Entropy Flow in Near-Critical Quantum Circuits

Abstract

Near-critical quantum circuits close to equilibrium are ideal physical systems for asymptotically large-scale quantum computers, because their low energy collective excitations evolve reversibly, effectively isolated from microscopic environmental fluctuations by the renormalization group. Entropy flows in near-critical quantum circuits near equilibrium as a locally conserved quantum current, obeying circuit laws analogous to the electric circuit laws. These “Kirchhoff laws” for entropy flow are the fundamental design constraints for asymptotically large-scale quantum computers. A quantum circuit made from a near-critical system (of conventional type) is described by a relativistic 1+1 dimensional relativistic quantum field theory on the circuit. The quantum entropy current near equilibrium is just the energy current divided by the temperature. The universal properties of the energy–momentum tensor constrain the entropy flow characteristics of the circuit components: the entropic conductivity of the quantum wires and the entropic admittance of the quantum circuit junctions. For example, near-critical quantum wires are always resistanceless inductors for entropy. A universal formula is derived for the entropic conductivity: $\sigma _{S}(\omega ) = iv^{2} {\mathcal {S}}/\omega T$, where $\omega$ is the frequency, T the temperature, ${\mathcal {S}}$ the equilibrium entropy density and v the velocity of “light”. The thermal conductivity is ${\mathbf {Re}}(T\sigma _{S}(\omega ))=\pi v^{2} {\mathcal {S}}\, \delta (\omega )$. The thermal Drude weight is, universally, $v^{2}{\mathcal {S}}$. This gives a way to measure the entropy density directly.

Appendices

Appendix 1: Equal-Time Commutators of $T_{t}^{t}(x,t)$ and $T_{x}^{t}(x,t)$

The universal equal-time commutators of the energy and momentum densities

$$\begin{aligned} \frac{i}{\hbar } {[}T_{t}^{t}(x',t), \,T_{t}^{t}(x,t){]}= & {} -\partial _{x}\delta (x'-x) 2 T_{t}^{x}(x,t) - \delta (x'-x) \partial _{x} T_{t}^{x}(x,t) \end{aligned}$$

(106)

$$\begin{aligned} \frac{i}{\hbar } {[}T_{x}^{t}(x',t), \,T_{x}^{t}(x,t){]}= & {} \partial _{x}\delta (x'-x) 2 T_{x}^{t}(x,t) + \delta (x'-x) \partial _{x} T_{x}^{t}(x,t) \end{aligned}$$

(107)

$$\begin{aligned} \frac{i}{\hbar } {[}T_{t}^{t}(x',t), \,T_{x}^{t}(x,t){]}= & {} - \frac{c_{ U V}}{6} \frac{\hbar v}{2\pi } \partial _{x}^{3} \delta (x'-x) +\partial _{x}\delta (x'-x) [T_{t}^{t}(x,t)-T_{x}^{x}(x,t)] \nonumber \\&\qquad \qquad {} -\delta (x'-x)\partial _{x}T_{x}^{x}(x,t) \end{aligned}$$

(108)

are derived here from the Ward identities for the operator product of two energy–momentum tensors. The number $c_{ U V}$ is the bulk conformal central charge at short-distance.

Make an infinitesimal local variation of the space-time metric, $g_{\mu \nu }\rightarrow g_{\mu \nu } + \delta g_{\mu \nu }(x,t)$, combined with an infinitesimal space-time transformation, $x^{\mu }\rightarrow x^{\mu }+\delta x^{\mu }(x,t)$. The combined change in the metric is

$$\begin{aligned} g_{\mu \nu }\rightarrow g_{\mu \nu } +\partial _{\mu }(\delta x_{\nu }) +\partial _{\nu }(\delta x_{\mu }) + \delta g_{\mu \nu } + \delta x^{\alpha }\partial _{\alpha }(\delta g_{\mu \nu }) + \partial _{\mu }(\delta x^{\alpha })\delta g_{\alpha \nu } + \partial _{\nu }(\delta x^{\alpha })\delta g_{\mu \alpha } \,. \end{aligned}$$

(109)

Vary $\ln Z$, keeping terms that are first order in $\delta x^{\mu }$ and in $\delta g_{\mu \nu }$, to obtain the Ward identity on the time-ordered product of two energy–momentum tensors:

$$\begin{aligned} \frac{i}{\hbar }\,\partial '_{\mu '} \, t\{ T_{\nu '}^{\mu '}(x',t') \,T_{\nu }^{\mu }(x,t) \}= & {} \partial _{\alpha }\left[ \delta (x'-x)\delta (t'-t)\right] \left( \delta ^{\alpha }_{\nu '} T_{\nu }^{\mu } -g_{\nu '\nu }g^{\alpha \beta }T_{\beta }^{\mu } -\delta ^{\mu }_{\nu '} T^{\alpha }_{\nu } \right) (x,t) \nonumber \\&\qquad + \delta (x'-x)\delta (t'-t) \partial _{\nu '}T_{\nu }^{\mu }(x,t) \,. \end{aligned}$$

(110)

Integrate both sides of the Ward identity over $t'$ from $t-\epsilon$ to $t+\epsilon$ to obtain:

$$\begin{aligned}&{ \frac{i}{\hbar }{[}T_{\nu '}^{t}(x',t), \,T_{\nu }^{\mu }(x,t) {]} + \partial _{x'}\, \int _{t-\epsilon }^{t+\epsilon } \mathrm {d}t' \,\, \frac{i}{\hbar }\, t\{ T_{\nu '}^{x}(x',t') \,T_{\nu }^{\mu }(x,t) \} =} \nonumber \\&\int _{t-\epsilon }^{t+\epsilon } \mathrm {d}t' \,\, \bigg \{ \partial _{\alpha }\left[ \delta (x'-x)\delta (t'-t)\right] \left( \delta ^{\alpha }_{\nu '} T_{\nu }^{\mu } -g_{\nu '\nu }g^{\alpha \beta }T_{\beta }^{\mu } -\delta ^{\mu }_{\nu '} T^{\alpha }_{\nu } \right) (x,t) \nonumber \\&\qquad \qquad \qquad {}+ \delta (x'-x)\delta (t'-t) \partial _{\nu '}T_{\nu }^{\mu }(x,t) \bigg \} \,. \end{aligned}$$

(111)

The time integral on the lhs picks out the contact terms in the time-ordered operator product. The energy–momentum tensor has scaling dimension 2, so the contribution of the contact terms has the form:

$$\begin{aligned}&{ \int _{t-\epsilon }^{t+\epsilon } \mathrm {d}t' \,\, \frac{i}{\hbar }\, t\{ T_{\nu '}^{\mu '}(x',t') \,T_{\nu }^{\mu }(x,t) \} = } \nonumber \\&\int _{t-\epsilon }^{t+\epsilon } \mathrm {d}t' \,\, \left[ C_{\nu '\nu }^{\mu '\mu \alpha \beta } \partial _{\alpha }\partial _{\beta } +B_{\nu '\nu }^{\mu '\mu \alpha }(x,t) \partial _{\alpha } + A_{\nu '\nu }^{\mu '\mu }(x,t) \right] \left[ \delta (x'-x)\delta (t'-t)\right] \quad \end{aligned}$$

(112)

for some operator-valued tensors A, B, C. The operators $C_{\nu '\nu }^{\mu '\mu \alpha \beta }$ have scaling dimension 0, so are multiples of the identity.

By (111) and (112), the equal-time commutators of the energy and momentum densities are:

$$\begin{aligned} \frac{i}{\hbar } {[}T^{t}_{\nu '}(x',t), \,T^{t}_{\nu }(x,t){]}= & {} c_{\nu '\nu }\partial _{x}^{3}\delta (x'-x) + b_{\nu '\nu }(x,t) \partial _{x}^{2}\delta (x'-x) \nonumber \\&{}+ \left( \delta _{\nu '}^{x} T_{\nu }^{t} - \delta _{\nu '}^{t} T_{\nu }^{x} -g_{\nu '\nu } T_{x}^{t} +a_{\nu '\nu } \right) (x,t) \partial _{x}\delta (x'-x) \nonumber \\&{}+\partial _{\nu '}T_{\nu }^{t}(x,t) \delta (x'-x) \end{aligned}$$

(113)

where $a_{\nu '\nu }=A^{tt}_{\nu '\nu }$, $b_{\nu '\nu }=B_{\nu '\nu }^{ttxx}$, and $c_{\nu '\nu }=C_{\nu '\nu }^{ttxx}$.

The antisymmetry of the commutators is equivalent to:

$$\begin{aligned} 0= & {} c_{\nu '\nu }-c_{\nu \nu '}\end{aligned}$$

(114)

$$\begin{aligned} 0= & {} b_{\nu '\nu }+b_{\nu \nu '} \end{aligned}$$

(115)

$$\begin{aligned} 0= & {} \partial _{x} (a_{xx}-2T^{t}_{x}) \end{aligned}$$

(116)

$$\begin{aligned} 0= & {} \partial _{x} a_{tt} \end{aligned}$$

(117)

$$\begin{aligned} 0= & {} \partial _{x} (a_{xt}+a_{tx}+T^{x}_{x}-T^{t}_{t}) \end{aligned}$$

(118)

$$\begin{aligned} 0= & {} a_{tx}-a_{xt} - 2 \partial _{x}b_{tx} -T_{t}^{t} - T_{x}^{x} \,. \end{aligned}$$

(119)

Therefore $b_{xx}=b_{tt}=0$, $b_{xt}=-b_{tx}$, and, up to multiples of the identity operator,

$$\begin{aligned} a_{xx}= & {} 2T^{t}_{x} \end{aligned}$$

(120)

$$\begin{aligned} a_{tt}= & {} 0 \end{aligned}$$

(121)

$$\begin{aligned} a_{xt}= & {} \partial _{x}b_{tx}- T^{x}_{x} \end{aligned}$$

(122)

$$\begin{aligned} a_{tx}= & {} \partial _{x}b_{tx} + T^{t}_{t} \,. \end{aligned}$$

(123)

Ignoring multiples of the identity operator for the time being, the only unknown is the operator $b_{tx}(x,t)$. The equal-time commutators are, up to multiples of the identity,

$$\begin{aligned} \frac{i}{\hbar } {[}T_{t}^{t}(x',t), \,T_{t}^{t}(x,t){]}= & {} -\partial _{x}\delta (x'-x) 2 T_{t}^{x}(x,t) - \delta (x'-x) \partial _{x} T_{t}^{x}(x,t) \end{aligned}$$

(124)

$$\begin{aligned} \frac{i}{\hbar } {[}T_{x}^{t}(x',t), \,T_{x}^{t}(x,t){]}= & {} \partial _{x}\delta (x'-x) 2 T_{x}^{t}(x,t) + \delta (x'-x) \partial _{x} T_{x}^{t}(x,t) \end{aligned}$$

(125)

$$\begin{aligned} \frac{i}{\hbar } {[}T_{t}^{t}(x',t), \,T_{x}^{t}(x,t){]}= & {} +\partial _{x}\delta (x'-x) (T_{t}^{t}-T_{x}^{x})(x,t) -\delta (x'-x)\partial _{x}T_{x}^{x}(x,t) \nonumber \\&+ \partial _{x} \left[ \partial _{x}\delta (x'-x) b_{tx}(x,t) \right] \,. \end{aligned}$$

(126)

Take the time derivative of both sides of (124). In the time derivative of (124), use (126) to evaluate the commutators. The equation that results is:

$$\begin{aligned} 0 = 2 \partial _{x}^{3}\delta (x'-x)b_{tx}(x,t) +3 \partial _{x}^{2}\delta (x'-x)\partial _{x}b_{tx}(x,t) +\partial _{x}\delta (x'-x)\partial _{x}^{2}b_{tx}(x,t) \,. \end{aligned}$$

(127)

So $b_{tx}(x,t)=0$.

Equations (124–126), with $b_{tx}(x,t)=0$, give the equal-time commutators up to multiples of the identity. These are exactly (106–108), up to multiples of the identity. So all that remains is to determine the multiples of the identity operator that appear in the equal-time commutators.

The terms proportional to the identity operator in (106–108) are determined by evaluating the expectation values of the equal-time commutators in the ground-state. The spectral representation of the ground-state two-point function of the energy–momentum tensor is: [9]

$$\begin{aligned}&\bigg \langle \big |\frac{i}{\hbar }\, t \{ T_{\nu '}^{\mu '}(x',t') \, T_{\nu }^{\mu }(x,t) \}\big | \bigg \rangle = \int _{0}^{\infty } \mathrm {d}(m^{2})\,\rho _{c}(m^{2}) \; G^{\mu '\mu }_{\nu '\nu }(x'-x,t'-t;m^{2})\quad \end{aligned}$$

(128)

$$\begin{aligned}&G^{\mu '\mu }_{\nu '\nu }(x,t;\mu ) = \frac{1}{(2\pi )^{2}} \int \int \mathrm {d}p_{x}\mathrm {d}p_{t} \,\, \mathrm {e}^{i p_{x}x +i p_{t}t} \frac{(p_{\nu '}p^{\mu '}-\delta _{\nu '}^{\mu '}p^{2}) (p_{\nu }p^{\mu }-\delta _{\nu }^{\mu }p^{2})}{p_{\mu }p^{\mu }+m^{2}+i\epsilon }.\nonumber \\ \end{aligned}$$

(129)

The conformal central charge in the short distance limit, $c_{ U V}$, is given by

$$\begin{aligned} \int \mathrm {d}(m^{2})\,\rho _{c}(m^{2}) =\frac{c_{ U V}}{6} \frac{\hbar v}{2\pi } \,. \end{aligned}$$

(130)

Extract the equal-time commutator from (128) by evaluating at $t'=t+\epsilon$ and at $t'=t-\epsilon$ and taking the difference:

$$\begin{aligned}&{ \big \langle 0| \, \frac{i}{\hbar }\,{[}T_{\nu '}^{\mu '}(x',t)\,T_{\nu }^{\mu }(x,t){]} \, |0\big \rangle = \int _{0}^{\infty } \mathrm {d}(m^{2})\,\rho _{c}(m^{2}) \,\,\frac{1}{2\pi } \int \mathrm {d}p_{x} \,\, \mathrm {e}^{i p_{x}(x'-x)} } \nonumber \\&\qquad \qquad \frac{1}{2\pi } \int \mathrm {d}p_{t} \,\, \frac{\mathrm {e}^{ip_{t}\epsilon } - \mathrm {e}^{-ip_{t}\epsilon }}{p_{\mu }p^{\mu }+m^{2}+i\epsilon } \left[ (p_{\nu '}p^{\mu '}-\delta _{\nu '}^{\mu '}p^{2}) (p_{\nu }p^{\mu }-\delta _{\nu }^{\mu }p^{2}) \right] \,. \end{aligned}$$

(131)

In particular,

$$\begin{aligned} \big \langle 0| \, \frac{i}{\hbar }{[}T_{t}^{t}(x',t)\,T_{t}^{t}(x,t){]} \, |0\big \rangle= & {} 0 \end{aligned}$$

(132)

$$\begin{aligned} \big \langle 0| \, \frac{i}{\hbar }{[}T_{x}^{t}(x',t)\,T_{x}^{t}(x,t){]} \, |0\big \rangle= & {} 0 \end{aligned}$$

(133)

$$\begin{aligned} \big \langle 0| \, \frac{i}{\hbar }{[}T_{t}^{t}(x',t)\,T_{x}^{t}(x,t){]} \, |0\big \rangle= & {} - \partial _{x}^{3} \delta (x'-x) \frac{c_{ U V}}{6} \frac{\hbar v}{2\pi } \,. \end{aligned}$$

(134)

This fixes the terms proportional to the identity operator in (106–108), finishing their derivation.

Appendix 2: $\sigma _{S}(\omega )= iv^{2} {\mathcal {S}}/\omega T$ from the Kubo formula

The Kubo formula for the entropy current induced in a wire by an entropic potential $\Delta V_{S}(x,t)$ is

$$\begin{aligned} \Delta \langle j_{S}(x_{2},t_{2}) \rangle= & {} \int _{-\infty }^{t_{2}}\mathrm {d}t_{1} \,\, \langle \frac{i}{\hbar }{[}\Delta H(t_{1}), j_{S}(x_{2},t_{2}){]} \rangle _{ eq } \nonumber \\= & {} \int _{-\infty }^{t_{2}}\mathrm {d}t_{1} \,\, \langle \frac{i}{\hbar }{[}\int \mathrm {d}x_{1}\, \Delta V_{S}(x_{1},t_{1}) \rho _{S}(x_{1},t_{1}), j_{S}(x_{2},t_{2}){]} \rangle _{ eq } \,. \end{aligned}$$

(135)

The Kubo formula is the solution of the time evolution equation, (49), in the linear response approximation.

For an alternating entropic potential, $\Delta V_{S}(x,t)=\mathrm {e}^{iqx-i\omega t}\Delta V_{S}(0,0)$, the induced current is

$$\begin{aligned} \Delta \langle j_{S}(x,t) \rangle =\sigma _{S}(q,\omega ) \Delta E_{S}(x,t) \end{aligned}$$

(136)

where $\Delta E_{S}(x,t) = -iq \Delta V_{S}(x,t)$. The Kubo formula for the entropic conductivity is

$$\begin{aligned} \sigma _{S}(q,\omega )= & {} \frac{i}{q} \int \mathrm {d}x_{1}\, \int _{-\infty }^{t_{2}}\mathrm {d}t_{1} \,\, \mathrm {e}^{i\omega (t_{2}- t_{1})-iq(x_{2}-x_{1})} \bigg \langle \frac{i}{\hbar }{[} \rho _{S}(x_{1},t_{1}), j_{S}(x_{2},t_{2}){]}\bigg \rangle _{eq} \nonumber \\= & {} k^{2}\beta ^{2} \frac{i}{q} \int \mathrm {d}x_{1}\, \int _{-\infty }^{t_{2}}\mathrm {d}t_{1} \,\, \mathrm {e}^{i\omega (t_{2}- t_{1})-iq(x_{2}-x_{1})}\nonumber \\&\qquad \qquad \qquad \qquad \qquad \bigg \langle \frac{i}{\hbar }{[} T_{t}^{t}(x_{1},t_{1}), T_{t}^{x}(x_{2},t_{2}){]}\bigg \rangle _{eq} \,. \end{aligned}$$

(137)

Introduce the Fourier transform of the energy–momentum tensor:

$$\begin{aligned} \tilde{T}_{\nu }^{\mu }(p,\eta ) = \int \mathrm {d}x \int \mathrm {d}t \; \mathrm {e}^{i(\eta t-p x)} T_{\nu }^{\mu }(x,t) \,. \end{aligned}$$

(138)

Write its two-point functions:

$$\begin{aligned} \langle \tilde{T}_{\nu '}^{\mu '}(p',\eta ') \, \tilde{T}_{\nu }^{\mu }(p,\eta ) \rangle _{eq}= (2\pi )^{2} \delta (p'+p) \delta (\eta '+\eta ) G_{\nu '\nu }^{\mu '\mu }(p,\eta ) \,. \end{aligned}$$

(139)

The equilibrium expectation values of the commutators are given by

$$\begin{aligned} \bigg \langle \frac{i}{\hbar }{[} \tilde{T}_{\nu '}^{\mu '}(p',\eta ') \, \tilde{T}_{\nu }^{\mu }(p,\eta ){]} \bigg \rangle _{eq} = (2\pi )^{2} \delta (p'+p) \delta (\eta '+\eta ) \frac{i}{\hbar }\left( 1-\mathrm {e}^{\beta \hbar \eta } \right) G_{\nu '\nu }^{\mu '\mu }(p,\eta ).\nonumber \\ \end{aligned}$$

(140)

The Kubo formula becomes

$$\begin{aligned} \sigma _{S}(q,\omega ) = \frac{k^{2}\beta ^{2}}{\hbar } \int \mathrm {d}\eta \; \frac{1}{\omega +i\epsilon -\eta } \left( 1-\mathrm {e}^{\beta \hbar \eta } \right) \frac{1}{iq} G_{tt}^{tx}(q,\eta ) \,. \end{aligned}$$

(141)

Conservation and symmetry of the energy–momentum tensor imply

$$\begin{aligned} \eta \tilde{T}_{t}^{x}(q,\eta ) = -v^{2} q \tilde{T}_{x}^{x}(q,\eta ) \end{aligned}$$

(142)

so

$$\begin{aligned} \frac{1}{q} G_{tt}^{tx}(q,\eta ) = - \frac{v^{2}}{\eta } G_{tx}^{tx}(q,\eta ) \end{aligned}$$

(143)

so

$$\begin{aligned} \sigma _{S}(q,\omega ) = k^{2}v^{2}\beta ^{3} \int \mathrm {d}\eta \; \frac{i}{\omega +i\epsilon -\eta } \left( 1-\mathrm {e}^{\beta \hbar \eta } \right) \frac{1}{\beta \hbar \eta } G_{tx}^{tx}(q,\eta ) \,. \end{aligned}$$

(144)

In the uniform limit, $q\rightarrow 0$,

$$\begin{aligned} \lim _{q\rightarrow 0} G_{tx}^{tx}(q,\eta ) = \delta (\eta ) \langle H_{0}\, T_{x}^{x}(x,t) \rangle _{ eq } = -\delta (\eta ) \frac{\partial }{\partial \beta }\langle T_{x}^{x}(x,t) \rangle _{ eq } \end{aligned}$$

(145)

so

$$\begin{aligned} \sigma _{S}(\omega ) =\lim _{q\rightarrow 0} \sigma _{S}(q,\omega ) = k^{2}v^{2}\beta ^{3} \frac{i}{\omega +i\epsilon } \frac{\partial }{\partial \beta }\langle T_{x}^{x}(x,t) \rangle _{ eq } \end{aligned}$$

(146)

The equilibrium entropy density is [see (75]:

$$\begin{aligned} {\mathcal {S}}= k\beta ^{2} \frac{\partial }{\partial \beta }\langle T_{x}^{x}(x,t) \rangle _{ eq } \end{aligned}$$

(147)

so

$$\begin{aligned} \sigma _{S}(\omega ) = \frac{i k\beta v^{2}{\mathcal {S}}}{\omega } \,. \end{aligned}$$

(148)

The thermal conductivity is

$$\begin{aligned} \kappa (\omega ,T) = {\mathbf {Re}}( T \sigma _{S}(\omega )) = \pi v^{2} {\mathcal {S}}\delta (\omega ) \,. \end{aligned}$$

(149)