# Can mixed states be treated in the second quantization formalism? [closed]

In the first quantization formalism, mixed states can be handled using density matrices.

When treating many-body quantum systems however, the second quantization formalism often comes handier, allowing for example to automatically take into account the symmetry of the particles.

Is it possible to use density matrices in second quantization? In other words, is there a way to handle mixed states in a sort-of-generalized second quantization framework?

To shed light on what exactly I'm asking, let's consider the Hong-Ou-Mandel effect. We assume that the two photons are both in the same vertical polarization state (as generated for example by a type I SPDC process). In the first quantization formalism we can write this state as: $$\frac{1}{\sqrt 2} \left( | 0,V,1,V\rangle + | 1,V,0,V\rangle \right), \tag{1}$$ where for example $|0,V,1,V\rangle$ represents the state with the first photon in the spatial mode 0 and vertical polarization mode, and the second photon in the spatial mode 1 and vertical polarization mode.

Let's now suppose that, due to some noise added to the second spatial channel, the two photons are not completely indistinguishable when they enter the symmetric beam splitter. As a toy example, we model this noisy channel saying that the photon in the second spatial channel is actually in the mixed state $$\rho_2 = \frac{1}{2}\left[p|V\rangle \langle V | + (1-p)|H\rangle\langle H |\right] = \frac{1}{2} \begin{pmatrix} 1-p&0\\0&p \end{pmatrix}.$$ Equivalently, I can say that the noisy channel acts on the basis states as: $$| 1, H \rangle \rightarrow | 1, H \rangle, \qquad\qquad |1,V\rangle\rightarrow|1,V\rangle \\ |2,H\rangle \rightarrow \sqrt p |2,H\rangle + e^{i \theta} \sqrt{1-p} |2,V\rangle, \qquad\qquad |2,V\rangle \rightarrow \sqrt p |2,V\rangle + e^{i \theta} \sqrt{1-p} |2,H\rangle,$$ where $\theta$ is a randomly changing phase. Applying this transformations to (1) we get the state (I put this automatically generated image to avoid typos in the latex typesetting):

It is now straightforward (though a lenghty calculation) to write the evolution of this state through the symmetric beam splitter, that is, through the evolution described by the unitary matrix $$\mathcal U_{\operatorname{BS}} \otimes \boldsymbol 1 \otimes \mathcal U_{\operatorname{BS}} \otimes \boldsymbol 1, \qquad \text{where } \qquad \mathcal U_{\operatorname{BS}} \equiv \frac{1}{\sqrt 2} \begin{pmatrix}1&1\\1&-1\end{pmatrix},\,\, \boldsymbol 1 \equiv \begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

The resulting final state is quite aweful, and it is probably not particularly useful to write here. The point is that with this final state we can easily get the corresponding density matrix $\rho_f$. This is the resulting density matrix (I hope it will be readable enough):

Being the phases $\theta$ randomly varying, we can just set to 0 the terms containing phase factors of the form $e^{i \theta}$, obtaining as a result the mixed output state we were looking for:

This density matrix is the result I was looking for, but now you can hopefully better see my point: to get this result I had to pass through the bra-ket formalism to properly symmetrize the (bosonic) states, and then use the "trick" of adding a randomly changing phase $\theta$, which allows me to do the calculations treating all states as pure and only at the last moment set the $e^{i \theta}$ factors to zero, thus properly accounting for the incoherent nature of the considered noise.

If it was possible to somehow treat density matrices in a second-quantization-like formalism, one could (possibly?) do the entire calculation in the density matrix formalism, starting from an initial state that would be written as something like $$\left[ \begin{pmatrix}1&0\\0&0\end{pmatrix} \otimes \begin{pmatrix}1&0\\0&1\end{pmatrix} \right] \otimes_S \left[ \begin{pmatrix}0&0\\0&1\end{pmatrix} \otimes \begin{pmatrix}1-p&0\\0&p\end{pmatrix} \right],$$ where $\otimes_S$ is some sort of "symmetrized tensor product", which automatically allows to take into account the fact that the two photons are indistinguishable (or to be exact, it does not make sense to ask "which of the two photons" went on the second spatial mode and which went in the first spatial mode). The question is then: how can this be done?

Note: this is not a duplicate of this apparently similar question: my question is specifically with regard to the problem of describing mixed states in second quantization (if that is even possible), which is specifically not covered in the linked question.

## closed as too broad by Norbert Schuch, Sebastian Riese, user36790, ACuriousMind♦, GertMar 2 '16 at 1:51

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• – glS Mar 1 '16 at 19:22
• @glS I don't mean to intrude, but came up on this ref., thought it might be useful to you: Quantum optics of the beam-splitter, quantum.physik.uni-potsdam.de/teaching/ss2013/qo2/…. Pedagogical intro on how to treat a beam splitter in second quantization. – udrv Mar 1 '16 at 20:43

We write the state of two indistinguishable photons (bosons) in the modes $0$ and $1$, in the second quantization formalism, as $$\lvert \psi \rangle = a_0^\dagger a_1^\dagger \lvert \text{vac} \rangle.$$ This immediately generalizes to the density matrix formalism, in which this case reads $$\rho = a_0^\dagger a_1^\dagger \lvert \text{vac} \rangle \langle \text{vac} \rvert a_0 a_1.$$ Taking the matrix elements of this operator correctly reproduces the density matrix $$\rho = \frac{1}{2}\begin{pmatrix}0&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0&0\end{pmatrix}.$$ Applying this to the calculation shown in the question, the density matrix of the initial state becomes $$\rho = a_{0,V}^\dagger a_{1,V}^\dagger \lvert \text{vac} \rangle \langle \text{vac} \rvert a_{0,V} a_{1,V},$$ while the state after the addition of incoherent noise is: $$\rho = p \,\, a_{0,V}^\dagger a_{1,V}^\dagger \lvert \text{vac} \rangle \langle \text{vac} \rvert a_{0,V} a_{1,V} + (1-p) \,\, a_{0,V}^\dagger a_{1,H}^\dagger \lvert \text{vac} \rangle \langle \text{vac} \rvert a_{0,V} a_{1,H}.$$ The evolution of this state is now straightforward, as every creation(destruction) operator evolves according to the usual evolution rules.
More generally, a mixed state $\sigma$ with elements $\sigma_{i,j}$ will have in "second quantization" the form: $$\hat \sigma = \sum_{i,j} \sigma_{i,j} \,\,\, \hat a_i^\dagger \lvert \text{vac} \rangle\langle \text{vec} \rvert \hat a_j.$$