In part 4 we’re going to go over the logical internals of some of these gates to see how each gate manupilates a qubit’s probability of being a 0 or 1.

*Note: this post is part of my series, Programming the Multiverse*

As we saw in Part 3, a qubit has a formula associated with it, called bra-ket notation, that describes its probability of being a 0 or 1:

\[\ket{\psi} = a\ket{0} + b\ket{1}\]When we apply gates to the qubits, we’re adjusting *a* and *b* and the squares of *a* and *b* have to add up to a 100% probability (or just 1 in math terms):

So how do we get at those *a*’s and *b*’s?

## The magic matrices

Let’s take a look at a simple gate, the Pauli-X gate.

This is essentially a NOT operation on a qubit (it’s a simpler version of our earlier CNOT, but without the control input). You give it a \(\ket{0}\) and get a \(\ket{1}\) and vice-versa.

If we dig inside the Pauli-X we discover that it’s hiding a magic matrix!

\[\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]\]What’s a matrix? If you never took linear algebra or have repressed those memories, a matrix is like a two-dimensional array that has *m* rows and *n* columns.

The cool thing about matrices is that you can do math on them. When we apply a Pauli-X gate to a qubit, we are doing some matrix multiplication.

But first, in order to do this matrix multiplication, we have to turn our qubit into a matrix.

So let’s take our \(\ket{0}\) qubit, which is represented by:

\[1\ket{0} + 0\ket{1}\]We can take those *a* and *b* terms and put them into a 2-row by 1-column matrix (also called a unit vector):

So placing our 1 and 0, we get:

\[\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\]Now we can do the matrix multiplication (also known as the **dot product**). We always place the qubit matrix on the right and the gate matrix on the left:

When we multiply, we take every row in the left-hand matrix, *A* and multiply it by each column in the right-hand matrix. Let’s start with row 1 and column 1, the result of which we’ll put into the top cell of a new 2x1 matrix:

Remember, our rows are labeled \(m\) and our columns are \(n\). To designate a cell, we’ll refer to a row and column from \(A\) as \(A_{mn}\). Our new 2x1 matrix we’ll call \(C\).

\[A \cdot B = \left[ \begin{array}{cc} \textcolor{red}{A_{1,1}} & \textcolor{red}{A_{1,2}} \\ A_{2,1} & A_{2,2} \end{array} \right] \cdot \left[ \begin{array}{c} \textcolor{red}{B_{1,1}} \\ \textcolor{red}{B_{2,1}} \end{array} \right] = \left[ \begin{array}{c} \textcolor{red}{C_{1,1}} \\ C_{2,1} \end{array} \right]\]To get the first cell in \(C\), we loop through each position in the row/column and multiply each cell in \(A\) with the matching cell in \(B\), then add them up:

\[C_{1,1} = (A_{1,1} \times B_{1,1}) + (A_{1,2} \times B_{2,1})\]To get the second cell in \(C\) we operate on row 2 of the gate matrix:

\[C_{2,1} = (A_{2,1} \times B_{1,1}) + (A_{2,2} \times B_{2,1})\]If you, like me, get sick of doing matrix multiplication by hand, you can use one of many online tools or use a programming language like Ruby or Python that have great matrix support.

OK, so when we multiply our input matrix by the gate matrix, we get:

\[\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \cdot \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 1 \end{array} \right]\]And we transform the resulting matrix back into bra-ket notation by putting \(C_{1,1}\) into *a* and \(C_{2,1}\) into *b*:

We’ve done it! We flipped the bit!

## More inputs, bigger matrices

We just did the math for a quantum gate with a single input, but let’s see how we do it with a gate with two inputs. Let’s work with the SWAP:

Out input vector will always be an \(m \times 1\) matrix, so we have to work some alchemy craft a new super qubit out of our two individual qubits. Our new super qubit represents the probability of getting every combination of states that can come out of our individual qubits. So given we have two qubits \(q_0\) and \(q_1\):

\[q_{0} = a\ket{0} + b\ket{1}\] \[q_{1} = c\ket{0} + d\ket{1}\]We do what’s called a **tensor product** of the two vectors. Now you can sound fancy at your next cocktail party and talk about tensors. Oooohhh! Tensors!!

Let’s set \(q_0\) and \(q_1\) to \(\ket{1}\) and \(\ket{0}\):

\[q_{0} = 0\ket{0} + 1\ket{1}\] \[q_{1} = 1\ket{0} + 0\ket{1}\] \[q_{0} \otimes q_{1} = (0 \times 1)\ket{00} + (0 \times 0)\ket{01} + (1 \times 1)\ket{10} + (1 \times 0)\ket{11}\]Which we gives us:

\[0\ket{00} + 0\ket{01} + 1\ket{10} + 0\ket{11}\]And we can put that into a vector matrix:

\[\left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right]\]We then multiply it times the magic matrix that represents a SWAP and get our new output vector (I’m not going to work through all the matrix multiplication again, that’s for you to do for “fun”):

\[\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \cdot \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right]\]Which we put back into our super tensor ultra qubit:

\[0\ket{00} + 1\ket{01} + 0\ket{10} + 0\ket{11}\]This means we have a 100% chance of a combination of 0 and 1, Which means our two qubits are:

\[q_0 = \ket{0}\] \[q_1 = \ket{1}\]And that’s the magic behind a Freaky Qubit Friday!

OK that’s enough math for now, and trust me, you won’t normally need to do any matrix multiplication by hand unless you need to deeply debug something. But it’s good to know why everything is happening. But eventually you’ll be able to just glance at a matrix and see some patterns, like identity matrices, that give you an idea of what’s going to happen when you apply it.