The title says it all.

My understanding is that a qubit is a superposition of $|0\rangle$ and $|1\rangle$, i.e. the answer to a binary question. So I imagine that specifying an event in spacetime would require an infinite sequence of qubits. I'm thinking of a binary search tree. Is there some other way to do it with a finite amount of qubits?

  • 3
    $\begingroup$ How many qubits does it take to screw in a lightbulb? $\endgroup$ Commented May 24, 2012 at 7:21

3 Answers 3


You'd never need a infinite sequence as of course there's a limit to the accuracy of the a position measurement. Given either by Planck's length (but no measurement gets near Planck length, because no imaging beam get near Plank's energy), in the non-Planckian regime, you can measure a position by scattering a beam of light energy E= hv, and get a position accurate to.

$$ \Delta x = \frac{\hbar}{2p} = \frac{\hbar c}{2E} $$

If you're starting in a box of volume V you already know the particle is in, then you need,

$$ B = \log_2 \left ( \frac{V}{\Delta x^3}\right) = 3\log_2 \left ( \frac{ VE^3 }{\hbar^3c^3} \right) $$

Bits of information to describe where you measured the particle to be.

Qubits are more complicated though, using qubit instead of digital bits, you can describe a distribution of the probabilities of the particle being at each location inside the volume. A measurement yielding qubits might does not need to collapse the wavefunction, so may describe the amplitude of each location.

  • $\begingroup$ I also think the last suggestion here is more useful than simply encoding a position "classically" in a quantum register. That is, try to make the measurement apparatus transform the measured state into a (continous) superpositioned state of |0> and |1>. However, be aware of the no-cloning theorem when you suggest that this could be done without "collapsing the wavefunction"! $\endgroup$
    – BjornW
    Commented Jun 5, 2011 at 11:16
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    $\begingroup$ @Dr BDO Adams: Measuring a quantum system without collapse into Qubits will collapse the wavefunction, or else it violates the no-cloning theorem. $\endgroup$
    – Ron Maimon
    Commented Nov 26, 2011 at 3:55

What is the binary search tree for? Let us assume you mean a classical spacetime physical event for which the specification of one (say qubit) measurement is insufficient. In twistor Minkowski space the most trivial spacetime event is a point, specified essentially by an element of SL(2,C). This requires a combination of spatial qubits X, Y, Z and a diagonal component for time ie. a quaternion. But the qubits here are the Pauli operators rather than the eigenvalues. If you want to specify 'complex' numbers with binary trees, as in the surreal number system, that is an interesting direction to go in, but perhaps not what you meant.


Alternatively you might take your starting point to be N quantum particles in a box, and try to encode the data needed to describe the quantum state. As Dr BDO Adams stated above, our qubits will be used to describe probability distributions of positions of particles, so this is really an elaboration on his answer.

Since we have a compact space (the box), we have discrete eigenstates. Rather than thinking about the position basis and Heisenberg, I'd suggest the energy eigenstate basis. Lets have the particles to be N distinguishable particles (not indistinguishable, for then we'd have to consider boson/fermion). Each particle would get a countably infinite sequence of complex numbers whose norms sum to 1.

Should we use one qubit to store one real number, and thus get a countably infinite number of qubits to store the full information of the state of each particle? Can we do better?

Lets start from a naive math argument. How much information is held in a countably infinite sequence of real numbers? We can make a bijection from countably infinite sequences of real numbers to just the real numbers. (First take the arctangent of each number in your sequence and divide by pi to get the numbers between 0 and 1. Write down the numbers as a decimal in a list one after another. Then form a new real number from all of those digits by starting in the upper left corner and zig-zag-ing down. This is a bijection up to .99999... ambiguities.)

So maybe we can take the whole state of the particle and just cram it into one qubit?

On second thought, that is a horrible horrible non-continuous way to store information. We should require some continuity in the way we store the information, so that a small movement of a particle makes a small change to our qubits. Continuous curves still allow us to compress information in horrible ways. (a space filling curve takes each point in a plane to a unique real number, but is fractal like.) We should make our information storage, which I'm considering here as a functor from the Hilbert space of 1-particle in a box to the Hilbert space of one or more qubits, to be a smooth functor.

Imagine the box is 2-dimensional. Imagine putting the box on a surface of a sphere. That would give you a correspondence between points in the box and qubits states on the bloch sphere, right? Each position eigenstate gets mapped to the qubit thats represented by a spin arrow that points to it on the edge of the sphere.

What's wrong with this picture? Linear superpositions of position eigenstates are not necessarily mapped to unique states. Orthogonal position states don't get mapped to orthogonal qubit states.

So here is what we really want: a linear, invertible, unitary (orthogonal states go to orthogonal states) map from the Hilbert space of the particle in the box to the Hilbert space of quibits. Invertible linear maps from infinite dimensional Hilbert spaces like that of the particle in the box will need infinite dimensions to go into, so infinite qubits as you said, to capture ALL of the information. If you don't need all the information, maybe you limit momentum to some large cutoff, than that just leads to using only finitely many of the energy eigenstates and mapping the others to 0. The number of energy eigenstates used will be the dimension of the Hilbert space, and the number of qubits you need for each particle. (A qubit has one complex degree of freedom, since it is specified by 2 complex coefficients but is normalized.).

I think Dr. BDO Adams means to use $\Delta x = \hbar / (2p) $ with p being the momentum of the particle, not the incoming light. The smallest $\Delta x$ will come with the largest $p$ you allow. Our answers match if you use the momentum cutoff p where he has placed p and the relation between Energy and momentum being that of the particles, $p^2/2m$.

I'm assuming that the uncertainty of position of the particle should be related to the uncertainty in its momentum. If you bombard it with light, you collapse the state into one whose uncertainty of position is now given by the momentum of the light. 1. You've changed the state. 2. Now the particle has been kicked up to the higher momentum of the light, and we have to raise our momentum cutoff. Better to just leave the state alone.

  • $\begingroup$ Context: Random late at night thoughts about this problem. $\endgroup$
    – bdubsair
    Commented Jul 28, 2011 at 6:03
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    $\begingroup$ meaningless. A qubit cannot store more than one bit. $\endgroup$ Commented Nov 16, 2012 at 10:04

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