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Possible States of n Qubits as opposed to classical bits

  1. Aug 13, 2011 #1
    I am reading an introduction to quantum computing and I have a question about one thing I dont understand.

    "In classical physics, the possible states of a system of n particles, whose individual states can be described by a vector in a two dimensional vector space, form a vector space of 2*n dimensions. However, in a quantum system the resulting state space is much larger; a system of n qubits has a state space of 2^n dimensions."

    Its the dimensions thing that confuses me. Lets say we have 3 classical bits of information. There are 2^3=8 possible unique configurations of these three bits, so there are 8 states.

    So the above asserts that these three bits have 2*3=6 dimensions, which makes sense because each bit has 2 possible states and thus there are six different numbers involved.

    Now consider three qubits. They apparently have 2^3=8 dimensions. Could someone explain this? And how many states do they have? I would have assumed that because of superposition they would have something like the original 8 states plus all combinations of these. Is that the right idea?
  2. jcsd
  3. Aug 13, 2011 #2
  4. Aug 13, 2011 #3
    Thank you, Ive read that but I will read it again.

    Ive clarified a few things for myself over the past while. The three qubits will have an infinite number of states because they can be in any quantum superposition of their 8 classical configurations, ie they are a linear combination of their basis vectors, of which, by what Ive written above, there are 8.

    So three classical bits have 8 states and 6 dimensions, three qubits have infinite states and 8 dimensions.

    I am still having trouble with the relationship between the 6 and 8 dimensions of classical versus quantum bits, respectively. Im remembering more and more linear algebra but there's something that I dont yet understand.

    The qubits are in any linear combination of these:

    And the classical bits as well only with the constraint that each coefficient must be 0 or 1 and must sum to 1. So why is this 6 and not 8 dimensions?
  5. Aug 14, 2011 #4


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    Your calculation for classical information is wrong.
    Each particle represent not a single bit, but two.
    We have 3 pairs of bits as individual states can be described by a vector in a two dimensional vector space. So the state space of n particles having 2-dim individual spaces is 2n-dimensional. In your case (3 particles) 64, not only 8, different values may be coded (first particle in one of four states, the same for 2nd and 3rd).

    Your calculation for qubits is right: using n qubits you may create 2^n orthogonal vectors, they define the total space.
  6. Aug 14, 2011 #5
    Thanks. I think see where I went wrong. I was thinking in terms of bits themselves and not particles. A bit could be described by a vector in a one-dimensional vector space, a particle by a vector in two-dimensions.

    So these two dimensions of each particle (x,y) would be physically something like up/down and right/left?
  7. Aug 14, 2011 #6


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    Almost, actually it is the polarisation, what is most commonly used representation in experiments: we use polarized photons, and one dimension of their state space is [0=vertical, 1=horizontal], the second is [0 = 45° left-slanted, 1 = 45° right-slanted].
    But if you use electrons the spin orientation up/down or left/right would be the right choice.
  8. Aug 14, 2011 #7
    I believe this is incorrect. For the photon, the two dimensions are horizontal and vertical. Arbitrary state X can be represented as a linear combination of the two: |X> = a|H> + b|V> where a and b are complex coefficients.

    Accordingly, directions slanted at 45 degrees are not 'another dimension' but a different choice of basis vectors. You can transform back and forth between them:
    |H> = c(|R45>+|L45>),
    |V> = c(|R45>-|L45>)

    Here c = sqrt(1/2) is a coefficient to keep the wavefunction normalized.

    You can also have complex coefficients, eg. left and right circular polarization:
    |L> = c(|H>+i|V>)
    |R> = c(|H>-i|V>)

    Note that |H> and |V> vectors are orthogonal in Hilbert space, meaning their scalar product <H|V>=0, but is has nothing to do with them being at 90 degrees in our 3D space, its is just a coincidence which causes a great deal of confusion. For electrons and other spin 1/2 particles, directions 'up' and 'down' along arbitrary axis are orthogonal and any other direction is expressed as a linear combination of the two (with complex coefficients):
    |+X> = c(|+Z> + |-Z>)
    |-X> = c(|+Z> - |-Z>)
    |+Y> = c(|+Z> + i|-Z>)
    |-Y> = c(|+Z> - i|-Z>)

    Anyway, back to the OP question. 1 qubit has 2 basis vectors and 2 complex coefficients. A system of N qubits will have 2^N basis vectors and with 2^N complex coefficients.
    (Actually, since the coefficients must be normalized and only relative phase shifts between them are important, one complex number is redundant and the correct answer is 2^N-1).

    Compare it with a system of classical particles in XY plane: each particle requires 4 real numbers for position and momentum, and the the entire system requires only 4N real numbers (or 2N complex numbers) to describe it.

    So simulating 32 billiard balls will require 128 real numbers or 512 bytes of memory (single-precision floats). Simulating a system of 32 qubits will require 2^32 complex numbers or 32GB of memory. Just think about it, let the mind boggle for a bit :)
  9. Aug 15, 2011 #8


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    Delta Kilo is of course right. Thanks for correcting me!
  10. Aug 15, 2011 #9
    Thank you both for the kind responses!
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