13
$\begingroup$

For a final project in my class, I decided to try to simulate a quantum computer and implement Grover's algorithm. I followed this excellently written blog post by Craig Gidney, and was successful in getting Grover's algorithm to work using matrix multiplication.

However, I wanted to figure out how to generate the quantum circuit that implements the oracle. The blog post by Craig Gidney intentionally skips over how this is done, and instead says that an oracle can always be represented as a matrix such as

$$ \pmatrix{ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 } $$

In this case, Grover's algorithm would be searching over a "database" of 4 items, and since the -1 is in the 3rd row, the entry we are searching for would be 3.

Using this matrix as the oracle, my implementation of Grover's algorithm works, with the correct "entry" coming up as the result with a >98% probability for all oracles I've tried.

The blog post links to this mathoverflow answer by Robin Kothari on how to actually go about constructing the oracle. Using it as a guide, I was also able to write code that finds the appropriate quantum circuit to simulate a classical circuit (which represents the search function $f$), and generates the matrix which performs

$$ \left | x \right \rangle \left | 0^{k} \right \rangle ↦ \left | f(x) \right \rangle \left | \mathrm{junk}(x) \right \rangle $$

From here, using the "uncomputation trick," it's easy to get

$$ \left | x \right \rangle \left | 0^{k} \right \rangle \left | b \right \rangle ↦ \left | x \right \rangle \left | 0^{k} \right \rangle \left | b \oplus f(x) \right \rangle $$

The resulting matrix should correspond to the oracle function.

According to the Robin Kothari's answer, my next step would be to set $\left | b \right \rangle$ to $\frac{1}{\sqrt{2}} \left ( \left | 0 \right \rangle - \left | 1 \right \rangle \right )$. As far as I understand, this is equivalent to setting $\left | b \right \rangle$ to $\left | 1 \right \rangle$, and then applying a Hadamard gate. This is because

$$ H (\left | 1 \right \rangle) = \pmatrix{0 & 1} \frac{1}{\sqrt{2}} \pmatrix{1 & 1 \\ 1 & -1}=\frac{1}{\sqrt{2}} \pmatrix{1 & -1} $$

However, when I use the matrix I generated and apply the necessary Hadamard gate, Grover's function does not work as before.

For example, let's say $f$ can be represented with the boolean expression $(x_{1} \land x_{2})$. This is a valid circuit because there is only one solution. My code simulates this with a single Toffoli gate which acts on a third qubit. Then a CNOT gate copies the state of the third qubit onto a fourth. Then the Toffoli gate is reversed. This is represented by the matrix

$$ \pmatrix{ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ } $$

As far as I can tell, this matrix does the trick. It maps $\left | 0000 \right \rangle ↦ \left | 0000 \right \rangle$, $\left | 0100 \right \rangle ↦ \left | 0100 \right \rangle$, and $\left | 1100 \right \rangle ↦ \left | 1101 \right \rangle$.

However, when I use this matrix as the oracle, the vector representing the state of the qubits comes out to be

$$ \frac{1}{2^{\frac{3}{2}}}\pmatrix{-1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 0 & -1 & -1 & 0} $$

Of course, when measured, this will not result in a high likelihood of the desired answer of $\left | 3 \right \rangle$.

What's the reason that my oracle matrix is in a different form than the one described in Craig Gidney's blog post? Did I miss a step in the process for generating it, or should they be different? If they should be different, why should they both work, and what did I miss?

$\endgroup$
6
$\begingroup$

The simplest way to flip the phase of a single amplitude is to use a Z gate with controls on every wire.

──•──
  │
──•──
  │
──•──
  │
──•──
  │
──•──
  │
──Z──

The above gate flips the phase of the all-on |111111> state.

Note that it doesn't matter which wire gets the Z gate instead of a control. Also note that HXH = Z and HZH = X, so you can always turn a many-controlled-not into a many-controlled-phase (or vice versa) by bracketing the gate (not the controls!) with Hadamard gates.

To flip the amplitude of a state besides the |1111...111> state, you can bracket X gates around the controls on any bits that should be zero instead of one:

───•───
   │
───•───
   │
─X─•─X─
   │
───•───
   │
───•───
   │
───Z───

The above gate flips the phase of the |110111> state.

If you're worried about using huge numbers of controls on a single gate, and want to restrict yourself to CNOT gates and single-qubit gates, or some finite set of gates, well then that's a bit more involved.

In the situation where you have some circuit that toggles some work bits to produce its output, and you want to find a specific output, just condition the toggling on the output (but beware multiple inputs giving the same output):

       C       C
    ┌────┐   ┌────┐
  ──┤In1 ├───┤In1 ├─
    │    │   │    │
  ──┤In2 ├───┤In2 ├─
    │    │   │    │
  ──┤In3 ├───┤In3 ├─
    │    │   │    │
  ──┤In4 ├───┤In4 ├─
    │    │   │    │
0 ──┤Out1├─•─┤Out1├─ 0
    │    │ │ │    │ 
0 ──┤Out2├─Z─┤Out2├─ 0
    └────┘   └────┘

The above circuit negates the phase of inputs that cause C to toggle both of the output bits. If only one input does that, it meets the criteria needed for (the simplest) Grover's algorithm.

Glad you liked the post.

$\endgroup$
2
  • 2
    $\begingroup$ Thanks for the answer! I understand that a Z gate with controls on every wire will produce the effect that's desired for the oracle (i.e. flipping the phase of the solution state). However, wouldn't you need to already know the solution to the classical circuit in order to know which qubit you need to put the Z gate on? Remember, I am not just trying to simulate the oracle with quantum gates, but actually construct it from a classical circuit. $\endgroup$ – Rhyzomatic May 3 '15 at 2:16
  • 2
    $\begingroup$ Oh. Well if you have a circuit C that toggles some ancilla bit when some condition is met on bits Bs (and that condition happens in exactly one case), then you can just do (Z on T) (C on Bs, T) (Z on T) (C on Bs, T). Note that the rest of the Grover search steps should only be done on the Bs, not on C, because C is not part of the search space. $\endgroup$ – Craig Gidney May 3 '15 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.