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: $$ \left<\mathcal O\right> = \frac{\int \mathcal D U\;e^{-S[U]} \mathcal O[U]}{\int \mathcal D U\;e^{-S[U]}}$$

Calculations that have sign problems:

• Real-time response (as opposed to thermal equilibrium)
• Hubbard model away from half-filling
• Finite density of relativistic fermions

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

Calculations without sign problems:

• Zero-fermion-density equation of state
• Some masses (pion)

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

• Some quantum algorithms
• Hamiltonian lattice gauge theory
• Quantum simulations of field theory
• Hybrid algorithm for near-term quantum computers

This is a spin from QM.

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. $$ \mathcal H = \mathrm{span}\{|00\cdots\rangle,|10\cdots\rangle,|01\cdots\rangle,\cdots\} $$

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

On two qubits, the controlled-not operation:

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.

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

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 $$ |x\rangle|0\rangle \rightarrow |x\rangle |f(x)\rangle $$

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 $$ e^{-i H t} \approx \left({\color{blue}e^{-i A \epsilon}} {\color{green}e^{-i B \epsilon}}\right)^{t / \epsilon} $$

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".

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 \(G\).

We can work with momentum eigenstates instead.

wikimedia

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

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

$$ \mathcal H = \mathbb C G \otimes \mathbb C G \otimes \cdots $$

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

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

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

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

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

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!

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

When performing time evolution, we would like \( e^{-i H \Delta t} e^{-i H \Delta t} \cdots \), but instead we get: $$ \color{blue}e^{-i H \Delta t} \color{red} U_{\mathrm{drift}} \color{blue}e^{-i H \Delta t} \color{red} U_{\mathrm{drift}} \color{blue}e^{-i H \Delta t} \color{red} U_{\mathrm{drift}} \cdots $$

Suppress unphysical states with an energy penalty.

Define \( H_{\mathrm{gauss}} \) by:

and add \( H_{\mathrm{gauss}} \) to the Hamiltonian used for evolution.

Paths which pass through the unphysical subspace will pick up phase cancellations and be supressed.

We don't need to! We only need \( U_{\mathrm{gauss}} \sim e^{-i H_{\mathrm{gauss}} t} \)

At long times (equivalently, large energies), this gives each unphysical state a random phase. Approximate by performing a random gauge transformation. $$ e^{-i H \Delta t} \phi(g_0) e^{-i H \Delta t} \phi(g_1) e^{-i H \Delta t} \cdots $$

Each \(\phi(g)\) blesses an unphysical state with a random phase.

Inelastic scattering: need a \((4\;\mathrm{fm})^3\) box (at least). (See: Jordan, Lee, Preskill (2011).)

With a lattice spacing of \(0.1\;\mathrm{fm}\), that's \(2 \times 10^5\) links, or \( 2 \times 10^6\) qubits. Yikes!

What can we do with a small volume? Hydrodynamics

Early in a heavy-ion collision, temperature is \(\sim 300\;\mathrm{MeV}\); physics well-described by hydrodynamics. $$\def\d{\mathrm{d}} \rho \frac{\d u_i}{\d t} + \partial_i p = {\color{blue}\eta} \left(\frac 1 3 \partial_i \partial_j u_j + \partial_j^2 u_i \right) + {\color{blue}\zeta}\cdots $$

The fireball is small (\(\sim 10\;\mathrm{fm}\))! This is good for us: even smaller volumes might behave hydrodynamically.

EFT coefficients (shear viscosity, ...) not known from first principles.

Too cold? "Whack and wait".

Too hot? You're on your own.

Advantages: simple, cheap.

Can we go cheaper?

State preparation dominates the gate cost of the calculation.

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.

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.)

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 \(\langle + | \mathcal O(t) | +\rangle - \langle - | \mathcal O(t) | -\rangle\).

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 \(\rho_{ii}\).

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 \(\rho\)).

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

Create a shear wave: \[ u_x = A \cos(k z) \] Near equilibrium, a shear wave decays as \[ A(t) \sim A(0) e^{-\frac{\eta k^2}{\rho} t} \]

(Discussed and endorsed by Hess (2001))

All told, for \(3 + 1\) \(SU(3)\) Yang-Mills, \(\sim 4 \times 10^4\) qubits

For \(2 + 1\) \(SU(2)\) gauge theory, \(\lesssim 1500\) qubits

Hydrodynamics of a (strongly interacting) \(2+1\) spin model could be \(\sim 100\) qubits.