I have a feeling this is a very basic question. I apologize if it is.

Using Dirac's notation, can every (mixed) density operator $\rho_A$ of system $A$ be written as the ket-bra (outer) product $|a_1 \rangle \langle a_2|$ for some vectors $|a_1\rangle , |a_2\rangle \ \in A$ ?

I ask this because in the book "Quantum Computation and Quantum Information" by Nielsen & Chuang, the definition of reduced density operator of operator $\rho^{AB}$ of systems $A$ and $B$ is given as so:

$$\rho^{A} =\mathrm{tr}_B(\rho^{AB}) = \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2| \right) = | a_1 \rangle \langle a_2 | \ \ \mathrm{tr} (|b_1 \rangle \langle b_2| )$$

which -- to me at least-- is implicitly asserting two things:

  1. Every operator $\rho^{AB} \in A \otimes B $ can be expressed as a tensor product $\rho_A \otimes \rho_B$, for some $\rho_A \in A, \ \rho_B \in B$, which I though was only true for separable $\rho^{AB}$ ,
  2. Every $A$ can be expressed as a product $|a_1 \rangle \langle a_2|$, and similarly for $B$.

So this second point is the one I'm asking.


Thanks to @NorbertSchuch for pointing out my mistake.

In the above question I wrongly merged two definitions of Nielsen and Chuang's book into one equation. The first one is the definition of the reduced density operator

$$\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$

The second one is the definition of the partial trace, which is defined (completely independently of the above definition) as

$$\mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2| \right) = | a_1 \rangle \langle a_2 | \ \ \mathrm{tr} (|b_1 \rangle \langle b_2| )$$

What the book doesn't say --which is what I was after-- is what the general definition of the partial trace is in terms of bra-ket notation.

I found the general definition of the partial trace here.

So, thanks to this and @CraigGidney, @WetSavannaAnimalakaRodVance, @ACuriousMind answers I now know that the answer is no.

  • $\begingroup$ I've now fixed the notation , sorry. $\endgroup$ – Hugo Nava Kopp May 26 '16 at 11:55
  • $\begingroup$ Are you sure you don't misquote Nielsen & Chuang? $\endgroup$ – Norbert Schuch May 26 '16 at 13:04
  • $\begingroup$ @NorbertSchuch nop, I just double checked now. This definition is on page 105, Ed. 2007. $\endgroup$ – Hugo Nava Kopp May 26 '16 at 14:31
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    $\begingroup$ I checked, and you are misquoting it. There is no such equation. You merged two equations into one above (without telling us!) Please re-read it. The first is the definition of the reduced density matrix, and the second the definition of a partial trace on a basis. --- Why do people always think the error is in the book and not in their interpretation, anyways? $\endgroup$ – Norbert Schuch May 26 '16 at 21:26
  • $\begingroup$ @NorbertSchuch Ohhhh man, I see my mistake now. Thanks for clarifying, I very much appreciate it. Just to be clear I didn't mean to mislead anyone nor did I think the book was wrong, I genuinely didn't understand the definition. Simple as that. But now you've answered my question. Many many thanks for this. Would you mind adding your comment as an answer so I can accept it as the answer? $\endgroup$ – Hugo Nava Kopp May 27 '16 at 10:39

You are misunderstanding (and thus misquoting) the book of Nielsen and Chuang. What they do is to first define the reduced density matrix $\rho_A$ of a bipartite state $\rho_{AB}$ as $$\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$ Then, they explain what a partial trace is: It is the linear map defined by its action $$ \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2| \right) = | a_1 \rangle \langle a_2 | \ \ \mathrm{tr} (|b_1 \rangle \langle b_2| )$$ on a (possibly overcomplete) basis $|a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2|$.

They never claim that the two equations above are equal, though.


Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of operators on a finite-dimensional Hilbert space $H_A$, then apply this property again to $A = H_A\otimes H_A^\ast$ to get a decomposition of such an operator in terms of bras and kets. Additionally, the equation you have written down doesn't make any sense - $\rho^A$ is supposed to be an operator on $A$, but $\langle a_1 \vert a_2 \rangle$ and $\operatorname{tr}(\lvert b_1 \rangle\langle b_2 \rvert)$ seem to be both numbers. You can find the proper definition of the partial trace in the Wikipedia article.

  • $\begingroup$ Yep, I messed up the notation on the rhs of the equation. I've fixed it now. $\endgroup$ – Hugo Nava Kopp May 26 '16 at 12:00
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    $\begingroup$ "You can find the proper definition of the partial trace in the Wikipedia article." -- Or in the book of Nielsen and Chuang, for that matter. $\endgroup$ – Norbert Schuch May 26 '16 at 21:32

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$.

Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for $p\not\in\{0,\,1\}$. But outer products of vectors are always singular, so no outer product will give you the density matrix.

Check this last statement carefully in the 2D case: the rows of $X\,Y^\dagger$ for column vectors $X$, $Y$ are all scaled versions of the row $Y^\dagger$ (the scaling factors are simply the components of $X$).

  • $\begingroup$ Yep, I messed up the last statement. I've fixed it now. $\endgroup$ – Hugo Nava Kopp May 26 '16 at 12:00

The answer to your question is NO.

The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$.

For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ can't be 0, since we need $xz=1$, and clearly $t$ can't be 0 since we need $yt=1$. Therefore there is no solution.

More abstractly, consider that the operator $|a\rangle\langle b|$ is the matrix that sends $b$ to $a$ when left-multiplying, but does nothing else (nothing else that isn't implied by linearity, to be technical). Any vectors perpendicular to $b$ get sent nowhere, and end up as 0. Some density operators don't send any vectors to 0, but an operator of the form $|a\rangle\langle b|$ can't achieve that even for a 2d space.


protected by Qmechanic May 26 '16 at 16:13

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