23]:
for m in result.measurement:
    plt.plot(times, m[:, 0].real, 'b', alpha=0.025)
    
plt.plot(times, result_ref.expect[1], 'k', lw=2);
plt.ylim(-15, 15)
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0].real, 'r', lw=2);


Using QuTiP built-in homodyne detection functions for $D_1$ and $D_2$


In [24]:
result = smesolve(H, rho0, times, [], c_ops, e_ops, ntraj=ntraj, nsubsteps=nsubsteps, solver="pc-euler",
                  method='homodyne', store_measurement=True)


10.0%. Run time:   2.19s. Est. time left: 00:00:00:19
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90.0%. Run time:  12.92s. Est. time left: 00:00:00:01
Total run time:  14.22s

In [25]:
plot_expectation_values([result, result_ref]);



In [26]:
for m in result.measurement:
    plt.plot(times, m[:, 0].real / np.sqrt(gamma), 'b', alpha=0.025)
    
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0].real / np.sqrt(gamma), 'r', lw=2);
plt.plot(times, result_ref.expect[1], 'k', lw=2)


Out[26]:
[<matplotlib.lines.Line2D at 0x7ff799003b38>]

Solve problem again, this time with a specified noise (from previous run)


In [27]:
result = smesolve(H, rho0, times, [], c_ops, e_ops, ntraj=ntraj, nsubsteps=nsubsteps, solver="pc-euler",
                  method='homodyne', store_measurement=True, noise=result.noise)


10.0%. Run time:   1.36s. Est. time left: 00:00:00:12
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90.0%. Run time:  12.64s. Est. time left: 00:00:00:01
Total run time:  16.34s

In [28]:
plot_expectation_values([result, result_ref]);



In [29]:
for m in result.measurement:
    plt.plot(times, m[:, 0].real / np.sqrt(gamma), 'b', alpha=0.025)
    
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0].real / np.sqrt(gamma), 'r', lw=2);
plt.plot(times, result_ref.expect[1], 'k', lw=2)


Out[29]:
[<matplotlib.lines.Line2D at 0x7ff798e2b860>]

Heterodyne detection


In [30]:
e_ops = [a.dag() * a, a + a.dag(), -1j * (a - a.dag())]

In [31]:
result_ref = mesolve(H, rho0, times, c_ops, e_ops)

Stochastic master equation for heterodyne in Milburn's formulation

$\displaystyle d\rho(t) = -i[H, \rho(t)]dt + \gamma\mathcal{D}[a]\rho(t) dt + \frac{1}{\sqrt{2}} dW_1(t) \sqrt{\gamma} \mathcal{H}[a] \rho(t) + \frac{1}{\sqrt{2}} dW_2(t) \sqrt{\gamma} \mathcal{H}[-ia] \rho(t)$

where $\mathcal{D}$ is the standard Lindblad dissipator superoperator, and $\mathcal{H}$ is defined as above, and $dW_i(t)$ is a normal distributed increment with $E[dW_i(t)] = \sqrt{dt}$.

In QuTiP format we have:

$\displaystyle d\rho(t) = -i[H, \rho(t)]dt + D_{1}[A]\rho(t) dt + D_{2}^{(1)}[A]\rho(t) dW_1 + D_{2}^{(2)}[A]\rho(t) dW_2$

where $A = \sqrt{\gamma} a$, so we can identify

$\displaystyle D_{1}[A]\rho = \gamma \mathcal{D}[a]\rho = \mathcal{D}[A]\rho$


In [32]:
#def d1_rho_func(A, rho_vec):
#    return A[7] * rho_vec

L = liouvillian(H, c_ops=c_ops).data
def d1_rho_func(t, rho_vec):
    return cy.spmv(L, rho_vec)

$D_{2}^{(1)}[A]\rho = \frac{1}{\sqrt{2}} \sqrt{\gamma} \mathcal{H}[a] \rho = \frac{1}{\sqrt{2}} \mathcal{H}[A] \rho = \frac{1}{\sqrt{2}}(A\rho + \rho A^\dagger - \mathrm{Tr}[A\rho + \rho A^\dagger] \rho) \rightarrow \frac{1}{\sqrt{2}} \left\{(A_L + A_R^\dagger)\rho_v - \mathrm{Tr}[(A_L + A_R^\dagger)\rho_v] \rho_v\right\}$

$D_{2}^{(2)}[A]\rho = \frac{1}{\sqrt{2}} \sqrt{\gamma} \mathcal{H}[-ia] \rho = \frac{1}{\sqrt{2}} \mathcal{H}[-iA] \rho = \frac{-i}{\sqrt{2}}(A\rho - \rho A^\dagger - \mathrm{Tr}[A\rho - \rho A^\dagger] \rho) \rightarrow \frac{-i}{\sqrt{2}} \left\{(A_L - A_R^\dagger)\rho_v - \mathrm{Tr}[(A_L - A_R^\dagger)\rho_v] \rho_v\right\}$


In [33]:
n_sump = spre(c_ops[0]) + spost(c_ops[0].dag())
n_sump_data = n_sump.data/np.sqrt(2)
n_summ = spre(c_ops[0]) - spost(c_ops[0].dag())
n_summ_data = -1.0j*n_summ.data/np.sqrt(2)

def d2_rho_func(A, rho_vec):
    out = np.zeros((2,len(rho_vec)),dtype=complex)

    e1 = cy.cy_expect_rho_vec(n_sump_data, rho_vec, False)
    out[0,:] += cy.spmv(n_sump_data, rho_vec) - e1 * rho_vec

    e1 = cy.cy_expect_rho_vec(n_summ_data, rho_vec, False)
    out[1,:] += cy.spmv(n_summ_data, rho_vec) - e1 * rho_vec

    return out



#def d2_rho_func(t, rho_vec):
#    e1 = cy.cy_expect_rho_vec(n_sum_data, rho_vec, False)
#    out = np.zeros((1,len(rho_vec)),dtype=complex)
#    out += cy.spmv(n_sum_data, rho_vec) - e1 * rho_vec
#    return out

In [34]:
result = general_stochastic(ket2dm(rho0), times, d1=d1_rho_func, d2=d2_rho_func, 
                  e_ops=[spre(op) for op in e_ops], solver="platen", # order=1
                  ntraj=ntraj, nsubsteps=nsubsteps, len_d2=2,
                  m_ops=[spre(a + a.dag()), (-1j)*spre(a - a.dag())],
                  dW_factors=[2/np.sqrt(gamma), 2/np.sqrt(gamma)],
                  store_measurement=True, map_func=parallel_map)


10.0%. Run time:  23.13s. Est. time left: 00:00:03:28
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100.0%. Run time: 165.88s. Est. time left: 00:00:00:00
Total run time: 165.96s

In [35]:
plot_expectation_values([result, result_ref])


Out[35]:
(<Figure size 576x288 with 3 Axes>,
 array([[<matplotlib.axes._subplots.AxesSubplot object at 0x7ff798d45668>],
        [<matplotlib.axes._subplots.AxesSubplot object at 0x7ff798e76d30>],
        [<matplotlib.axes._subplots.AxesSubplot object at 0x7ff79981beb8>]],
       dtype=object))

In [36]:
for m in result.measurement:
    plt.plot(times, m[:, 0].real, 'r', alpha=0.025)
    plt.plot(times, m[:, 1].real, 'b', alpha=0.025)

plt.ylim(-20, 20)
    
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0].real, 'r', lw=2);
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,1].real, 'b', lw=2);

plt.plot(times, result_ref.expect[1], 'k', lw=2);
plt.plot(times, result_ref.expect[2], 'k', lw=2);


Using QuTiP built-in heterodyne detection functions for $D_1$ and $D_2$


In [37]:
result = smesolve(H, rho0, times, [], c_ops, e_ops, ntraj=ntraj, nsubsteps=nsubsteps, solver="milstein", # order=1
                  method='heterodyne', store_measurement=True)


10.0%. Run time:   1.90s. Est. time left: 00:00:00:17
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80.0%. Run time:  20.96s. Est. time left: 00:00:00:05
90.0%. Run time:  23.19s. Est. time left: 00:00:00:02
Total run time:  26.30s

In [38]:
plot_expectation_values([result, result_ref]);



In [39]:
for m in result.measurement:
    plt.plot(times, m[:, 0, 0].real / np.sqrt(gamma), 'r', alpha=0.025)
    plt.plot(times, m[:, 0, 1].real / np.sqrt(gamma), 'b', alpha=0.025)
    
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0,0].real / np.sqrt(gamma), 'r', lw=2);
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0,1].real / np.sqrt(gamma), 'b', lw=2);

plt.plot(times, result_ref.expect[1], 'k', lw=2);
plt.plot(times, result_ref.expect[2], 'k', lw=2);


Solve problem again, this time with a specified noise (from previous run)


In [40]:
result = smesolve(H, rho0, times, [], c_ops, e_ops, ntraj=ntraj, nsubsteps=nsubsteps, solver="milstein", # order=1
                  method='heterodyne', store_measurement=True, noise=result.noise)


10.0%. Run time:   2.00s. Est. time left: 00:00:00:18
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80.0%. Run time:  16.94s. Est. time left: 00:00:00:04
90.0%. Run time:  18.78s. Est. time left: 00:00:00:02
Total run time:  20.38s

In [41]:
plot_expectation_values([result, result_ref]);



In [42]:
for m in result.measurement:
    plt.plot(times, m[:, 0, 0].real / np.sqrt(gamma), 'r', alpha=0.025)
    plt.plot(times, m[:, 0, 1].real / np.sqrt(gamma), 'b', alpha=0.025)
    
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0,0].real / np.sqrt(gamma), 'r', lw=2);
plt.plot(times, np.array(result.measurement).mean(axis=0)[:,0,1].real / np.sqrt(gamma), 'b', lw=2);

plt.plot(times, result_ref.expect[1], 'k', lw=2);
plt.plot(times, result_ref.expect[2], 'k', lw=2);

plt.axis('tight')
plt.ylim(-25, 25);


Software version


In [43]:
from qutip.ipynbtools import version_table

version_table()


Out[43]:
SoftwareVersion
QuTiP4.4.0.dev0+21a52e95
Numpy1.16.0
SciPy1.2.0
matplotlib3.0.2
Cython0.29.3
Number of CPUs2
BLAS InfoOPENBLAS
IPython7.2.0
Python3.6.7 (default, Oct 22 2018, 11:32:17) [GCC 8.2.0]
OSposix [linux]
Fri Feb 01 13:43:38 2019 JST

In [ ]: