Quantum Simulation of Gauge Theories

Scott Lawrence
in collaboration with Andrei Alexandru, Paulo Bedaque, Siddhartha Harmalkar,
Hersh Kumar, Henry Lamm, Neill Warrington, Yukari Yamauchi
"The best model of a cat is another cat"
Norbert Wiener

Neutron stars and little bangs

Thomas McCauley/CERN
$$ \mathscr L_{\mathrm{\color{red}Q\color{green}C\color{blue}D}} = - \frac 1 4 F_{\mu\nu} F^{\mu\nu} + \bar \psi \left(i \gamma^\mu D_\mu - m\right)\psi $$

The lattice, briefly

QCD is a non-perturbative theory. (There's no small parameter about which to Taylor expand.)

At every link lives a unitary matrix. $$ U_{yx} \in SU(3) $$

The action is a sum over all `plaquettes'. $$ S = \sum_P \mathrm{Re\,Tr\,} P $$

This is a theory of with a finite number of degrees of freedom. Expectation values are given by: $$ \left<\mathcal O\right> = \frac{\int \mathcal D U\;e^{-S[U]} \mathcal O[U]}{\int \mathcal D U\;e^{-S[U]}}$$

A sign problem

$$ \int f \approx 1 \pm 0.5$$
$$ \int f \approx 0 \pm 10^4 $$

$$ \langle x_1|e^{-i H t}|x_0\rangle = \int_{x(0)=x_0}^{x(1)=x_1} \mathcal D x(t) \;e^{i S[x]} $$

What calculations have a sign problem?

Calculations that have sign problems:

Closely related, but mild: determining the mass of a proton

Calculations without sign problems:


The best model of a quantum system is another quantum system.
Norbert Wiener, almost
A quantum computer is a quantum system evolved in real-time.

Set up an analogy between the quantum computer and the system to be simulated, and treat the computer like a (perfectly controlled) laboratory.

From Bit to Qubit

$$ |0\rangle,|1\rangle,\ldots,\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right) $$

This is a spin from QM.

A Quantum Computer

What's the Hilbert space? $$ \mathcal H = \mathcal H_1 \otimes \cdots \otimes \mathcal H_1 \;\text{ where } \mathcal H_1 = \mathrm{span}\{|0\rangle,|1\rangle\} $$

In other words, superpositions of all possible bitstrings. $$ \mathcal H = \mathrm{span}\{|00\cdots\rangle,|10\cdots\rangle,|01\cdots\rangle,\cdots\} $$

$$ \left({\color{blue}|0\rangle + i|1\rangle}\right)\left({\color{green}|0\rangle+|1\rangle}\right) = |{\color{blue}0\color{green}0}\rangle + |{\color{blue}0\color{green}1}\rangle + i|{\color{blue}1\color{green}0}\rangle + i|{\color{blue}1\color{green}1}\rangle $$ Not entangled
$$ |{\color{blue}0}{\color{green}0}\rangle + |{\color{blue}1}{\color{green}1}\rangle $$ Entangled


All operations must be unitary (conservation of probability). On one qubit:

$$ T = \left( \begin{matrix} 1 & 0\\ 0 & e^{i \pi / 4}\\ \end{matrix} \right) $$
$$ H = \frac 1 {\sqrt{2}}\left( \begin{matrix} 1 & 1\\ 1 & -1\\ \end{matrix} \right) $$

On two qubits, the controlled-not operation:

$$ CX = \left( \begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ \end{matrix} \right) \left( \begin{matrix} |00\rangle\\ |01\rangle\\ |10\rangle\\ |11\rangle\\ \end{matrix} \right) $$

These fundamental gates are sufficient to construct any unitary we want.


In principle, we measure any Hermitian operator. In practice, we measure \(\sigma_z\) acting on each qubit. $$\Psi = \alpha|0\rangle + \beta|1\rangle$$

Each measurement yields \(1\) or \(0\). Probability of \(1\) is \(\langle\Psi|1\rangle\langle 1|\Psi\rangle = |\beta|^2\).

Thus, we require many measurements for a precise result. This is "shot noise".

Hermitian operators are exactly those which may appear as terms in the Hamiltonian.

State of the art

Current best: \(\sim 50\) qubits.

Each qubit can undergo \(\lesssim 10\) operations before decohering.

Are large processors worth it?

What is a classical computer?

A quantum computer constantly being measured
\(|01\rangle\) is okay; \(\left[|10\rangle + |01\rangle\right]\) gets destroyed

This restricts the set of possible operations, as well. We only have permutations:

\[\left( \begin{matrix} 1& 0 & 0 & 0\\ 0& 1 & 0 & 0\\ 0& 0 & 0 & 1\\ 0& 0 & 1 & 0\\ \end{matrix} \right) \] and not \[ \left( \begin{matrix} -i & 0\\ 0 & i\\ \end{matrix} \right) \]

Classical algorithms are quantum algorithms

Any classical circuit can be re-labelled as a quantum circuit.

Example: two-bit adder $$ \begin{align} |00\rangle |00\rangle &\rightarrow |00\rangle |00\rangle \\ |01\rangle |00\rangle &\rightarrow |01\rangle |01\rangle \\ |10\rangle |00\rangle &\rightarrow |10\rangle |01\rangle \\ |11\rangle |00\rangle &\rightarrow |11\rangle |10\rangle \\ \end{align} $$

In general, given a classical circuit for \(f(x)\), we can obtain $$ |x\rangle|0\rangle \rightarrow |x\rangle |f(x)\rangle $$

Suzuki-Trotter decomposition

$$ H = {\color{blue}A} + {\color{green}B} $$

Assume we can evolve under \(A\) and \(B\). How to get evolution under \(H\)? $$ e^{-i H \epsilon} \approx {\color{blue}e^{-i A \epsilon}} {\color{green}e^{-i B \epsilon}} $$

So, time-evolve by rapidly alternating between two Hamiltonians $$ e^{-i H t} \approx \left({\color{blue}e^{-i A \epsilon}} {\color{green}e^{-i B \epsilon}}\right)^{t / \epsilon} $$

Field theories, the finite way

Each lattice site has a degree of freedom with Hilbert space \(\mathcal H_1\). The whole system has Hilbert space $$ \mathcal H = \mathcal H_1 \otimes \mathcal H_1 \otimes \cdots $$

Some Hamiltonian \(H\) couples the different lattice sites. For a spin system, we might have $$ H = \sum_{\langle i j\rangle} \sigma_z(i) \sigma_z(j) + \sum_i \sigma_x(i) $$

When correlations are large, the lattice structure is irrelevant, hence "field theory".

Quantum mechanics on a group

The Hamiltonian of a free particle moving on \(G = SU(3)\): $$ H = -\nabla_\ell^2 $$

Hilbert space is \(\mathbb C G\), the space of complex functions on \(G\).

$$ \left|\left( \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \right)\right>, \left|\left( \begin{matrix} 0 & 1 & 0\\ 0 & 0 & i\\ -i & 0 & 0\\ \end{matrix} \right)\right>, \left|\left( \begin{matrix} 0 & 0 & 1\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0\\ \end{matrix} \right)\right> $$

We can work with momentum eigenstates instead.


Gauge theories

$$ H = -\beta_K\sum_\ell \nabla_\ell^2 + \beta_P\sum_P \mathrm{Re\,Tr\,}P $$

Gauge symmetry

$$ U_{ij} \mapsto V_j U_{ij} V_i^\dagger $$
$$ \begin{align} &\mathrm{Tr\,} U_{14}^\dagger U_{45}^\dagger U_{25} U_{12}\\ \mapsto& \mathrm{Tr\,}U_{14}^\dagger U_{45}^\dagger U_{25} V^\dagger_2 V_2 U_{12} \end{align} $$

The Hamiltonian is gauge-invariant. $$ H = -\beta_K\sum_\ell \nabla_\ell^2 + \beta_P\sum_P \mathrm{Re\,Tr\,}P $$

Hilbert space

$$ \mathcal H = \mathbb C G \otimes \mathbb C G \otimes \cdots $$
Only gauge-invariant states are physical.
$$\color{red}|U_{12}\rangle $$ $$\color{green} \int \mathrm{d} V_1 \mathrm{d} V_2 |V^\dagger_{2} U_{12} V_1\rangle $$

Here's a projection operator: $$ P |U_{12}\cdots\rangle = \int(\mathrm{d}V_1 \mathrm{d}V_2\cdots)|(V^\dagger_2 U_{12} V_1)\cdots\rangle $$

Example: \(\mathbb Z_2\) gauge theory

$$ H = \sigma_x(a) + \sigma_x(b) + \sigma_z(a) \sigma_z(b) $$

Four states, naively: \(|00\rangle\), \(|01\rangle\), \(|10\rangle\), \(|11\rangle\). But these are not gauge-invariant!

Gauge transformation takes \(00 \leftrightarrow 11\) and \(01 \leftrightarrow 10\). $$ \color{green} |00\rangle + |11\rangle \;\text{ and }\; |01\rangle + |10\rangle $$

Simulating a field theory: general principles

We need a mapping between the Hilbert spaces. Locality is nice. $$ \mathcal H_1 \leftrightarrow \mathcal H_1^{\mathrm{QC}} $$

This induces a "nice" map \(\mathcal H \leftrightarrow \mathcal H^{\mathrm{QC}}\)

Implement \(e^{-i H t}\) via Suzuki-Trotter

Time evolution

$$ H = {\color{red}\overbrace{\sum_L \nabla^2_L}^{H_K}} + {\color{blue}\overbrace{\sum_P \mathrm{Re\,Tr\,} P}^{H_V}} $$

One link only

Diagonal in Fourier space

Mutually commuting terms


Four links

Diagonal (in our basis)

Mutually commuting terms

\[ e^{-i H t} \approx \left[{\color{red}\left( e^{-i \nabla_1^2 \epsilon} e^{-i \nabla_2^2 \epsilon} \cdots \right)} {\color{blue}\left( e^{-i \epsilon \mathrm{Re\,Tr\,} P_1} e^{-i \epsilon \mathrm{Re\,Tr\,} P_2} \cdots \right)} \right]^{t/\epsilon} \]

Example: \(\mathbb Z_2\) gauge theory (again)

$$ H = \sigma_x(a) + \sigma_x(b) + \sigma_z(a) \sigma_z(b) $$

Valentiner gauge theory

We'd like to simulate \(SU(3)\), but \(\mathbb C SU(3)\) is infinite-dimensional.

\[ \color{red}2^Q < \infty \]
We can approximate \(SU(3)\) by a finite subgroup.
\[ S(1080) < SU(3) \]

Is this approximation any good?

After adding an extra term to the action... $$ S=-\sum_p \left(\frac{\beta_0}{3}\mathrm{Re\,Tr\,} U_p +\beta_1\mathrm{Re\,Tr\,} U_p^2\right) $$

Construct a dimensionless quantity from: Wilson flow, critical temperature

Particle masses not yet measured...


Gauge invariance

The physical Hilbert space is not \(\mathbb C S(1080) \otimes \mathbb C S(1080) \otimes\cdots\).

But, the Hilbert space on the quantum computer is isomorphic to that space!

Time evolution is gauge-invariant.
$$ [P, H] = 0 $$

If we start in a gauge-invariant state, we stay gauge-invariant.

The hadronic tensor

\[ W^{\mu\nu}(q) = \int {\mathrm d}x\;e^{iqx} \left<e^{i H x^0}J^\mu(\vec x) e^{-i H x^0}J^\nu(\vec 0)\right>_{\mathrm{proton}} \]

The hadronic tensor captures nonperturbative (in QCD coupling) information about the proton. For electron-proton scattering, to leading order in \(\alpha_{\mathrm{QED}}\): \[ \frac{{\mathrm d}^2\sigma}{\mathrm{d} x\;\mathrm{d} y} = \frac{\alpha^2 y}{Q^4} L_{\mu\nu} W^{\mu\nu} \]

Preparing an interesting state

Naive state preparation: couple to a heat bath and cool the system. Expensive!

Alternative: adiabatic state preparation

Other proposals:

Hard to analyze, impossible to test

Adiabatic theorem

Take a time-varying (slowly) Hamiltonian \(H(t)\).

Prepare an eigenstate of \(H(0)\), with a gap of \(\Delta\).

When \(\dot H / \Delta^2 \ll 0\), time-evolution will keep us in the eigenstate.

Time needed to prepare ground state: \(\Delta^{-2}\)

Making a proton

Restrict to a certain sector of Hilbert space:

  • Free fermions and glue (massive)
  • Ground state exactly prepared
  • Small gap (\(O(\frac 1 V)\))
  • Hadrons
  • Large gap (\(m_\pi\))
Total circuit size: \(O(V^3)\)

Measuring the hadronic tensor

\[ W^{\mu\nu}(q) = \int {\mathrm d}x\;e^{iqx} \left<e^{i H x^0}J^\mu(\vec x) e^{-i H x^0}J^\nu(\vec 0)\right>_{\mathrm{proton}} \]

How do we measure \(e^{i H t}J_0(x)e^{-i H t} J_0(0)\)? \[ H(t) = H_0 + \epsilon \delta(t) J^\nu(\vec 0) \]

Linear response: differentiate with respect to \(\epsilon\) \[ \frac{\partial}{\partial \epsilon} \langle e^{i H t} J^\mu(\vec x) e^{-i H t}\rangle \]

Another target: transport coefficients

Shear/bulk viscosity, conductivity, etc... $$ \eta = \frac T V \int \mathrm{d}^4 x\; \langle \pi_{ij}(x) \pi_{ij}(0) \rangle $$ $$ \sigma \sim \int \mathrm{d}t\;\langle j(t) j(0)\rangle $$

The expectation value still has the form of linear response: $$ H(t) = H_0 + \epsilon \delta(t) \pi_{xy}(0) $$

This algorithm looks a lot like a physical experiment!

Avoiding state preparation

State preparation dominates the gate cost of the calculation.

Can we skip state preparation?

Classical lattice methods are very good at simulating thermodynamics; can't do real-time.

Quantum computers simulate real-time evolution easily; thermodynamics can be expensive.

Hybrid algorithm: thermodynamics classically, evolution quantumly

The Euclidean lattice

Classical thermodynamics

Boltzmann factor \(p_i \propto e^{-\beta E_i}\)

Partition function \(Z = \sum_i e^{-\beta E_i}\)

Expectation values \(\langle \mathcal O\rangle = \sum_i p_i \mathcal O_i\)

Quantum thermodynamics

Density matrix \(\rho = e^{-\beta H}\)

Partition function \(Z = \mathrm{Tr\,}\rho\)

Expectation values \(\langle \mathcal O\rangle = Z^{-1} \mathrm{Tr\,} \mathcal O \rho\)

Several ways of turning \(\rho\) into a probability distribution to sample. We could take the diagonal $$ \langle \mathcal O \rangle = \frac{\sum_i \rho_{ii} \mathcal O_{ii}}{\sum_i \rho_{ii}} $$

Equality holds when \(\mathcal O\) is diagonal. (Common in lattice QCD.)

The real-time oracle

We want \(\langle \mathcal O(t)\rangle\); the operator isn't diagonal! $$ \langle \mathcal O(t)\rangle = \frac{\sum_{ij} \rho_{ij} \mathcal O(t)_{ji}}{\sum_{i}\rho_{ii}} = \left(\frac{\sum_{ij} \rho_{ij} \mathcal O(t)_{ji}}{\sum_{ij} \rho_{ij}}\right)\color{red}\underbrace{\left(\frac{\sum_{ij} \rho_{ij}}{\sum_{i} \rho_{ii}}\right)}_{\mathrm{Normalization}} $$

Therefore: sample pairs of states \(|i\rangle\langle j|\), distributed by \(\rho_{ij}\).

How to compute \(\mathcal O(t)_{ij}\)? This part is done on a quantum computer.

Preparing a basis state \(|i\rangle\) is cheap. Start with the states $$ |+\rangle = |i\rangle + |j\rangle\;\mathrm{ and }\; |-\rangle = |i\rangle - |j\rangle $$ and look at \(\langle + | \mathcal O(t) | +\rangle - \langle - | \mathcal O(t) | -\rangle\).

How to sample

How do we sample \(\rho_{ii}\)?

Even computing \(\rho_{ij}\) is hard! Turns out, sampling is easier: $$ \mathrm{Tr\,} e^{-\beta H} = \sum_{i,j,k,\ldots} \color{green}\left(e^{-\delta H}\right)_{ij} \left(e^{-\delta H}\right)_{jk} \cdots $$

For small \(\delta\), the summand is easily computed. This is a joint distribution over \((i,j,k,\ldots)\). Marginalized, it approximates \(\rho_{ii}\).

Diagonal operators $$ \rho_{ii} \approx \left(e^{-\delta H}\right)_{ij} \left(e^{-\delta H}\right)_{jk} \left(e^{-\delta H}\right)_{ki} $$

(Periodic boundary conditions)

Arbitrary operators $$ \rho_{i\ell} \approx \left(e^{-\delta H}\right)_{ij} \left(e^{-\delta H}\right)_{jk} \left(e^{-\delta H}\right)_{k\ell} $$

(Open boundary conditions)

How much do we save?

One step of time evolution requires \(O(V)\) operations.

With state prep, need to evolve for \(O(T + \Delta^{-2})\) steps.

Characteristically, \(\Delta \sim V^{1/3}\). Per measurement: $$ \color{red} O(V^{5/3}) $$

Without state prep, each quantum computation is now $$ \color{green} O(V T) $$

At the cost of needing to perform many more computations (sampling \(\rho\)).

A big advantage for noisy quantum processors.

The catch

Signal-to-noise problem!
$$ \langle (\mathcal O_{ij})^2\rangle \gg \langle (\mathcal O)_{ij}\rangle^2 $$

The QC calculation is shorter, but we need many of them.

Probably only useful for near-term, noisy processors.
Thanks for watching!
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