On the Small Mass Limit of Quantum Brownian Motion with Inhomogeneous Damping and Diffusion

Abstract

We study the small mass limit (or: the Smoluchowski–Kramers limit) of a class of quantum Brownian motions with inhomogeneous damping and diffusion. For Ohmic bath spectral density with a Lorentz–Drude cutoff, we derive the Heisenberg–Langevin equations for the particle’s observables using a quantum stochastic calculus approach. We set the mass of the particle to equal $m = m_{0} \epsilon$, the reduced Planck constant to equal $\hbar = \epsilon$ and the cutoff frequency to equal $\varLambda = E_{\varLambda }/\epsilon$, where $m_0$ and $E_{\varLambda }$ are positive constants, so that the particle’s de Broglie wavelength and the largest energy scale of the bath are fixed as $\epsilon \rightarrow 0$. We study the limit as $\epsilon \rightarrow 0$ of the rescaled model and derive a limiting equation for the (slow) particle’s position variable. We find that the limiting equation contains several drift correction terms, the quantum noise-induced drifts, including terms of purely quantum nature, with no classical counterparts.

Appendices

Appendices

1.1 A Derivation of Heisenberg Equations for Particle’s Observables

In this appendix we derive Eqs. (8)–(9). Let

$$\begin{aligned} b(\omega )= & {} \sqrt{\frac{\omega }{2\hbar }}\left( x(\omega )+\frac{i}{\omega }p(\omega ) \right) , \ \ \ \ b^{\dagger }(\omega ) = \sqrt{\frac{\omega }{2\hbar }}\left( x(\omega )-\frac{i}{\omega }p(\omega ) \right) , \end{aligned}$$

(123)

$$\begin{aligned}&[x(\omega ),p(\omega ')] = i\hbar \delta (\omega -\omega ')I, \end{aligned}$$

(124)

where we have normalized the masses of all bath oscillators.

The Heisenberg equation of motion gives

$$\begin{aligned} \dot{X}(t)&= \frac{i}{\hbar } [H, X(t)] = \frac{P(t)}{m}, \end{aligned}$$

(125)

$$\begin{aligned} \dot{P}(t)&= \frac{i}{\hbar } [H, P(t)] \nonumber \\&= -U'(X(t)) + f'(X(t)) \int _{\mathbb {R}^{+}} d\omega c(\omega ) \sqrt{\frac{2\omega }{\hbar }} x_{t}(\omega ) - 2f(X(t)) f'(X(t)) \int _{\mathbb {R}^{+}} r(\omega ) d\omega , \end{aligned}$$

(126)

$$\begin{aligned} \dot{x}_{t}(\omega )&= \frac{i}{\hbar } [H, x_{t}(\omega )] = p_{t}(\omega ), \ \ \omega \in \mathbb {R}^{+}, \end{aligned}$$

(127)

$$\begin{aligned} \dot{p}_{t}(\omega )&= \frac{i}{\hbar } [H, p_{t}(\omega )] = - \omega ^2 x_{t}(\omega ) + \sqrt{\frac{2\omega }{\hbar }} c(\omega ) f(X(t)), \ \ \omega \in \mathbb {R}^{+}, \end{aligned}$$

(128)

where $r(\omega ) = |c(\omega )|^2/(\hbar \omega )$ and $f'(X)= [f(X),P ]/(i\hbar )$.

Next we eliminate the bath degrees of freedom from the equations for X(t) and P(t). Solving for $x_{t}(\omega )$, $\omega \in \mathbb {R}^{+}$, gives:

$$\begin{aligned} x_{t}(\omega ) = \underbrace{x_{0}(\omega ) \cos (\omega t) + p_{0}(\omega ) \frac{\sin (\omega t)}{\omega }}_{x^{0}_{t}(\omega )} + \int _{0}^{t} \frac{\sin ( \omega (t-s))}{\omega } \sqrt{\frac{2\omega }{\hbar }} c(\omega ) f(X(s)) ds. \end{aligned}$$

(129)

Substituting this into the equation for P(t) results in:

$$\begin{aligned} \dot{P}(t)&= -U'(X(t)) + f'(X(t)) \int _{\mathbb {R}^{+}} d\omega c(\omega ) \sqrt{\frac{2\omega }{\hbar }} x^{0}_{t}(\omega ) \nonumber \\&\quad + \frac{2}{\hbar } f'(X(t)) \int _{\mathbb {R}^{+}} d\omega |c(\omega )|^2 \int _{0}^{t} ds \sin (\omega (t-s)) f(X(s))\nonumber \\&\quad - 2 f(X(t)) f'(X(t)) \int _{\mathbb {R}^{+}} d\omega r(\omega ). \end{aligned}$$

(130)

Using integration by parts, we obtain

$$\begin{aligned} \int _{0}^{t} ds \sin ( \omega (t-s)) f(X(s)) = \frac{f(X(t))}{\omega } - f(X)\frac{\cos (\omega t)}{\omega } - \int _{0}^{t} \frac{\cos (\omega (t-s))}{\omega } \frac{d}{ds}\left( f(X(s)) \right) ds \end{aligned}$$

(131)

and therefore,

$$\begin{aligned} \dot{P}(t)= & {} -U'(X(t)) + f'(X(t)) \underbrace{\int _{\mathbb {R}^{+}} d\omega c(\omega ) (b^{\dagger }_t(\omega ) + b_t(\omega ) ) }_{\zeta (t)} \nonumber \\&- f'(X(t)) \int _{0}^{t} ds \underbrace{\int _{\mathbb {R}^{+}} d\omega 2 r(\omega ) \cos (\omega (t-s))}_{\kappa (t-s)} \frac{d}{ds}\left( f(X(s)) \right) \nonumber \\&- f'(X(t)) f(X) \underbrace{\int _{\mathbb {R}^{+}} d\omega 2r(\omega ) \cos (\omega t)}_{\kappa (t)}, \end{aligned}$$

(132)

where

$$\begin{aligned} \frac{d}{ds}\left( f(X(s)) \right) = \frac{i}{\hbar }[H, f(X(s))] = \frac{\{ f'(X(s)), P(s)\}}{2 m}, \end{aligned}$$

(133)

$b_t(\omega ) = b(\omega )e^{-i\omega t}$, $b_t^{\dagger }(\omega ) = b^{\dagger }(\omega ) e^{i\omega t}$ and $\{\cdot , \cdot \}$ denotes anti-commutator.

1.2 B Solving the Operator Lyapunov Equation

We outline the derivation of the solution, $\varvec{\bar{J}}$, to the operator Lyapunov equation:

$$\begin{aligned} \varvec{\hat{\gamma }}(\bar{X}(s)) \varvec{\bar{J}} + \varvec{\bar{J}} \varvec{\hat{\gamma }}(\bar{X}(s))^{T} = \varvec{\sigma } \varvec{\sigma }^{T}, \end{aligned}$$

(134)

where $\varvec{\hat{\gamma }}$ and $\varvec{\sigma }$ are block operator matrices, defined in Sect. 6. First, we observe that upon taking transpose on both sides of the equation, we have $\varvec{\hat{\gamma }}(\bar{X}(s)) \varvec{\bar{J}}^{T} + \varvec{\bar{J}}^{T} \varvec{\hat{\gamma }}(\bar{X}(s))^{T} = \varvec{\sigma } \varvec{\sigma }^{T}$, so uniqueness of the solution implies $\varvec{\bar{J}} = \varvec{\bar{J}}^{T}$, i.e. $J_{k,l} = J_{l,k}$ for all k, l.

We write $\varvec{\bar{J}}$ in the block-structure form:

$$\begin{aligned} \varvec{\bar{J}} = \left[ \begin{array}{cc} \varvec{J}_1 &{} \varvec{J}_2 \\ \varvec{J}_2^{T} &{} \varvec{J}_4 \end{array} \right] , \end{aligned}$$

(135)

where

$$\begin{aligned} \varvec{J}_1 = \left[ \begin{array}{cc} J_{1,1} &{} J_{1,2} \\ J_{1,2} &{} J_{2,2} \end{array} \right] , \ \ \ \varvec{J}_2 = \left[ \begin{array}{ccc} J_{1,3} &{} J_{1,4} &{} \cdots \\ J_{2,3} &{} J_{2,4} &{} \cdots \end{array} \right] \ \text { and } \ \varvec{J}_4 = \left[ \begin{array}{ccc} J_{3,3} &{} J_{3,4} &{} \cdots \\ J_{4,3} &{} J_{4,4} &{} \cdots \\ \vdots &{} \vdots &{} \ddots \end{array} \right] . \end{aligned}$$

(136)

Working out the matrix multiplications of the block operator matrices in the equation gives

$$\begin{aligned} \varvec{J}_4 = \frac{1}{2} \varvec{D}^{-1} \varvec{\varSigma }^2, \end{aligned}$$

(137)

which is a diagonal block operator matrix, and the following Sylvester-type equations:

$$\begin{aligned} \varvec{A} \varvec{J}_2 + \varvec{J}_2 \varvec{D}&= - \frac{1}{2} \varvec{B} \varvec{D}^{-1} \varvec{\varSigma }^2, \end{aligned}$$

(138)

$$\begin{aligned} \varvec{A} \varvec{J}_1 + \varvec{J}_1 \varvec{A}^{T}&= -\varvec{B} \varvec{J}_2^{T} - \varvec{J}_2 \varvec{B}^{T}. \end{aligned}$$

(139)

Equation (138) gives a system of linear equations for $J_{1,n+3}$ and $J_{2,n+3}$, for $n=0,1,\ldots$:

$$\begin{aligned} \frac{f'}{m_0} J_{2,n+3} + a_n J_{1,n+3}&= \frac{f'}{2m_0} \frac{\varSigma _n^2}{a_n}, \end{aligned}$$

(140)

$$\begin{aligned} -a_0 f' J_{1,n+3} + (a_0 + a_n) J_{2,n+3}&= 0, \end{aligned}$$

(141)

which has the solution:

$$\begin{aligned} J_{2,n+3}&= \frac{\varSigma _n^2}{2m_0} \frac{a_0}{a_n^2(a_0+a_n)} \left[ I+\frac{a_0}{m_0 a_n(a_0+a_n)} (f')^2 \right] ^{-1} (f')^2, \end{aligned}$$

(142)

$$\begin{aligned} J_{1,n+3}&= \frac{\varSigma _n^2}{2m_0 a_n^2} \left[ I+\frac{a_0}{m_0 a_n(a_0+a_n)} (f')^2 \right] ^{-1} f', \end{aligned}$$

(143)

where we have used the fact that $h(X)g(X) = g(X) h(X)$ for any functions g, h. Similarly, Eq. (139) gives:

$$\begin{aligned} J_{1,2}&= \sum _{n=0}^{\infty } J_{1,n+3}, \ \ \ J_{2,2} = \frac{1}{2} \left\{ f', \sum _{n=0}^{\infty } J_{1,n+3} \right\} , \end{aligned}$$

(144)

$$\begin{aligned} J_{1,1}&= \left( (f')^{-1} + \frac{1}{m_0 a_0} f'\right) \sum _{n=0}^{\infty } J_{1,n+3} - \frac{1}{m_0 a_0} \sum _{n=0}^{\infty } J_{2,n+3}. \end{aligned}$$

(145)

Substituting the expressions for $J_{2,n+3}$ and $J_{1,n+3}$ from (142) to (143) into the above equation gives the formula for $J_{1,2}$, $J_{2,2}$ and $J_{1,1}$. In particular,

$$\begin{aligned} J_{1,1} = \sum _{n=0}^{\infty } \left\{ \frac{\varSigma _n^2}{2m_0 a_n^2} \left[ I+\frac{a_n}{m_0 a_0(a_0+a_n)} (f')^2 \right] \left[ I + \frac{a_0}{m_0 a_n(a_0+a_n)} (f')^2 \right] ^{-1} \right\} . \end{aligned}$$

(146)

We remark that upon taking the limit, the contributions involving the $J_{1,2}$ and $J_{1,3}$ cancel each other, as in the classical situation, and so the contributions coming from $J_{1,n}$ ($n \ge 4$) are indeed correction drift terms induced by purely quantum noises.