this is a very simple question, but apparently one that has no simple answer, at least from standard quantum mechanics theory

I'm trying to figure the number of simple quantum states (microstates) of the allowed configuration space of a two-level quantum system, ignoring for now mixed states. So if i have a state like:

$$ \Psi = \psi_0 | 0 \rangle + \psi_1 | 1 \rangle $$

removing all redundancies from the representation, we are left

$$ \psi_0 = \rho_0 $$ $$ \psi_1 = \rho_1 e^{-i \theta_1}$$ $$ \rho_0^2 + \rho_1^2 = 1$$

Now, this state space looks like $[0,1] \times S_1$. I've seen discussions that quantum states take some volume in phase space, but i want to understand that in detail: so in this case,

Question: how many distinct states are in this configuration space? is this a finite set? infinite but numerable set? nonnumerable?

Following Steve suggestion, lets write the Shannon entropy of this wave function:

$$ S( \Psi ) = - \rho_0 \ln \rho_0 - \rho_1 \ln \rho_1 $$

There is some conflict when i see this entropy; it cannot derive from any counting state scheme, in particular because it completely ignores the relative phase factor between the basis states, and as we all we know by now, states with different relative phase factors are physically different; otherwise there would be no interference terms, and no quantum mechanics to speak of!

  • $\begingroup$ Do you mean to say "$\theta_1$ is a randomly-chosen number between 0 and 2pi? If so, you are talking about a mixed state, not a pure state. If not, what is $\theta_1$? You should say how you think entropy is defined ... are you trying to use the Von Neumann entropy formula or something different? $\endgroup$ Aug 13, 2012 at 21:38
  • $\begingroup$ @SteveB, no, relative phase factors between states exist in pure states, the proof is that i use the state vector directly, hence the density matrix is evidently factorizable $\endgroup$ Aug 13, 2012 at 22:10
  • $\begingroup$ regarding entropy; it sounds like i would be missing the interesting step here: in classical mechanics you count states on an ensemble and then obtain entropy as the logarithm of that, while on QM you sort of cheat and skip that step and go directly to define some entropy. But i guess i could work your suggestion in reverse: compute the Shannon entropy of the wave function and ask: what number density gives as logarithm this entropy? $\endgroup$ Aug 13, 2012 at 22:13
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    $\begingroup$ @Christoph, now, i'm not saying otherwise; a pure quantum state is a single microstate - and i agree. The question i did was different; how many of those microstates are on the above two-level system? $\endgroup$ Aug 13, 2012 at 23:25
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    $\begingroup$ @diffeomorphism: the qubit state space $P_1\mathbb C$ is diffeomorphic to $S^2$ via Hopf fibration, see Bloch sphere $\endgroup$
    – Christoph
    Aug 13, 2012 at 23:42

3 Answers 3


There are several different notions of microstates or distinguishability that might be relevant to your question.

Coarse-graining of phase space into Planck cells.

Consider two classical variables $x$ and $p$ with $x \sim x+x_0$ and $p \sim p+p_0$. You can think of this system as describing a particle that lives on a circle of radius $x_0$ and where momentum is also defined only up to multiples of $p_0$. A physical realization of this situation is provided by a particle moving on a discrete lattice with periodic boundary conditions.

We can write the classical Lagrangian in the suggestive form $L = \dot{x} p - H$ which means that $p$ is conjugate to $x$. The volume of phase space is finite and given by $x_0 p_0$. Dividing phase space into Planck cells of volume $h$ gives us $N = x_0 p_0 / h$ states. This is the semiclassical estimate.

Now we turn to the full quantum theory. Consider the operators $X = e^{i 2\pi x/x_0} $ and $P = e^{i 2\pi p/p_0}$. These operators satisfy $PX = XP e^{i 4\pi^2 \hbar/(x_0 p_0)} = XP e^{ 2\pi i /N}$. Starting with a state $|0\rangle$ satisfying $P|0\rangle = |0\rangle$ we can create new states by acting with $X$. The state $X^n |0\rangle$ satisfies $P X^n |0\rangle = e^{ 2 \pi i n/N} X^n |0\rangle$, so the state $X^N |0\rangle$ is proportional to $|0\rangle$. Thus we have constructed precisely $N$ states of the form $\{|0\rangle, X|0\rangle, ..., X^{N-1} |0\rangle \}$ using the exact quantum operator algebra. What we have shown is that the semiclassical estimate gives the correct full quantum state count in this example.

One answer to your original question is then that a two level system can be understand as two Planck cells of phase space.

Counting partially orthogonal quantum states.

Of course, there are an infinite number of quantum states as parameterized by the Bloch sphere - each point is slightly different. However, most of these states are not orthogonal. We can ask how many different choices of parameters give states that are $\epsilon$-orthogonal e.g the size of the largest set $\{|\psi_i \rangle \}$ with $|\langle \psi_i | \psi_j \rangle| \leq \epsilon$ for $i \neq j$. Obviously for $\epsilon \rightarrow 0$ we get something like the dimension of the Hilbert space. For example, the one parameter family of states $|\theta\rangle = \cos{\theta} |0 \rangle + \sin{\theta}|1\rangle$ satisfies $\langle \theta | \theta'\rangle = \cos{(\theta-\theta')}$. If $\epsilon = 0$ then we can only choose $\theta=0,\pi/2$. On the other hand, if $\epsilon = 1- \delta$ then we have $\cos{(\Delta \theta)} < 1 - \delta$ ($\Delta \theta$ is the spacing between neighboring $\epsilon$-orthogonal states) or $\Delta \theta > \sqrt{2 \delta}$ ($\Delta \theta$ assumed small). This implies that we have roughly $2\pi/ \Delta \theta$ $\epsilon$-orthogonal states. You can see that as we permit more and more nearly parallel states, the number of $\epsilon$-orthogonal states increases.

A general estimate for the number $\epsilon$-orthogonal states in a $D$ dimensional Hilbert space is $(1-\epsilon)^{-D/2}$ for $1-\epsilon$ small. Roughly speaking, we're covering up the $D$ dimensional space with little spherical chuncks of radius $\sqrt{1-\epsilon}$ and volume $(1-\epsilon)^{D/2}$. To be a bit more precise, the volume of a $D$ dimensional sphere of unit radius is roughly $\frac{2 \pi^{(D+1)/2}}{\Gamma((D+1)/2)}$ while the volume of a little chunk is $\frac{2 \pi^{D/2}}{D \Gamma(D/2)} (1-\epsilon)^{D/2}$. Packing the chunks in as tightly as possible gives roughly $(1-\epsilon)^{-D/2}$ states. This is the origin of the statement that in a system of $n$ qubits, where $D=2^n$, the number of distinguishable states grows doubly-exponentially in $n$ i.e. like $(1-\epsilon)^{-2^n}$.

The notion of $\epsilon$-orthogonal states then provides another way to quantify the number of states in Hilbert space. We probably don't want to say the number of states is physically infinite, since nearby states are almost identical as regards physical properties. On the other hand, requiring strict orthogonality might also be to stringent a requirement.

von Neumann entropy.

Quite generally, the von Neumann entropy of any mixed state of a single qubit is bounded by $\ln{2}$ (or $1$ using base two log). The entropy is another good measure of microstates in many senses. One example: in the context of quantum communication, Holevo's theorem is a result that shows that a single qubit isn't worth more than one classical bit in certain communication protocols even though the wavefunction formally requires an infinite amount of information to specify the complex amplitudes.

  • $\begingroup$ this is the closest i've seen to an answer of the concern i tried to express on my question, so kudos +1.. I'm still unconvinced that partially orthogonal states makes up for this, since it seems a bit arbitrary to me as an state enumeration scheme, i guess i'll have to read and think about it a little more $\endgroup$ Sep 7, 2012 at 20:32

It is misleading to write $\rho_i$ for the components of $\psi$, and as they are complex numbers, you cannot use these in the formula for entropy.

The space of wave functions is (not $[0,1]\times S^1$ but) the Poincare sphere (or Bloch sphere) S^2, parameterized by quaternions (corresponding to points on the complexified circle).

The probabilities are $\rho_j=|\psi_j|^2$ and add up to 1. The Shannon entropy applies with this definition of $\rho$. (Note that a pure state has zero entropy in the statistical mechanics sense only if it is an eigenstate of the Hamiltonian, and if the probabilities considered are in the corresponding eigenbasis.)

Edit: In quantum statistical mechanics, an equilibrium state has a density matrix commuting with the Hamiltonian; the diagonal elements in the eigenbasis determine the probabilities. Only eigenstates are eligible as accessible states in the entropy formula.

The only pure states that commute with the Hamiltonian are the states $\rho=\psi\psi^*$ with an eigenstate $\psi$; in these states all but one probability vanish, and the entropy is zero. This situation applies for a 2-state system, the Hamiltonian is $H=\pmatrix{E_0 & 0\\0 & E_1}$ in the eigenbasis of $H$, and $E_0<E_1$ is assumed. In this case, precisely two eigenstates exist, hence precisely two states are accessible, except when the system is already in one of the pure eigenstates.

An arbitrary state with a nondiagonal density matrix is not in equilibrium, hence the assignment of probabilities, while formally possible, lacks a simple interpretation. Different pure states may or may not have the same probabilities; they cannot be distinguished solely on the basis of probabilities (or entropy). A nonequilibrium pure state always has a positive entropy, since all eigenstates to which it is not orthogonal are accessible.

  • $\begingroup$ fair enough, but i don't see accounted the fact that each wavefunction is a different state, or at least i don't understand where it becomes accounted this fact in computing entropy, or why is not relevant, because intuitively, each eigenstate is a different state, and if i give a quantum system enough energy to be in a superposition of eigen states, the Shannon entropy is 1 because it is pure, but how can the number of accessible states be one? it is a superposition of two eigenstates after all, so it can become any of them two $\endgroup$ Sep 3, 2012 at 19:46
  • $\begingroup$ @diffeomorphism: I edited my answer to reflect your comments. $\endgroup$ Sep 4, 2012 at 7:05

The number of states in this two-level Hilbert space is infinite and nonnumerable. I think this is very clear, because $\rho_0=\sqrt{1-\rho_1^2}\in[0,1]$ and $\theta_1\in[0,2\pi]$ are two continuous variables. The question is akin to "how many function values of $\sin(x)$ with $x\in[0,\pi]$?".

  • $\begingroup$ Please construct a density matrix with entropy more than $k_B \ln 2$. Because $S=k_B \ln \Omega$ and you believe $\Omega$ can be infinitely large. $\endgroup$
    – Siyuan Ren
    Aug 22, 2012 at 1:50
  • $\begingroup$ While at the moment I cannot give a strict argument, from phenomenology it is obvious that something is wrong here. Each degree of freedom adds to the entropy. Now a linear spring has one degree, so it can only contribute one unit of entropy. Now connect a second orthogonal spring to your object, this should double the degrees of freedom but according to your argument the degrees of freedom, and therefore the entropy, are now infinite. $\endgroup$
    – Alexander
    Sep 4, 2012 at 12:07

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