What is our quantum computer?

What is a universal quantum computer?

|\psi\rangle = a | 0 \rangle + b | 1\rangle

1 qubit

2 qubits

3 qubits

4 qubits

And you get the idea lah

|\psi\rangle = U |00\ldots0\rangle \Rightarrow U \approx U_c

we will approximate via circuits

It's all about the group of unitary matrices

U(n) = \{ M\in \mathbb C^{n\times n}: M^{-1} = M^\dagger\}

M,M' \in U(n) \Rightarrow (M M')(MM')^\dagger = M M'(M')^\dagger M^\dagger

= M M^\dagger = I

M,M' \in U(n) \Rightarrow M M'\in U(n)

M \in U(n) \Rightarrow \exist H: M = e^{i H}

What is a quantum algorithm?

\hat H \mapsto \hat \mathcal H_\ell = \hat \mathcal U_\ell^\dagger \hat H \hat\mathcal U_\ell

\partial_\ell \hat \mathcal H_\ell = [[A(\hat \mathcal H_\ell),B(\hat \mathcal H_\ell)], \hat \mathcal H_\ell]

0

C

What is a quantum computer?

It's an experimental setup A which includes a quantum system B and that setup A allows to manipulate the quantum state of B.

What is a quantum algorithm?

0

C

What is a universal quantum computer?

It's an experimental setup which includes a quantum system and that setup allows to manipulate its quantum state.

It's an experimental setup which includes few-level quantum systems and that setup allows to manipulate the quantum state using gates.

These gates form a universal gate set which means that if you apply sufficiently many you will be able to reach any desired quantum state.

What are quantum gates?

|\psi\rangle = a | 0 \rangle + b | 1\rangle

|+\rangle = \frac 1 {\sqrt 2} (| 0 \rangle + | 1\rangle)

What are quantum gates?

|\psi\rangle = | 0 \rangle

Z|0\rangle = | 0\rangle

Z|\mu\rangle = (-1)^\mu| \mu\rangle

Useful notation

Exercise: By evaluating the determinant, prove that it's impossible to apply Z in the lab

What are quantum gates?

|\psi\rangle = | 0 \rangle

|+\rangle = \frac 1 {\sqrt 2} (| 0 \rangle + | 1\rangle)

Use the quantum computer to transform

into

Trivial quantum algorithm solves it:

Apply the Haddamard gate

H = \frac 1 {\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 &- 1\end{pmatrix}

0

H

What are all single qubit quantum gates?

What are 2 qubit quantum gates?

|\psi\rangle = | 0 \rangle \otimes |0\rangle

|+\rangle \otimes |+\rangle = \frac 1 { 2} (| 0 0\rangle + | 01\rangle+|10\rangle + | 11\rangle)

Use the quantum computer to transform

into

Apply the Haddamard gate on each qubit

H\otimes H

0

H

0

H

What are quantum gates?

|\psi\rangle = | 0 \rangle \otimes |0\rangle

|\Phi\rangle = \frac 1 { \sqrt 2} (| 0 0\rangle + | 11\rangle)

Use the quantum computer to transform

into

Apply the Haddamard gate on each qubit

H\otimes 1

and then apply the controlled-not gate

\text{CNOT} = |0\rangle\langle 0|\otimes 1 + |1\rangle\langle1 |\otimes X

H

0

+

Universal gate set "Ising + single qubit"

g_{XX}(t) = e^{-it X_iX_{i+1} }

g_X(i) = e^{-itX_i}

g_Z(i) = e^{-itZ_i }

0

https://arxiv.org/abs/quant-ph/0307190

https://arxiv.org/abs/quant-ph/0109064

https://arxiv.org/abs/quant-ph/0106064

Universal gate set #2

CZ

g_X = e^{-itX_i}

g_Z = e^{-itZ_i }

https://arxiv.org/abs/quant-ph/0308006v3

= \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

https://arxiv.org/abs/quant-ph/9505018

https://arxiv.org/abs/quant-ph/9503016

Universal gate set #3: Surface code

CZ

T_i = e^{-i\pi/4 X_i}

Z_i

= \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

X_i

https://arxiv.org/abs/quant-ph/0110143

About CZ

I\otimes Z = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& -1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

Z\otimes I = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&-1&0 \\ 0&0&0 &- 1\end{pmatrix}

Z\otimes Z = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& -1 &0 & 0\\ 0&0&-1&0 \\ 0&0&0 & 1\end{pmatrix}

About CZ

I\otimes Z = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& -1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

Z\otimes I = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&-1&0 \\ 0&0&0 &- 1\end{pmatrix}

Z\otimes Z = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& -1 &0 & 0\\ 0&0&-1&0 \\ 0&0&0 & 1\end{pmatrix}

2 I\otimes I -Z\otimes I- I\otimes Z + Z\otimes Z = \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

Exercise: derive this using the Hilbert-Schmidt scalar product.

Implementing CZ

\text{CZ} = e^{i \frac \pi 2 ( Z\otimes I + I\otimes Z - Z\otimes Z)}

g_{XX}(t) = e^{-it X_iX_{i+1} }

g_X(i) = e^{-itX_i}

g_Z(i) = e^{-itZ_i }

0

Exercise: Use commutation relations and BCH formula to implement CZ with the Ising gate set

All 2 qubit operations

CNOT

g_X = e^{-itX_i}

g_Z = e^{-itZ_i }

https://arxiv.org/abs/quant-ph/0308006v3

About CNOT

+

\text{CNOT} = |0\rangle\langle 0|\otimes 1 + |1\rangle\langle1 |\otimes X

\text{CZ} = |0\rangle\langle 0|\otimes 1 + |1\rangle\langle1 |\otimes Z

= \begin{pmatrix} 1 & 0 &0 & 0\\ 0& 1 &0 & 0\\ 0&0&1&0 \\ 0&0&0 &- 1\end{pmatrix}

= (1\otimes H) CNOT (1\otimes H)

Swap gate from CNOT

SWAP = \begin{bmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{bmatrix}

https://arxiv.org/abs/quant-ph/0308006v3

SWAP=\frac{I\otimes I +X\otimes X+Y\otimes Y+Z\otimes Z}{2}

Universality is generic:

https://arxiv.org/abs/1404.7822

https://arxiv.org/abs/quant-ph/0303048

(Paywall link)

Universal gate set "Ising + single qubit"

g_{XX}(t) = e^{-it X_iX_{i+1} }

g_X(i) = e^{-itX_i}

g_Z(i) = e^{-itZ_i }

0

How can we use it to implement any unitary?

Universal quantum computation

How can we use it to implement any unitary?

U(n) = \{ M\in \mathbb C^{n\times n}: M^{-1} = M^\dagger\}

SU(n) = \{ M\in U(n): det(M)=1\}

= \{ exp(i H): H=H^\dagger, tr(H)=0\}

Implement its Hamiltonian

Implementing Hamiltonians

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}} \langle Z(\mu)X(\nu),H\rangle Z(\mu)X(\nu)

H + H' = H^\dagger +(H')^\dagger =(H +H')^\dagger

Hermitian matrices are a vector space

W_{\mu,\nu} = Z(\mu)X(\nu)

Is a basis

Orthonormal

expansion

In the exam* you will prove

W_{\mu,\nu} = Z(\mu)X(\nu)

Is a basis

Z(\mu) =\otimes_{k=1}^L(\sigma^z)^{\mu_k}

X(\nu) =\otimes_{k=1}^L(\sigma^x)^{\nu_k}

How to get

Z_i =I^{\otimes (i-1)} \sigma^z\otimes I^{L-i}

e_i= (0,0,\ldots 1 @ i, 0\ldots)

Z(e_i) = Z_i

* This is super important, so what if it was on some exam? Could you prove it?

In the exam you will prove

W_{\mu,\nu} = Z(\mu)X(\nu)

Is a basis

Z(\mu) =\otimes_{k=1}^L(\sigma^z)^{\mu_k}

X(\nu) =\otimes_{k=1}^L(\sigma^x)^{\nu_k}

How to get

Y_i =I^{\otimes (i-1)} \sigma^y\otimes I^{L-i}

e_i= (0,0,\ldots 1 @ i, 0\ldots)

Z(e_i)X(e_i) \propto Y_i

Universal quantum computation

SU(n) = \{ exp(i H): H=H^\dagger, tr(H)=0\}

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}} \langle Z(\mu)X(\nu),H\rangle Z(\mu)X(\nu)

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}}H_{\mu,\nu} Z(\mu)X(\nu)

Universal quantum computation

SU(n) = \{ exp(i H): H=H^\dagger, tr(H)=0\}

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}}H_{\mu,\nu} Z(\mu)X(\nu)

U=e^{it H} \equiv e^{i H'}

U=e^{it H} \neq \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}e^{iH_{\mu,\nu} Z(\mu)X(\nu)}\right )

Trotter-Suzuki decomposition

\hat H_\text{TFIM} = \sum_{i=1}^{L-1}X_iX_{i+1}+\sum_{i=1}^L Z_i\\ \equiv \hat H_X +\hat H_Z

|\psi(t)\rangle = e^{-it \hat H_\text{TFIM}}{|0\rangle}^{\otimes L}

|\psi(t)\rangle = \left(e^{-i\frac tN \hat H_X}e^{-i\frac tN \hat H_Z}\right)^N{|0\rangle}^{\otimes L} + \hat E{|0\rangle}^{\otimes L}

g_{XX}(t) = e^{-it X_iX_{i+1} }

g_X(i) = e^{-itX_i}

g_Z(i) = e^{-itZ_i }

0

Trotter-Suzuki decomposition

\|\hat E\| = \|e^{-it(\hat H_X +\hat H_Z)} - \left(e^{-i\frac tN \hat H_X}e^{-i\frac tN \hat H_Z}\right)^N\|

\le \frac{t^2}N\|[\hat H_X,\hat H_Z]\|

Why does it work?

BCH formula

e^{-it( \hat H_X +\hat H_Z)}=e^{-it\hat H_X }e^{-it\hat H_Z}e^{-i\frac {t^2}2[\hat H_X ,\hat H_Z]}\ldots

\|\mathbb 1 - e^{-i\frac {t^2}2[\hat H_X ,\hat H_Z]}\| \le \frac {t^2}2 \|[ \hat H_X ,\hat H_Z]\|

Conclusion: For short evolution time we're happy

How to implement Trotter-Suzuki?

e^{-it\hat H_X}=\prod_{i=1}^{L-1}e^{-it X_iX_{i+1} }

g_X(i) = e^{-itX_i X_{i+1}}

g_Z(i) = e^{-itZ_i }

Use Solovay-Kitaev algorithm to compile these gates but usually they are the primitive gates

https://arxiv.org/abs/quant-ph/0505030

(e^{-i\frac tN \hat H_X}e^{-i\frac tN \hat H_Z})^N{|0\rangle}^{\otimes L}

0

Exercise: Local error bound

Q_x = e^{i(1-x)A}e^{i(1-x)B}e^{ix(A+B)}

https://arxiv.org/abs/1312.1695

Q_1 = e^{i(A+B)}

Q_0 = e^{i A}e^{i B}

\partial_x Q_x = e^{i(1-x)A}\left(-A-B+ A_B+B\right)e^{i(1-x)B} e^{ix(A+B)}

A_B = e^{i(1-x)B}A e^{-i(1-x)B} = A + i e^{i\xi_x B}[A,B] e^{-i\xi_x B}

\|Q_0-Q_1\| =\| \int_0^1 dx \partial_x Q_x\| = \|[A,B]\|

https://arxiv.org/abs/1901.00564

Exercise: Non-commutative identity

X^m - Y^m = \sum_{j=0}^{m-1}X^{m-1-j}(X-Y)Y^j

\le \sum_{j=0}^{N-1}\|e^{-i (N-1-j)\frac tN \hat H_X}(e^{-i\frac tN \hat H_X}-e^{-i\frac tN \hat H_Z})e^{-ij\frac tN \hat H_Z}\|

\|e^{-it(\hat H_X +\hat H_Z)} - \left(e^{-i\frac tN \hat H_X}e^{-i\frac tN \hat H_Z}\right)^N\|

\le \sum_{j=0}^{N-1}\|e^{-i \frac tN \hat H_X}-e^{-i\frac tN \hat H_Z}\| \le \frac{t^2}N \|[H_X,H_Z]\|

cf.:

x^m -y ^m = (x-y)\sum_{j=0}^{m-1}x^{m-1-j}y^j

Application to physics:

Universal quantum computation

SU(n) = \{ exp(i H): H=H^\dagger, tr(H)=0\}

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}}H_{\mu,\nu} Z(\mu)X(\nu)

U=e^{it H} \neq \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}e^{iH_{\mu,\nu} Z(\mu)X(\nu)}\right )

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}e^{i\frac 1N H_{\mu,\nu} Z(\mu)X(\nu)}\right )^N

Constructive quantum compilation

SU(n) = \{ exp(i H): H=H^\dagger, tr(H)=0\}

H = \sum_{\mu,\nu\in\{0,1\}^{\times L}}H_{\mu,\nu} Z(\mu)X(\nu)

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}e^{i\frac 1N H_{\mu,\nu} Z(\mu)X(\nu)}\right )^N

How to get

Clifford circuits

W_{\mu,\nu} = Z(\mu)X(\nu)

For every

There exists

C_{\mu,\nu} \in U(n)

Such that

C_{\mu,\nu} W_{\mu,\nu} C_{\mu,\nu}^\dagger= Z_1

Super crucial exam formula

H' = UHU^\dagger

e^{iH'} = Ue^{iH}U^\dagger

\Rightarrow

C_{\mu,\nu} \in U(n)

C_{\mu,\nu} W_{\mu,\nu} C_{\mu,\nu}^\dagger= Z_1

U \in U(n)

Super crucial exclam formula

\Rightarrow

e^{iW_{\mu,\nu}'} =C_{\mu,\nu}^\dagger e^{iZ_1}C_{\mu,\nu}

Super cool exciting formula

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}e^{i\frac 1N H_{\mu,\nu} Z(\mu)X(\nu)}\right )^N

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}C_{\mu,\nu}^\dagger e^{i\frac 1 N H_{\mu,\nu}Z_1}C_{\mu,\nu}\right )^N

Super cool exclam formula

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}C_{\mu,\nu}^\dagger e^{i\frac 1 N H_{\mu,\nu}\sigma^z}\otimes I^{\otimes L-1}C_{\mu,\nu}\right )^N

In this (insane) model: all we ever do is make infinitesimally small rotations on qubit 1 and then distribute that onto all the other qubits using Clifford operations

Surely, there must be better ways?!

Addendum: Clifford circuits

W_{\mu,\nu} = Z(\mu)X(\nu)

For every

There exists

C_{\mu,\nu} \in U(n)

Such that

C_{\mu,\nu} W_{\mu,\nu} C_{\mu,\nu}^\dagger= Z_1

Addendum: formal definition Clifford circuits

W_{\mu,\nu,q} = q Z(\mu)X(\nu)

W_{\mu,\nu,q} W_{\mu',\nu',q}= q'' Z(\mu'')X(\nu'')=W_{\mu'',\nu'',q''}

This is called the Pauli group

\mathcal P(L)

Addendum: formal definition Clifford circuits

W_{\mu,\nu,q} = q Z(\mu)X(\nu)

Pauli operators have essentially the same spectrum

Z(\mu) =\otimes_{k=1}^L(\sigma^z)^{\mu_k}

X(\nu) =\otimes_{k=1}^L(\sigma^x)^{\nu_k}

Z(\mu)|\mu\rangle = \otimes Z^{\mu_k}|\mu_k\rangle= \otimes (-1)^{\mu_k}|\mu_k\rangle

X(\mu)|\mu\rangle_+ = \otimes X^{\mu_k}|\mu_k\rangle_+= \otimes (-1)^{\mu_k}|\mu_k\rangle_+

There must be unitary operators mapping them to each other

U_{\mu,\nu,q}^\dagger W_{\mu,\nu,q} U_{\mu,\nu,q} W_{\mu',\nu',q}

Addendum: formal definition Clifford circuits

W_{\mu,\nu,q} = q Z(\mu)X(\nu)

C(L) = \{ U\in U(L) \text{ s.t. } U\mathcal P(L)U^\dagger = \mathcal P(L)\}

All you need is single qubit and CNOTs

U=e^{it H} = \lim_{N\rightarrow \infty} \left( \prod_{\mu,\nu\in\{0,1\}^{\times L}}C_{\mu,\nu}^\dagger e^{i\frac 1 N H_{\mu,\nu}\sigma^z}\otimes I^{\otimes L-1}C_{\mu,\nu}\right )^N

|\psi(t)\rangle = e^{-it \hat H}|\psi(0)\rangle

How to compute it on a laptop?

How to compute it on a quantum computer?

Hamiltonian simulation

|\psi(t)\rangle = e^{-it \hat H}|\psi(0)\rangle

How to compute it on a laptop?

For qubits, your laptop can do ~13 spins at finite temperature and ~25 spins for a pure state (use sparsity)

\begin{pmatrix}1\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}

At the end of the day:

https://arxiv.org/abs/1311.0169

https://arxiv.org/abs/1411.0660

https://arxiv.org/abs/1408.5341

https://arxiv.org/abs/1901.05824

Workarounds:

Hamiltonian simulation

|\psi(t)\rangle = e^{-it \hat H}|\psi(0)\rangle

How to compute it on a quantum computer?

Use quantum algorithms 'Hamiltonian simulation'

Trotter-Suzuki

Linear combination of unitaries

Qubitization

Randomized compiler

https://arxiv.org/abs/1610.06546v3

https://arxiv.org/abs/1811.08017

https://arxiv.org/abs/1901.00564

https://arxiv.org/abs/quant-ph/0209131

Hamiltonian simulation

Truncated series

https://arxiv.org/abs/1412.4687

https://www.quantamagazine.org/a-short-guide-to-hard-problems-20180716/

https://en.wikipedia.org/wiki/P_versus_NP_problem

P: Runs easily

BPP: Often runs easily

BQP: Often quantums easily

NP: Optimizes easily

https://www.quantamagazine.org/a-short-guide-to-hard-problems-20180716/

QMA

https://en.wikipedia.org/wiki/P_versus_NP_problem

P: Runs easily

BPP: Often runs easily

BQP: Often quantums easily

NP: Optimizes easily

Key idea for post-Trotter methods

C(U,V) = |0\rangle\langle 0| \otimes U + |1\rangle\langle 1| \otimes V

Step 1: Show that it's unitary

C(U,V) \, C(U,V)^\dagger = |0\rangle\langle 0| \otimes UU^\dagger + |1\rangle\langle 1| \otimes V V^\dagger = \mathbb 1

https://arxiv.org/abs/1412.4687

https://arxiv.org/abs/1202.5822

Key idea for post-Trotter methods

C(U,V) = |0\rangle\langle 0| \otimes U + |1\rangle\langle 1| \otimes V

Step 2: Apply to flag qubit in superposition

Key idea for post-Trotter methods

C(U,V) = |0\rangle\langle 0| \otimes U + |1\rangle\langle 1| \otimes V

Step 3: Consider what happens if applied to superposition:

(R_x(\theta)\otimes 1)C(U,V) |+\rangle \otimes |\psi\rangle =

R_x(\theta) |0\rangle = c |0\rangle+ s |1\rangle

R_x(\theta) |1\rangle = c |1\rangle -s |0\rangle

|0\rangle \otimes (cU -s V) |\psi\rangle + |1\rangle \otimes (cV +sU) |\psi\rangle

Key idea for post-Trotter methods

C(U,V) = |0\rangle\langle 0| \otimes U + |1\rangle\langle 1| \otimes V

Step 4: Assume flag is measured with outcome 1 and discard it

(P_+\otimes 1)(R_x(\theta)\otimes 1)C(U,V) |+\rangle \otimes |\psi\rangle \propto (cU+s V) |\psi\rangle

R_x(\theta) |0\rangle = c |0\rangle+ s |1\rangle

R_x(\theta) |1\rangle = c |1\rangle -s |0\rangle

Conclusion: We can (probabilistically) apply (normalized) sums of unitary operators

Key idea for reliable post-Trotter methods

R = 1 - 2|\phi\rangle\langle \phi|

Grover reflector

Step 1: Show that it's unitary

https://arxiv.org/abs/1412.4687

...

Key idea for reliable post-Trotter methods

R = 1 - 2|\phi\rangle\langle \phi|

Grover reflector

Step 2: Consider applying it to a state overlapping with it

https://arxiv.org/abs/1412.4687

R(|\phi\rangle + |\phi^\perp\rangle) \propto - |\phi\rangle+ |\phi^\perp\rangle

Key idea for reliable post-Trotter methods

R = 1 - 2|\phi\rangle\langle \phi|

Grover reflector

Step 3: Reflect around the linear combination of unitaries

https://arxiv.org/abs/1412.4687

This is also called oblivious amplitude amplification, and the crux is in making this efficiently and obliviously i.e. without knowing or destroying the reflector state

R_{c,s} = Q(1-|\phi\rangle\langle \phi|)Q^\dagger

Gaussian quantum simulators

https://arxiv.org/abs/2004.04174

https://arxiv.org/abs/2201.00049

https://arxiv.org/abs/1101.2659

https://arxiv.org/abs/2003.01808

Gaussian quantum simulators

https://arxiv.org/abs/2004.04174

https://arxiv.org/abs/2201.00049

https://arxiv.org/abs/1101.2659

https://arxiv.org/abs/2003.01808

Gaussian quantum simulators

https://arxiv.org/abs/2004.04174

https://arxiv.org/abs/2201.00049

https://arxiv.org/abs/1101.2659

https://arxiv.org/abs/2003.01808

How?

Ultra-cold 1d gases

Inside: atoms

Outside: wavepackets

hydrodynamics

Energy of phonons

E

\hat H_\text{TLL} = \int \text{d}z\left[ \frac{\hbar^2n_{GP}}{2m}\partial_z\hat\varphi^2+g\delta\hat\varrho^2\right]

\delta\varrho

\partial_z\varphi

\ll

\delta\varrho

\ll

Tomonaga-Luttinger liquid

Quantum field refrigerators in the TLL model:

System

Piston

Bath

Bath with excitations

System cooled down

Breaking of the Huygens-Fresnel principle

in the inhomogenous TLL model:

https://arxiv.org/abs/2104.07751

https://arxiv.org/abs/2006.01177

Why?

Why develop continuous field

quantum simulators?

Representation theory: Quantum information?
Continuum limits: BQP and QMA or more?
Are nanowires computationally hard to simulate?

What do we know is difficult?

SM

Fundamental

Universal

Effective

Why develop continuous field

quantum simulators?

Representation theory: Quantum information?
Continuum limits: BQP and QMA or more?
Are nanowires computationally hard to simulate?

What do we know is difficult?

SM

Fundamental

Universal

Effective

Non-thermal

steady states

Sine-Gordon

thermal states

Atomtronics

Generalized hydrodynamics

Recurrences

Some highlights:

Schweigler, Schmiedmayer,

et. al, arxiv:1505.03126

Langen, Schmiedmayer,

et. al, arxiv:1411.7185

Rauer, Schmiedmayer,

et. al, arxiv:1705.08231

Schweigler, Schmiedmayer,

et. al, arxiv:2003.01808

Aidelsburger, Beugnon,

et. al, arXiv:1705.02650

Eckel, Campbell,

et. al, arXiv:1406.1095

Møller, Schmiedmayer,

et. al, arXiv:2006.08577

Malvania, Weiss,

et. al, arxiv:2009.06651

Schemmer, Bouchoule,

et. al, arxiv:1810.07170

Interferometry measures velocities

van Nieuwkerk, Schmiedmayer, Essler, arXiv:1806.02626

Schumm, Schmiedmayer, Kruger, et al., arXiv:quant-ph/0507047

\partial_z\hat \varphi(z) \approx\frac{\hat \varphi(z)-\hat \varphi(z+\Delta z)}{\Delta z}

\partial_z\hat \varphi(z) \approx\frac{\Delta\hat\varphi(z)}{\Delta z}

\Delta z

Tomography

Tomography for phonons

\hat H_\text{TLL} = \int \text{d}z\left[ \frac{\hbar^2n_{GP}}{2m}\partial_z\hat\varphi^2+g\delta\hat\varrho^2\right]

\hat H_\text{TLL} = \sum_{k>0} \frac {\hbar \omega_k}2 (\phi_k^2+\delta\hat\rho_k^2) + g\hat\rho_0^2

\langle\hat \phi_k^2(t)\rangle= \cos^2(\omega_k t) \langle \phi_k^2(0)\rangle \\ \quad\quad\quad\quad\quad+\sin^2(\omega_k t) \langle \delta\hat \rho_k^2(0)\rangle

\delta \hat\rho

\langle\hat \phi^2(t)\rangle

\langle\delta\hat \rho^2(0)\rangle =0?

\langle \hat \phi^2(0)\rangle

\hat \phi

Tomography for phonons

\hat H_\text{TLL} = \int \text{d}z\left[ \frac{\hbar^2n_{GP}}{2m}\partial_z\hat\varphi^2+g\delta\hat\varrho^2\right]

\hat H_\text{TLL} = \sum_{k>0} \frac {\hbar \omega_k}2 (\phi_k^2+\delta\hat\rho_k^2) + g\hat\rho_0^2

\langle\hat \phi_k^2(t)\rangle= \cos^2(\omega_k t) \langle \phi_k^2(0)\rangle \\ \quad\quad\quad\quad\quad+\sin^2(\omega_k t) \langle \delta\hat \rho_k^2(0)\rangle

\delta \hat\rho

\langle\hat \phi^2(t)\rangle

\langle\delta\hat \rho^2(0)\rangle =0?

\langle \hat \phi^2(0)\rangle

\hat \phi

https://arxiv.org/abs/2108.07829

https://arxiv.org/abs/1807.04567

https://arxiv.org/abs/2005.09000

What are eigenmodes?

\hat H_\text{TLL} = \int \text{d}z\left[ \frac{\hbar^2n_{GP}}{2m}\partial_z\hat\varphi^2+g\delta\hat\varrho^2\right]

\hat H_\text{TLL} = \sum_{k>0} \frac {\hbar \omega_k}2 (\phi_k^2+\delta\hat\rho_k^2) + g\hat\rho_0^2

\hat \phi_k = \int_0^L dz \cos(\pi k z/L)\hat \varphi(z)

\int_0^L dz \cos(\pi k z/L)\cos(\pi k'z/L) = \delta_{k,k'}

https://arxiv.org/abs/2104.07751

Transmutation

\langle\hat \phi_k^2(t)\rangle= \cos^2(\omega_k t) \langle \hat\phi_k^2(0)\rangle +\sin^2(\omega_k t) \langle \delta\hat \rho_k^2(0)\rangle

\langle\hat \phi_k^2(t)\rangle= \langle \hat\phi_k^2(0)\rangle

\langle\hat \phi_k^2(t)\rangle= \langle \delta\hat \rho_k^2(0)\rangle

t=0:

t=\frac{\pi}{2\omega_k}:

https://arxiv.org/abs/2108.07829

\delta \hat\rho

\langle\hat \phi^2(t)\rangle

\langle\delta\hat \rho^2(0)\rangle =0?

\langle \hat \phi^2(0)\rangle

\hat \phi

Tomography

A_{1,2}= \sin^2(\omega_k t_1 )

A_{1,1}= \cos^2(\omega_k t_1)

\langle\hat \phi_k^2(t)\rangle= \cos^2(\omega_k t) \langle \hat\phi_k^2(0)\rangle+ \sin^2(\omega_k t) \langle \delta\hat \rho_k^2(0)\rangle

A_{3,2}= \sin^2(\omega_k t_3 )

A_{3,1}= \cos^2(\omega_k t_3)

A_{2,2}= \sin^2(\omega_k t_2 )

A_{2,1}= \cos^2(\omega_k t_2)

A \begin{pmatrix} \langle \hat\phi_k^2(0)\rangle \\\langle \delta\hat \rho_k^2(0)\rangle \end{pmatrix} =\begin{pmatrix} \langle \hat\phi_k^2(t_1)\rangle \\\langle \hat\phi_k^2(t_2)\rangle \\\langle \hat\phi_k^2(t_3)\rangle \end{pmatrix}

\|Aq - d\| = \text{min}

(This formalism: Tomography for many modes)

Tomography Klein-Gordon thermal state after quench

\hat H_\text{KG}=\hat H_\text{TLL}+J\int \mathrm{d}z \,n_{GP} \hat \varphi^2

https://arxiv.org/abs/2108.07829

https://arxiv.org/abs/1807.04567

Extracting physical properties

https://arxiv.org/abs/1807.04567

Extracting physical properties

https://arxiv.org/abs/1807.04567

https://arxiv.org/abs/1705.08231

Extracting physical properties

https://arxiv.org/abs/1807.04567

Tomography for optical lattices

https://arxiv.org/abs/2005.09000

Material science?

0

C

arXiv:2206.11772

arxiv:1912.06076

arxiv:2005.09000

arxiv:1807.04567

Diagonalization quantum algorithm

DSF of Rydberg arrays

Phonon tomography

Optical lattice tomography

\langle\hat \phi_k^2(t)\rangle= \cos^2(\omega_k t) \langle \phi_k^2(0)\rangle \\ \quad\quad\quad\quad\quad+\sin^2(\omega_k t) \langle \delta\hat \rho_k^2(0)\rangle

\delta \hat\rho

\langle\hat \phi^2(t)\rangle

\langle\delta\hat \rho^2(0)\rangle =0?

\langle \hat \phi^2(0)\rangle

\hat \phi