# Lecture 005 - Quantum Gates

## Quantum Gates

We model all instructions (gates) as

Name Interpret F
$I$ Add $0$ to $Ans$ $F() = 0$
Not, $X$ Add $1$ to $Ans$ $F() = 1$
Controlled-Not, $CNOT$, $XNOT$ Add $x_1$ to $Ans$ $F(x_1) = x_1$
$C_0NOT$ Add $NOT x_1$ to $Ans$ $F(x_1) = \lnot x_1$
Controlled-CNot $CCNOT$, Toffoli Add $x_1 \land x_2$ to $Ans$ $F(x_1, x_2) = x_1 \land x_2$
N/A Add $x_1 \lor x_2$ to $Ans$ $F(x_1, x_2) = x_1 \lor x_2$
Swap Swap $A$ with $B$ L/R Shift on 2 bits

## Computability

Any truth table can be built by circuit

• with $n \cdot m \cdot 2^n$ gates (by brute force: use $NOT$ to make input all $0$s, combine them using $OR$, if target is $1$, then use another $NOT$, else do nothing)

• or with $m \cdot 2^n/n$ gates (C. Shannon)

Any circuit can be converted to reversible manner, except for creating zero-ed input.

Theorem: $\forall F: \{0, 1\}^n \to \{0, 1\}^m$ if it is computable with $G$ classical gates, then you can convert to reversible quantum gates with $n+m+G$ qubits and $G+n$ gates, assuming all extra $G$ bits are initialized to $0$, left in a garbage state.

Theorem: reduction from turing machine to gates is only polynomial slowdown.

DeMorgan's Law: you only need NOT and one of AND or OR gate to perform all deterministic binary operation (Note that TM-complete also requires random access infinite memory)

• one should be able to construct a NAND gate, so you can then build a XOR gate. With XOR and AND you can build a half-adder. Combine half-adders to build a full-adder.

• Any NAND-based circuit can be rebuilt using NORs exclusively and vice versa.

CCNOT(X,Y,Z): "Add (X AND Y) To Z", equivalent of AND gate in quantum, but if we allow Z to be 1, it is NAND gate (because we are adding, instead of setting like classical computer).

• inverse is itself

• it is an AND gate when Z is 0

• it is an NAND gate when Z is 1

• it is a COPY Y Z when X is 1 and Z is 0

## "Garbage" Collection

### Garbage Problem

Why get rid of trash?

• wasting spacetime to keep temporary variables from last computation, since they can only be used once set to zero

• garbage prevents cancelation of amplitudes in joined quantum state

Example: say we have a qubit $a$ along with some other qubits $q_0, ..., q_n$

When all $q_0, ..., q_n$ are deterministic, then we might have the following state: (result of $H(0)$)

\begin{align*} \text{Amplitude}(0, 0...0)=&\sqrt{1/2}\\ \text{Amplitude}(1, 0...0)=&\sqrt{1/2}\\ \end{align*}

Then we do a $H$ gate, we have: (result of $H(H(0)) = 0$)

\begin{align*} \text{Amplitude}(0, 0...0)=&\frac{1}{2}+\frac{1}{2}=1\\ \text{Amplitude}(1, 0...0)=&\frac{1}{2}-\frac{1}{2}=0\\ \text{Probability}(0, 0...0)=&1^2=1\\ \text{Probability}(1, 0...0)=&0^2=0\\ \end{align*}

which is all good.

However, if at least one of $q_0, ..., q_n$ are non-deterministic garbage, then the amplitude will perhaps look like: (where the first and last bit of trash is entangled with the qubit we care about)

\begin{align*} \text{Amplitude}(0, 1...0)=&\sqrt{1/2}\\ \text{Amplitude}(1, 0...1)=&\sqrt{1/2}\\ \end{align*}

Then we do a $H$ gate, we have

\begin{align*} \text{Amplitude}(0, 1...0)=&\frac{1}{2}\\ \text{Amplitude}(1, 0...1)=&\frac{1}{2}\\ \text{Amplitude}(0, 0...0)=&\frac{1}{2}\\ \text{Amplitude}(1, 0...0)=&-\frac{1}{2}\\ \text{Probability}(0, 1...0)=&\left(\frac{1}{2}\right)^2=\frac{1}{4}\\ \text{Probability}(1, 0...1)=&\left(\frac{1}{2}\right)^2=\frac{1}{4}\\ \text{Probability}(0, 0...0)=&\left(\frac{1}{2}\right)^2=\frac{1}{4}\\ \text{Probability}(1, 0...0)=&\left(-\frac{1}{2}\right)^2=\frac{1}{4}\\ \end{align*}

Note that none of the states above can be merged, therefore we observe $0$ and $1$ half of the time.

### Garbage Removal

Easy. We just reverse (inverse) all the instruction except for the last where we add the answer.

• Precondition: all tmp variables initialized to zero

• Intermediate: answer is good, but tmp variables are garbage

• Postcondition: answer is good, all tmp variables set to zero

We can always transform classical circuit with $G$ gates to deterministic, reversible, garbage-free gates with $\leq 2 \cdot G$ gates.

### Measureing

When we measure, we use one particle, say $|m\rangle$, to interact with another partical $|v\rangle$ we want to measure. If we don't keep track of $|m\rangle$, it will look like we have measured $|v\rangle$. But what it really does is $|m\rangle$ becomes entangled with $|v\rangle$, and since we lost track of $|m\rangle$, it becomes garbage. In math, you can write:

Say you have a qubit you want to measure:

|v\rangle = |Rot_{0^\circ}\rangle + |Rot_{90^\circ}\rangle = |\uparrow\rangle + |\downarrow\rangle = |Rot_{45^\circ}\rangle

Then you collide $|v\rangle$ with $|m\rangle$, and it produces some maximally entangled particle, say:

|vm\rangle = |Rot_{0^\circ}\rangle|Rot_{0^\circ}\rangle + |Rot_{90^\circ}\rangle|Rot_{90^\circ}\rangle

Let us have a measurement device that measure only qubit $|v\rangle$ in $|Rot_{45^\circ}\rangle, |Rot_{135^\circ}\rangle$, then we have

\begin{align*} |Rot_{90^\circ}\rangle =& |Rot_{45^\circ}\rangle + |Rot_{135^\circ}\rangle = |\leftarrow\rangle + |\rightarrow\rangle\\ |Rot_{0^\circ}\rangle =& |Rot_{45^\circ}\rangle - |Rot_{135^\circ}\rangle = |\leftarrow\rangle - |\rightarrow\rangle\\ \end{align*}

So we subsitute above into $|vm\rangle$:

\begin{align*} |vm\rangle =& |Rot_{0^\circ}\rangle|Rot_{0^\circ}\rangle + |Rot_{90^\circ}\rangle|Rot_{90^\circ}\rangle\\ =&(|Rot_{45^\circ}\rangle - |Rot_{135^\circ}\rangle)|Rot_{0^\circ}\rangle + (|Rot_{45^\circ}\rangle + |Rot_{135^\circ}\rangle)|Rot_{90^\circ}\rangle\\ =&|Rot_{45^\circ}\rangle|Rot_{0^\circ}\rangle - |Rot_{135^\circ}\rangle|Rot_{0^\circ}\rangle + (|Rot_{45^\circ}\rangle|Rot_{90^\circ}\rangle + |Rot_{135^\circ}\rangle|Rot_{90^\circ}\rangle\\ \end{align*}

Now, see we have $50\%$ chance to measure to $|Rot_{45^\circ}\rangle$ or $|Rot_{135^\circ}\rangle$. But clearly, if we just measure unentangled $|v\rangle$, we get $100\%$ chance to $|Rot_{45^\circ}\rangle$

Takeaway: the "measurement" happen because you don't have access to $|m\rangle$ while trying to understand $|v\rangle$. (inspired by this video)

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