Since a quantum information lecture today I have been wondering what does it really mean for a state to be in superposition? Is this something that is answerable?

This is what we learnt (or what I gathered :) ):

A classical bit is always in a state 0 or 1. Sometimes there exists a degree of uncertainty and so probabilities are assigned to either state but in reality it still is 0 or 1 right. However a qubit can be in a state 0, 1 or a superposition whereby this superposition is fundamentally different from the probability mixture for a classical bit. But how can this be so? Surely at any given time a system can in reality either be in the 0 or 1? Does it have something to do with the interference properties of the system?

If you could answer my questions and explain the fundamental differences between the qubit and the classical bit I would really appreciate it. Thanks!

  • $\begingroup$ No, in the reality, objects with 2 distinguishable states may be found in any superposition $\alpha|0\rangle+\beta|0\rangle$. All these superpositions are equally genuine and equally allowed, because of the so-called superposition principle. It is not true that they're objectively required to be in 0 or 1, after all. Indeed, there's a fundamental difference between bits and qubits but our Universe is a quantum world, so it fundamentally respects the rules of qubits, not classical bits, however counterintuitive it may look to a layman. $\endgroup$ Commented Oct 8, 2012 at 18:19
  • $\begingroup$ @Luboš that sounds like material for an answer $\endgroup$
    – David Z
    Commented Oct 8, 2012 at 18:54

3 Answers 3


Surely at any given time a system can in reality either be in the 0 or 1?

Actually, no, that is not true. A quantum system can be in a state which is neither $\lvert 0\rangle$ nor $\lvert 1\rangle$; this is not possible with a classical bit. However, this state can be mathematically described as a linear combination of $\lvert 0\rangle$ and $\lvert 1\rangle$.

Consider the analogy of a traditional magnetic compass.

A traditional compass

When the compass needle points north, that is like a qubit being in the state $\lvert 0\rangle$, and when the compass needle points east, that is like a qubit being in the state $\lvert 1\rangle$. But a compass needle can also point northeast. The direction northeast is neither north nor east, but it is a superposition of equal parts north and east: if you add a north-pointing vector and an east-pointing vector of equal magnitude, you will get a vector that points northeast. Similarly, the qubit state $\frac{1}{\sqrt{2}}(\lvert 0\rangle + \lvert 1\rangle)$ is neither $\lvert 0\rangle$ nor $\lvert 1\rangle$, but it is a superposition of equal parts $\lvert 0\rangle$ and $\lvert 1\rangle$.

  • $\begingroup$ I like the simplicity of the analogy and it makes sense. Thanks for your answer $\endgroup$
    – Physbox
    Commented Oct 9, 2012 at 11:48
  • $\begingroup$ Glad it helped. I actually like this analogy myself because it is much more accurate than most analogies in physics - the mathematics of compass directions is actually almost the same as that of qubits, the main difference being that qubits can have complex coefficients. $\endgroup$
    – David Z
    Commented Oct 9, 2012 at 16:32
  • $\begingroup$ But on the Bloch sphere, which is an isomorphic representation of the qubit, $|0\rangle$ and $|1\rangle$ are antipodes, not at right angle... $\endgroup$ Commented Oct 10, 2012 at 11:39
  • $\begingroup$ Sure, so what? Just because there exists one isomorphic representation of the qubit space where $\lvert 0\rangle$ and $\lvert 1\rangle$ are mapped to opposite vectors doesn't mean that will be the case for every isomorphic representation. And in the simplest representation, state space itself, $\lvert 0\rangle$ and $\lvert 1\rangle$ are orthogonal. $\endgroup$
    – David Z
    Commented Oct 10, 2012 at 13:59

The state of a classical bit is one of the numbers {0,1}. Each measurement reveals the value of the state.

The state of a qubit is a point on the 3-dimensional Bloch sphere.
Each measurement arrangement selects a particular point; the measurement then gives a value 1 with probability $\cos^2(\theta/2)$ and 0 with probability $\sin^2(\theta/2)$, where $\theta$ is the angle between the vectors from zero to the state point and the measurement point.

In a coordinate representation, the qubit is described with respect to a particular measurement point (the north pole). The corresponding classical bit consists of the north pole and the south pole only. All other qubit states are superpositions of the two poles.


bits are either zero or one if created specificially or measured. when bits are randomly generated, they are only a probability distribution until measured. See http://www.princeton.edu/~pear/ and their research with randomly generated bits.

take a look:


  • $\begingroup$ to the skeptics, please explain the double slit experiment results data. $\endgroup$
    – patrick n
    Commented May 23, 2013 at 3:20

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