How many combinations can 'n' qubits have?


With $4$ bits there are $2^4=16$ combinations. How many combinations can you have with $4$ qubits (assuming they are all superimposed)?


As per here, the argument is that if a qubit can hold more information when it's measurement(probability) is collapsed to give a $|1\rangle$ or a $|0\rangle$ (assuming measurement is without errors). For a superimposed qubit holding two values at the same time depending on how it has been encoded i.e. as per the parallel measurement (probably in a quantum gate) we could get $|00\rangle$, $|01\rangle $, $|10\rangle$, $11\rangle$. I am seeing this as if observed in parallelism 4 qubits would yield a combination ($2^8$) of 8 classical bits (i.e. n qubits having $2^{n+n}$ number of combinations in parallel). Is this correct?


Posted 2019-09-25T16:48:11.080

Reputation: 179

1Hi @LiNKer! Welcome to QCSE! This question is a little broad right now, because it's unclear what's being asked and what you know of. Nonetheless, $4$ qubits can also be in a uniform superposition of all $2^{4}=16$ basis states; a difference between a "normal pc" (what is called a classical computer) and a quantum computer is that the bits in a classical computer are either $0$ or $1$, while for a quantum computer the qubits can be in a superposition of $0$ and $1$. Can you consider revising your question to provide more clarity on what you know of and where your question comes from? – Mark S – 2019-09-25T17:18:45.980

Hi Mark, thanks for the suggestion. made the edit – LiNKeR – 2019-09-25T17:30:51.307


Possible duplicate of Understanding (theoretical) computing power of quantum computers

– ahelwer – 2019-09-25T20:37:07.293



This depends on precisely what you mean by "combinations". Let's go back to just one qubit to be clear. For one qubit, there are two classical states, $|0\rangle$ and $|1\rangle$. These are really the only things you can count. Of course, as a qubit, you are allowed any superposition of the form $$ \alpha|0\rangle+\beta|1\rangle,\qquad |\alpha|^2+|\beta|^2=1. $$ This is a continuum of possibilties that you cannot count.

Similarly for four qubits, there are $2^4=16$ classical states, $|0000\rangle, |0001\rangle,\ldots,|1111\rangle$, and this is the only thing you can really count. Again, because it's a quantum system, you can have an arbitrary superposition $$ \sum_{x\in\{0,1\}^4}c_x|x\rangle,\qquad\sum_{x\in\{0,1\}^4}|c_x|^2=1, $$ but it is also a continuum of possibilities that you cannot count.


Posted 2019-09-25T16:48:11.080

Reputation: 35 722


Well, it's still $2^4 = 16$ I believe. The point of superposition is that these 16 combinations can encode 16 inputs, and with something called "quantum parallelism" they can be calculated simutaneously. Even though they don't give out all $16$ results simultaneouly, the result can be manipulated in various ways to give you what you need while drastically reduce the running time, compare to a similar process done on a classical computer.

Kim Dong

Posted 2019-09-25T16:48:11.080

Reputation: 337

If I got the first part: 4 superimposed qubits still produce 16 combinations as 4 classical bits. – LiNKeR – 2019-09-25T18:31:39.343

I am lost with how each combination from the 16 can now encode 16 inputs – LiNKeR – 2019-09-25T18:33:45.517

Yes, each encode only 1 input, usually, if your algorithm isn't doing anything special – Kim Dong – 2019-09-26T08:09:53.123

just went through this again, great thanks for the help – LiNKeR – 2019-09-28T19:09:04.000


As other have said, you can see 4 qubits as having either 16 different combinations, or as continuum-many superpositions of these combinations. (Continuum-many means "the cardinality of the real numbers") However you can also ask about the dimension of the space, whcich is a different notion of size.

A single qbit is described by a vector in a 2-dimensional Hilbert space, so that you need 2 complex numbers to describe it. Equivalently, you need 4 real numbers. However, since the norm of the qubit has to be 1, you in fact only need 3 real numbers. But also a global phase is irrelevent ! So only 2 real numbers are needed to describe a qubit. That's why you can position it on the Bloch sphere, which is a real two dimensional space.

With 4 qubits you're in a Hilbert space of dimension 2*4 = 8. Hence you need at first glance 16 real numbers. But in fact, considering that a global phase doesn't matter and the norm must be 1, you only need 14 real dimensions to describe it.


Posted 2019-09-25T16:48:11.080

Reputation: 116


In addition to @D. Tran's answer and for anyone out there (just in case), this did it for me.

The first mistake I made was that I kept on thinking superimposed qubits could only hold 0s and 1s as pairs (2 states) i.e. say for $n=3$ qubits I thought it was equivalent of ##-##-## (e.g. 01-11-01 which is just $2^{n+n}$ combinations at max) but it turned out a superimposed qubit can represent multiple states and each of these can be represented by complex numbers on a Bloch sphere.

The second thing that got me confused again was that I thought because $n$ qubits could store all combinations ($2^n$) for $n$ bits at once (from the video in the question's EDIT), it meant this was the limit of $n$ qubits and thus I was trying so hard to relate the confusion and the mistake.

At least now that I am aware qubits can represent multiple states which can employ "quantum parallelism" to store all combinations of $n$ classical bits I'd argue a single qubit can be used to represent information for more complex-numbers/digits while a bit can only represent a binary digit.


Posted 2019-09-25T16:48:11.080

Reputation: 179