# What “states” can a qubit have if it doesn't have values?

I was told that qubits don't have a value, they have a "state". What does that mean? What are the different "states" that a qubit can have (like bits can be either 1 or 0)?

It strongly depends on your definition of have a value.

What you might be referring to is the fact that qubits (and more in general quantum systems) can be in a state that, when measured in specific ways, gives probabilistic outcomes. Here, with state I don't mean anything technical. A physical state is here meant as simply whatever set of property characterises the system under consideration.

For example, if the quantum system under consideration is a qubit in the following superposition state: $$\frac{1}{\sqrt2}(\lvert0\rangle+\lvert1\rangle),$$ then measuring in the so-called computational basis results $$50\%$$ of the time in finding the outcome corresponding to $$\lvert0\rangle$$, and $$50\%$$ of the time in finding the outcome corresponding to $$\lvert1\rangle$$.

A possible physical realisation of this situation is a photon that is sent through a balanced beamsplitter. In this case, $$\lvert0\rangle+\lvert1\rangle$$ describes the fact that the photon is in a superposition of the two output ports of the beamsplitter. Measuring in the computational basis then corresponds here to "asking in which of the two output ports the photon came out from", which you can do by putting photon detectors in the two output ports and seeing which one "clicks".

What are the different "states" that a qubit can have (like bits can be either 1 or 0)?

Yes. It should be noted however that, like bits, what "$$0$$" and "$$1$$" corresponds to physically depends on the situation. In the example above, "$$0$$" meant "photon on the first output port" while "$$1$$" meant "photon on the second output port". In a different situation "$$0$$" and "$$1$$" might instead represent two different states of the polarisation of the photon. Or one might be studying an entirely different physical scenario which has nothing to do with light.

Moreover, there are many situations in which a system can be in more than two possible states (in fact, arguably, almost all situations are like this). In this case one talks of a qudit (or in case of continuous systems sometimes of qumodes). For example, in the quantum optical scenario with the beamsplitter that I used above, if there are instead many output ports, then the output state should be described as a more general superposition of the form $$\sum_{k=1}^N c_k \lvert k\rangle$$, with $$N$$ the number of output ports of the interferometer.

You might also have a look at the answers to this related question: How can a quantum logic gate produce a value other than $$0$$ or $$1$$?.