# The Quantum Virtual Machine (QVM)¶

The Rigetti Quantum Virtual Machine is an implementation of the Quantum Abstract Machine from A Practical Quantum Instruction Set Architecture. [1] It is implemented in ANSI Common LISP and executes programs specified in the Quantum Instruction Language (Quil). Quil is an opinionated quantum instruction language: its basic belief is that in the near term quantum computers will operate as coprocessors, working in concert with traditional CPUs. This means that Quil is designed to execute on a Quantum Abstract Machine that has a shared classical/quantum architecture at its core. The QVM is a wavefunction simulation of unitary evolution with classical control flow and shared quantum classical memory.

Most API keys give access to the QVM with up to 30 qubits. If you would like access to more qubits or help running larger jobs, then contact us at support@rigetti.com. On request we may also provide access to a QVM that allows persistent wavefunction memory between different programs as well as direct access to the wavefunction memory (wrapped as a numpy array) from python.

## Using the QVM¶

The QVM is available in pyQuil via the api module.

from pyquil.api import QVMConnection
qvm = QVMConnection()


One executes quantum programs on the QVM using two paradigms: the .run(...) method, and the .wavefunction(...) method. The former closely mirrors how one will execute programs on a real QPU (see Using the QPU), while the latter takes advantage of the virtual machine, and allows direct access to the wavefunction. These two methods are described in the following two sections. (For information on constructing quantum programs, please refer back to The Basics: Programs and Gates.)

### The .run(...) method¶

program = Program(X(0), MEASURE(0, 0))
results = qvm.run(program, trials=1)
# results = [[1]]


The .run(...) method takes numerous arguments, several of which are optional. The most important are

1. the program to be executed on the QVM,
2. the classical_addresses which to be returned from the QVM (not included above; by default, these are set to the addresses used in the program’s MEASURE instructions), and
3. the number of trials to be executed on the machine.

The results returned are a list of lists of integers. In the above case, that’s

[[1]]


Let’s unpack this. The outer list is an enumeration over the trials; if you set trials=1 then len(results) should equal 1.

The inner list, on the other hand, is an enumeration over the results stored in the classical addresses. We see that the result of this program is that the classical register [0] now stores the state of qubit 0, which should be 1 after an $$X$$-gate. We can of course ask for more classical registers:

qvm.run(p, [0, 1, 2])

[[1, 0, 0]]


The classical registers are initialized to zero, so registers [1] and [2] come out as zero. If we stored the measurement in a different classical register we would obtain:

p = Program()   # clear the old program
p.inst(X(0)).measure(0, 1)
qvm.run(p, [0, 1, 2])

[[0, 1, 0]]


We can also run programs multiple times and accumulate all the results in a single list.

coin_flip = Program().inst(H(0)).measure(0, 0)
num_flips = 5
qvm.run(coin_flip, [0], num_flips)

[[0], [1], [0], [1], [0]]


Try running the above code several times. You will see that you will, with very high probability, get different results each time.

### The .wavefunction(...) method¶

The QVM is a virtual machine. As such, we can directly inspect the wavefunction of a program, even without measurements, using the .wavefunction(...) method:

coin_flip = Program().inst(H(0))
qvm.wavefunction(coin_flip)

<pyquil.wavefunction.Wavefunction at 0x1088a2c10>


The return value is a Wavefunction object that stores the amplitudes of the quantum state at the conclusion of the program. We can print this object

coin_flip = Program().inst(H(0))
wavefunction = qvm.wavefunction(coin_flip)
print(wavefunction)

(0.7071067812+0j)|0> + (0.7071067812+0j)|1>


To see the amplitudes listed as a sum of computational basis states. We can index into those amplitudes directly or look at a dictionary of associated outcome probabilities.

assert wavefunction[0] == 1 / np.sqrt(2)
# The amplitudes are stored as a numpy array on the Wavefunction object
print(wavefunction.amplitudes)
prob_dict = wavefunction.get_outcome_probs() # extracts the probabilities of outcomes as a dict
print(prob_dict)
prob_dict.keys() # these stores the bitstring outcomes
assert len(wavefunction) == 1 # gives the number of qubits

[ 0.70710678+0.j  0.70710678+0.j]
{'1': 0.49999999999999989, '0': 0.49999999999999989}


The result from a wavefunction call also contains an optional amount of classical memory to check:

coin_flip = Program().inst(H(0)).measure(0,0)
classical_mem = wavefunction.classical_memory


Additionally, we can pass a random seed to the Connection object. This allows us to reliably reproduce measurement results for the purpose of testing:

seeded_cxn = api.QVMConnection(random_seed=17)
print(seeded_cxn.run(Program(H(0)).measure(0, 0), [0], 20))

seeded_cxn = api.QVMConnection(random_seed=17)
# This will give identical output to the above
print(seeded_cxn.run(Program(H(0)).measure(0, 0), [0], 20))


It is important to remember that this wavefunction method is just a useful debugging tool for small quantum systems, and it cannot be feasibly obtained on a quantum processor.

## Multi-Qubit Basis Enumeration¶

The Rigetti QVM enumerates bitstrings such that qubit 0 is the least significant bit (LSB) and therefore on the right end of a bitstring as shown in the table below which contains some examples.

bitstring qubit_(n-1) qubit_2 qubit_1 qubit_0
1…101 1 1 0 1
0…110 0 1 1 0

This convention is counter to that often found in the quantum computing literature where bitstrings are often ordered such that the lowest-index qubit is on the left. The vector representation of a wavefunction assumes the “canonical” ordering of basis elements. I.e., for two qubits this order is 00, 01, 10, 11. In the typical Dirac notation for quantum states, the tensor product of two different degrees of freedom is not always explicitly understood as having a fixed order of those degrees of freedom. This is in contrast to the kronecker product between matrices which uses the same mathematical symbol and is clearly not commutative. This, however, becomes important when writing things down as coefficient vectors or matrices:

$\begin{split}\ket{0}_0 \otimes \ket{1}_1 = \ket{1}_1 \otimes \ket{0}_0 = \ket{10}_{1,0} \equiv \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}\end{split}$

As a consequence there arise some subtle but important differences in the ordering of wavefunction and multi-qubit gate matrix coefficients. According to our conventions the matrix

$\begin{split}U_{\rm CNOT(1,0)} \equiv \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\end{split}$

corresponds to the Quil instruction CNOT(1, 0) which is counter to how most other people in the field order their tensor product factors (or more specifically their kronecker products). In this convention CNOT(0, 1) is given by

$\begin{split}U_{\rm CNOT(0,1)} \equiv \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}\end{split}$

For additional information why we decided on this basis ordering check out our note Someone shouts, “|01000>!” Who is Excited? [2].

## Examples of Quantum Programs¶

To create intuition for a new class of algorithms, that will run on Quantum Virtual Machines (QVM), it is useful (and fun) to play with the abstraction that the software provides.

A broad class of programs that can easily be implemented on a QVM are generalizations of Game Theory to incorporate Quantum Strategies.

### Meyer-Penny Game¶

A conceptually simple example that falls into this class is the Meyer-Penny Game. The game goes as follows: The Starship Enterprise, during one of its deep-space missions, is facing an immediate calamity, when a powerful alien suddenly appears on the bridge. The alien, named Q, offers to help Picard, the captain of the Enterprise, under the condition that Picard beats Q in a simple game of penny flips.

The rules: Picard is to place a penny Heads up into an opaque box. Then Picard and Q take turns to flip or not flip the penny without being able to see it; first Q then P then Q again. After this the penny is revealed; Q wins if it shows Heads (H), while Tails (T) makes Picard the winner.

Picard quickly estimates that his chance of winning is 50% and agrees to play the game. He loses the first round and insists on playing again. To his surprise Q agrees, and they continue playing several rounds more, each of which Picard loses. How is that possible?

What Picard did not anticipate is that Q has access to quantum tools. Instead of flipping the penny, Q puts the penny into a superposition of Heads and Tails proportional to the quantum state $$|H\rangle+|T\rangle$$. Then no matter whether Picard flips the penny or not, it will stay in a superposition (though the relative sign might change). In the third step Q undoes the superposition and always finds the penny to shows Heads.

To simulate the game we first construct the corresponding quantum circuit, which takes two qubits – one to simulate Picard’s choice whether or not to flip the penny and the other to represent the penny. The initial state for all Qubits is $$|0\rangle (= |T\rangle)$$. To simulate Picard’s decision, we assume that he chooses randomly whether or not to flip the coin, in agreement with the optimal strategy for the classic penny-flip game. This random choice can be created by putting one qubit into an equal superposition, e.g. with the Hadamard gate H, and then measure its state. The measurement will show Heads or Tails with equal probability p=0.5.

To simulate the penny flip game we take the second qubit and put it into its excited state $$|1\rangle (= |H\rangle)$$ by applying the X (or NOT) gate. Q’s first move is to apply the Hadamard gate H. Picard’s decision about the flip is simulated as a CNOT operation where the control bit is the outcome of the random number generator described above. Finally Q applies a Hadamard gate again, before we measure the outcome. The full circuit is shown in the figure below.

First we import all the necessary tools:

from pyquil.quil import Program
import pyquil.api as api

from pyquil.gates import I, H, X
qvm = api.QVMConnection()


Then we need to define two registers that will be used for the measurement of Picard’s decision bit and the final answer of the penny tossing game.

picard_register = 1


Moreover we need to encode the two different actions of Picard, which conceptually is equivalent to an if-else control flow as:

then_branch = Program(X(0))
else_branch = Program(I(0))


and then wire it all up into the overall measurement circuit:

prog = (Program()
# Prepare Qubits in Heads state or superposition, respectively
.inst(X(0), H(1))
# Q puts the penny into a superposition
.inst(H(0))
# Picard makes a decision and acts accordingly
.measure(1, picard_register)
.if_then(picard_register, then_branch, else_branch)
# Q undoes his superposition operation
.inst(H(0))
# The outcome is recorded into the answer register


Finally we play the game several times

qvm.run(prog, [0, 1], trials=10)


and record the register outputs as

[[1, 1],
[1, 1],
[1, 0],
[1, 0],
[1, 0],
[1, 0],
[1, 1],
[1, 1],
[1, 0],
[1, 0]]


Remember that the first number is the outcome of the game (value of the answer_register) whereas the second number is the outcome of Picard’s decision (value of the picard_register).

Indeed, no matter what Picard does, Q will always win!

### Exercises¶

#### Prisoner’s Dilemma¶

A classic strategy game is the prisoner’s dilemma where two prisoners get the minimal penalty if they collaborate and stay silent, get zero penalty if one of them defects and the other collaborates (incurring maximum penalty) and get intermediate penalty if they both defect. This game has an equilibrium where both defect and incur intermediate penalty.

However, things change dramatically when we allow for quantum strategies leading to the Quantum Prisoner’s Dilemma.

Can you design a program that simulates this game?