I was reading the book by Nielsen & Chuang. I got the part about why we use the density operators. And then I got to the section of theorem 2.6. It says roughly this thing:-

The sets $|{\tilde\psi_i} \rangle$ and $|{\tilde\phi_j} \rangle$generate the same density matrix if and only if $|{\tilde\psi_i} \rangle= u_{i,j} |{\tilde\phi_j} \rangle$, (2.166) where $u_{i,j}$ is a unitary matrix of complex numbers, with indices $i$ and $j$, and we ‘pad’ whichever set of vectors $|{\tilde\psi_i} \rangle$ or $|{\tilde\phi_j} \rangle$ is smaller with additional vectors 0 so that the two sets have the same number of elements.

Ok. I will leave all this tech talk aside and now the main thing is it tries to say is, if I can find two sets of vectors that generate the same density operator, it would be related in some way.

Or if we think a bit more we can say that suppose we find one set of $\phi$ then from that if we apply a group of unitary operators, we can reach another set which will have the same state representation or same state? (Yes state because it will have the same density operator).

Is this why this theorem got a place in this book? I am missing the idea behind this theorem. Alright, this is what I thought but what do you people think?

If you see I am saying something wrong, please correct me.

  • 1
    $\begingroup$ this is pretty much the same statement I prove in the second part of this answer $\endgroup$
    – glS
    Commented Dec 25, 2020 at 23:35

2 Answers 2


The key point is that two different ensembles can give rise to the same density matrix. And, given knowledge of only the density matrix, we cannot assign a unique ensemble to it.

Let's consider a simple example, the density matrix corresponding to the maximally mixed state for a qubit is $\frac{\mathbb{I}}{2}$, which can be expressed via two different ensembles, say, \begin{align} \frac{1}{2} | 0 \rangle \langle 0 | + \frac{1}{2} | 1 \rangle \langle 1 | = \frac{\mathbb{I}}{2} = \frac{1}{2} | + \rangle \langle + | + \frac{1}{2} | - \rangle \langle - | . \end{align}

Should we, then, think of the state $\frac{\mathbb{I}}{2}$ as states $\{ | 0 \rangle, | 1 \rangle \}$ with equal probability or states $\{ | + \rangle, \{ - \} \}$ with equal probability? The point is, we cannot distinguish these two ensembles.

In fact, the maximally mixed state can be written as an equal convex combination of any two orthogonal pure states of the qubit. Therefore, there are infinitely many ensembles that correspond to the density matrix $\frac{\mathbb{I}}{2}$.

How much freedom is there in the choice of these ensembles, this is what Theorem 2.6 talks about; that these sets of states are related via a unitary $\left|\tilde{\psi}_{i}\right\rangle=\sum_{j} u_{i j}\left|\tilde{\varphi}_{j}\right\rangle$, where $u_{ij}$ is a unitary.

  • $\begingroup$ Sounds good. Let's see if there is any other viewpoint. Thanks for answering $\endgroup$
    – user27286
    Commented Dec 9, 2020 at 16:12
  • $\begingroup$ I don't understand this. Any two states related by a unitary will result in the same density matrix, but all quantum state evolution is governed by unitary transformations, so are all quantum states indistinguishable in density matric form? There must be a big flaw in my understanding of quantum mechanics somewhere. $\endgroup$ Commented May 22 at 16:08

The general form of this statement is that every decomposition of a state $\rho$ into a convex sum of rank-1 operators (i.e. pure states) can be written in the form $\rho = \sum_k u_k u_k^\dagger$ where the (not necessarily normalised) vectors $u_k$ are the columns of the matrix $\sqrt\rho V^\dagger$, for some isometry $V$.

Note that such decompositions amount to writing $\rho=UU^\dagger$ with $U\equiv \sum_k u_k e_k^\dagger$ (matrix whose columns are $u_k$), and thus for $U=\sqrt\rho V^\dagger$ we clearly have $\rho=\sqrt\rho V^\dagger V\sqrt\rho=\sqrt\rho\sqrt\rho=\rho$. This shows that the columns of $\sqrt\rho V^\dagger$ for any isometry $V$ provide a pure state decomposition. To show that all pure state decompositions have this form you can use the more general statements about uniqueness of singular value decompositions, which I discussed e.g. in this math.SE answer. The gist of it being that $AA^\dagger=BB^\dagger$ implies $A=BV^\dagger$ when $A$ has more rows than $B$.

For example, if $\rho=I/2$ and $U$ is any 2x2 unitary, then we get decompositions of the form $I/2=\frac12(|u_1\rangle\langle u_1|+|u_2\rangle\langle u_2|)$ for any pair of orthonormal vectors $|u_i\rangle$. More generally, if $V$ is an isometry, such as $$V=\frac{1}{\sqrt3}\begin{pmatrix}1&1\\1 & \omega_3 \\ 1 & \omega_3^2\end{pmatrix},$$ then we get the (non-maximal) decomposition $$\frac I2 = \frac13\left(\mathbb{P}\left(\frac{|0\rangle+|1\rangle}{\sqrt2}\right) + \mathbb{P}\left(\frac{|0\rangle+\omega_3|1\rangle}{\sqrt2}\right) + \mathbb{P}\left(\frac{|0\rangle+\omega_3^2|1\rangle}{\sqrt2}\right)\right),$$ where $\omega_3\equiv \exp(2\pi i/3)$.

It's worth stressing that these are simply different ways to express the same physical state. We are not "evolving" the state like we would with something like $V\rho V^\dagger$. The physical reason behind this is that mixed states can in general be "explained" in many equivalent ways. For example, $\rho=I/2$ is the state obtained if you are given either $|0\rangle$ or $|1\rangle$ with $50/50$ probability. The existence of different decompositions recognises that it is impossible to tell apart receiving, say, $|0\rangle$ or $|1\rangle$, rather than $|+\rangle$ or $|-\rangle$. While these two scenarios might correspond to different "physical preparations" (someone needed to prepare the actual states of course), quantum mechanics is such that it is impossible to retrieve the information about which of the two ensembles was prepared.


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