2
$\begingroup$

I am a CS major trying to decipher quantum computing. I have done some elementary study on qubits and I always seem to get lost at the "infinitely many" states that the qubit can have. And I really want to understand what it actually means. My understanding is as follows and please correct me if I am wrong,

A classical bit has two states 0 and 1, which means n bits can represent 2^n different states, though each state is individually expressed. A Qubit could have "infinitely many" states between 0 and 1. This means that n qubits can represent m^n states, Where 'm' is the base and m > 2. This would mean we can represent larger numbers using fewer qubits when compared to the classical bit. If this is so, how does this lead to an increase in computation speeds?

If I am completely off track, what is the actual interpretation of a qubit?

$\endgroup$
1
  • $\begingroup$ An "ok" start is: en.wikipedia.org/wiki/Qubit. But, that quickly devolves into an overly technical and not very useful exposition. (One of the answers, given below, is essentially a re-pasting of the Wikipedia information.) To understand quBits, you are going to have to delve into quantum mechanics and understand superposition. A discussion of "infinite states" has stepped past fundamental information, and won't be very helpful. $\endgroup$ – Thomas Bitonti Mar 10 at 20:09
4
$\begingroup$

A qubit is a vector in a 2-state complex space. Its vector form is something like Qubit = $ax + by$, which by convention in Dirac notation is $|Q\rangle = a|0\rangle + b|1\rangle$, where $|0\rangle$ and $|1\rangle$ represent linearly independent basis vectors and the coefficients $a,b$ are complex numbers. $a^2$ and $b^2$ represent the probability of measuring 0 or 1, and they can take on any complex value as long as $|a|^2 + |b|^2 = 1$ (which must be true because we only have one qubit).

In this framework, the qubit acts like a vector pointing from the origin which naturally evolves or rotates around the infinitely many surface points of a ‘Bloch Sphere’ except when it’s measured it only ever tells you 0 or 1 (which represent the north and south poles of the sphere).

In short, a qubit is a vector.

The ‘infinitely many’ language comes from the ability of $a,b$ to take on any complex value so long as their squares equal 1. Measuring a qubit (meaning when you read it, touch it, look at it, record its value etc) only gives you 0 or 1, not both and not a decimal in between.

If you also think of the qubit as a wave, then the goal of quantum computing is to take a superposition of all possible results of a qubit string (000, 001, 010, 011 etc), weight the components appropriately to constructively and destructively interfere with one another, so that at the very end you’re biased to get the right result... with a nonzero probability that you will get the wrong answer even if you do everything right.

Incidentally a string of three qubits, you’ll notice, is the tensor product of three qubit vectors. This creates a complex space with $2^3 = 8$ basis vectors which our 3-string qubit can happily explore.

$\endgroup$
0
$\begingroup$

A QuBit is an element of a quantum circuit.

Like regular bits, a QuBit has two states, 0 or 1. Unlike regular bits, a QuBit can be in a superposition of these states.

Point 1: This gives us a terminology problem: What should be meant by "state" when talking about quantum systems. This is a problem only if classical terminology is stretched too far. Accept that new understandings are needed for quantum systems, and don't get stuck in terminology issues.

Point 2: An understanding of "superposition" is needed. That would be a long topic to discuss by itself, and is not discussed here.

A single QuBit system isn't very useful. Using an operator to put a single QuBit into a state which assigns equal probability to either state then examining the state of the QuBit gives you the same result as a fair coin toss.

Point 3: "Operator" is another specific technical term. Loosely speaking, in a single bit system, operators might be "1" (set to the 1 state), "0" (set to the 0 state), "1/2" (set to the 1 state superimposed with the 0 state, with equal probabilities). Operators are more interesting when talking about more than one QuBit.

In a quantum circuit which has multiple QuBits, the states of QuBits can be entangled with each other. The circuit is programmed by applying quantum operators to pairs of QuBits. This puts the entire circuit in an entangled state. At this point, meaningful computation becomes possible.

Point 4: "Entangled" gives us the last technical term. This is another long topic which is not discussed here.

That's a start. To understand how computation is done using QuBits, all of the above should be understood. That will require understanding a number of computer science concepts as well as a number of concepts from quantum physics.

Point 5: Wait! What about complex numbers and probability amplitudes? Or 'ket notation?

All of that is important when talking concretely about superposition, entanglement, and operators. But one step at a time. This answer is to provide a brief description of what is a QuBit.

$\endgroup$

Your Answer

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