# How can a quantum logic gate produce a value other than $0$ or $1$?

In classical computation, a bit can have the value of either $$1$$ or $$0$$ and one can apply a logic gate to this bit. As far as I understand, in classical computation, no matter what gate is used, the value must still be either $$1$$ or $$0$$.

I understand that before measurement, a qubit can exist in a superposition of states, but that once it has been measured, it too must either produce the value of $$1$$ or $$0$$. However, it appears that there are many quantum logic gates that will operate on a qubit and produce a value that is not necessarily $$1$$ or $$0$$ (i.e. it the Z gate will change the value of $$1$$ to $$-1$$).

So, are we simply changing the basis? Furthermore what is the advantage or having all of these values that a qubit can be in, rather than simply the binary?

• I would say, intuitively, if we regard a simple 'computation step' as a map $S\times G\rightarrow S$, where $S$ is the 'state space' and $G$ is the 'operation space', an 'algorithm' is then a curve in the state space. With all these values that are not possible in classical computation, we have a larger 'state' space. So we can achieve more complex/efficient 'curves' so that we have a more powerful computation system. – XXDD Jul 20 '16 at 2:59

I think you've been fooled by notation. The value inside the $|\rangle$ is a state; a label. The value outside the $|\rangle$ is a weighting; an amplitude. You're negating the label instead of the weight.

The Z gate doesn't produce a $|-1\rangle$, but it can produce a $-|1\rangle$. Said another way, the Z gate doesn't produce a $|-True\rangle$ but it can turn a $|True\rangle$ into a $-|True\rangle$.

Quantum gates only ever affect the weights.

But that's okay, the state of $n$ qubits is defined by $2^n$ weights. So even simple operations on the qubits are doing huge operations on the weights, and when the operation you want happens to line up with those huge operations you can get quite a lot of time savings.

A qubit is in a mixture of the state "0" and the state "1." We write this mixture as $\alpha | 0 \rangle + \beta | 1 \rangle$. Here, $\alpha$ and $\beta$ are two complex numbers, subject to some constraints. It is not the case that a $Z$ gate will suddenly give us a new state $| - 1 \rangle$ -- the qubit does not possess such a state (it's a qubit, it has two states by definition). Quantum gates act on the $\alpha$ and $\beta$ numbers to perform operations; the $Z$ gate you mention will change $\beta \to - \beta$.

As for "what good are all these extra in-between states" I have to say that it's kind of too broad of a question. Quantum computing and quantum information in general depend on states working this way--I recommend reading more about potential applications of quantum computing and communication to learn about what possibilities are opened by this framework. A technology called "quantum key distribution" can be fairly layman-accessible and relies on the face that a qubit can be put in an indeterminate state.

• Not to be pedantic, I wouldn't use the term mixture for a superposition as mixture already has a very precise meaning. – Chris2807 Jul 19 '16 at 18:18
• I would say, intuitively, if we regard a simple basic 'computation step' as a map S×G→S, where S is the 'state space' and G is the 'operation space', an 'algorithm' is then a curve in the state space. With all these values that are not possible in classical computation, we have a larger 'state' space. So we can achieve more complex/efficient 'curves' to connect an initial state(inputs) with a final state (outputs), so that we have a more powerful computation system. This is why the QC is superior to the classical computers. – XXDD Jul 20 '16 at 14:42