2
$\begingroup$

I read this at a book;

The difference between bits and qubits is that a qubit can be in a state other than $|0\rangle$ or $|1\rangle$. It is also possible to form linear combinations of states, often called superpositions: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$.

What I concluded was that not all states are superpositions from way of written of the authors, am I right or I'm misunderstanding the authors? And if I am right, what is an example of a state rather than superpositions?

$\endgroup$
4
$\begingroup$

You conclude correctly. Not all physical states available to a qubit are pure superpositions, and it can also occupy states known as mixed states which are halfway between a superposition of $|0⟩$ and $|1⟩$, and a simple probabilistic mixture between the two.

States of the form $$|\psi⟩=\alpha|0⟩+\beta|0⟩\tag1$$ are known as pure states, and these are the (pure) superpositions. When first teaching quantum mechanics we focus on these, because (i) these encapsulate the ways in which QM is different from classical physics, and (ii) they're much easier to deal with than mixed states.

The state $(1)$ is often phrased as saying that the system is somehow "both" in $|0⟩$ and $|1⟩$ at the same time. This naive understanding is not 'wrong' but only because it doesn't really mean much - what does it even mean? The first thing that comes up is that if you actually look at it, it has a probability $p$ of being in $|0⟩$ and a probability $1-p$ of being in $|1⟩$.

The problem of that explanation is that as described, the state is not all that magical. It's perfectly possible to produce, within classical physics, a box which will have zeros $p$ of the time and ones $1-p$ of the time, simply by flipping coins before one closes the boxes. A superposition, on the other hand, is something beyond this. Rephrasing the state a bit, you can write it as $$|\psi⟩=\sqrt p|0⟩+e^{i\phi}\sqrt{1-p}|0⟩,\tag2$$ where the probabilities are explicitly spelled out, but there is another ingredient: the relative phase between the two components, $\phi$. If the state really is in a superposition, then there are experiments you can do between the two which will make the two components interfere in a way which is sinusoidal in $\phi$.

One good way of thinking about mixed states is as superposition states where the information about this phase is somewhat uncertain. When that happens, the interference pattern gets washed out a bit, and the fringes are less distinct than for a pure state. In the worst case scenario, we've got no information at all about the phase, and it can be $\phi=0$ in one realization and $\phi=\pi$ in the next. In this case, the peaks of one realization will be on the troughs of the next, and on average you will see no interference. This worst case is indistinguishable from a classical, probabilistic mixture.

To correctly describe mixed states, you need to move away from the wavefunction as the descriptor of the state of the system, and use density matrices. The system's density matrix is a hermitian, positive operator $\hat\rho$ which obeys $\operatorname{Tr}(\hat\rho)=1$, and which gives you the expectation value of any system observable $\hat A$ via $$ \left\langle\hat A\right\rangle=\operatorname{Tr}\left(\hat\rho\hat A\right). $$ For a pure state, the density matrix is equal to $\hat\rho=|\psi⟩⟨\psi|$. For a probabilistic mixture of pure states $|\psi_n\rangle$ with probabilities $p_n$ (where $\sum_np_n=1$), the density matrix is $\hat \rho=\sum_n|\psi_n⟩⟨\psi_n|$. Finally, the most general density matrix of the density matrix of a qubit, in the $\{|0⟩,|1⟩\}$ basis, is given by $$ \hat\rho=\begin{pmatrix}p&ce^{-i\phi}\sqrt{p(1-p)}\\ ce^{i\phi}\sqrt{p(1-p)}&1-p\end{pmatrix} $$ Here $p$ and $\phi$ are as before, and you have a new variable: the degree of coherence, $c$, which equals $1$ for a pure state, $0$ for a classical, probabilistic mixture, and in general somewhere between the two.

$\endgroup$
  • 3
    $\begingroup$ I'm not sure I completely understood the question, but it might be worth adding a note about the fact that the notion of superposition is basis-dependent, in the sense that $|0\rangle$ is a superposition if we consider a basis other than $\{|0\rangle, |1\rangle\}$. I suppose in the context of the question, this basis is "special" in the sense that it's "the" measurement basis, but we could just as easily have a different basis as the measurement basis. $\endgroup$ – march Oct 9 '15 at 22:52
1
$\begingroup$

Every qubit can be written as a superposition of the basis states $|0\rangle$ and $|1\rangle$.

If you want to work with states described by more than 1 qubit you will need to work in a different Hilbert space. For instance suppose we call the 1 qubit space $\mathcal{H}$ then the 2-qubit hilbert space will be $\mathcal{H} \otimes \mathcal{H}$. The states in this new space will not be superpostions of $|0\rangle$ and $|1\rangle$; but rather superpositions of $|00\rangle$, $|10\rangle$, $|01\rangle$, and $|11\rangle$.

$\endgroup$
  • $\begingroup$ So for one bit, we don't have any state other than liner combination of 0 and 1? $\endgroup$ – AmirHosein SadeghiManesh Oct 9 '15 at 19:32
  • 2
    $\begingroup$ @AmirHoseinSadeghiManesh, If we are talking pure quantum mechanics then yes. Quantum statistical mechancis allows some uncertainty about which quantum state the qubit is actually in; then we would have "mixed states" which are described by a density matrix. $\endgroup$ – Spencer Oct 9 '15 at 19:38
  • $\begingroup$ @AmirHoseinSadeghiManesh, $|+\rangle$ is a superposition. I'm not sure what the definition of "halfway" is. $\endgroup$ – Spencer Oct 9 '15 at 21:57
  • $\begingroup$ Yes, indeed mixed states are not necessarily superposition. $\endgroup$ – AmirHosein SadeghiManesh Nov 13 '15 at 15:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.