# The effect of Quantum Decoherence on density operators

Suppose we have a qubit in state $| \Psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle$

Suppose we expose this to decoherence, which we will express as the state $| R \rangle$ such that $$| 0 \rangle| R \rangle \rightarrow | 0 \rangle| R_0 \rangle$$ $$| 1 \rangle| R \rangle \rightarrow | 1 \rangle| R_1 \rangle$$

Where $| R \rangle, | R_0 \rangle$ and $| R_1 \rangle$ are all normalised states.

I'm trying to work how the density operator of the qubit changes when we apply $R$. If we consider the density operator for $|\Psi \rangle$ $$\rho=|\Psi \rangle \langle\Psi |= \Big( \begin{matrix} \alpha^2 & \alpha \beta \\ \alpha \beta & \beta^2 \end{matrix} \Big)$$

Assuming alpha and beta are real.

Next, we apply $|R\rangle$ to out qubit, $$\rho=|\Psi \rangle \langle\Psi | \rightarrow |\Psi \rangle |R\rangle \langle\Psi | \langle R|$$ $$=\alpha^2 |0 \rangle |R_0\rangle \langle 0 | \langle R_0|+\alpha\beta( |0 \rangle |R_0\rangle \langle 1 | \langle R_1|+ |1 \rangle |R_1\rangle \langle 0 | \langle R_0| )+\beta^2 |1 \rangle |R_1\rangle \langle 1 | \langle R_1|$$

Next would we take the reduced density operator of $|\Psi \rangle$ to find it's density operator? This would produce a $2 \times 2$ density operator, which I'm hoping to express in terms of $\rho=|\Psi \rangle \langle\Psi |$

Is this the correct way to think about the relation between density operators and reduced density operators for entangled states? What about unentangled states?

Also would anyone be able to briefly explain what decoherence is and why we can describe it as such a operator $|R \rangle$? Any help would be greatly appreciated.

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Presumably you mean $|1\rangle|R\rangle\rightarrow|1\rangle|R_1\rangle$ at the start. Otherwise your maths is inconsistent, and the reduced density operator will simply be $|0\rangle\langle0|$. –  Emilio Pisanty Mar 2 '13 at 16:18
@EmilioPisanty: Thanks for pointing that out! –  Freeman Mar 2 '13 at 16:20

First, let me restore the complex conjugation that was omitted without a good reason: $$\rho_{\rm pure}=|\Psi \rangle \langle\Psi |= \Big( \begin{matrix} |\alpha|^2 & \alpha^* \beta \\ \alpha \beta^* & |\beta|^2 \end{matrix} \Big)$$ Now, let us use a prettier (inverse) ordering of the tensor factors for the bra vectors. You meant: $$\rho=|\Psi \rangle \langle\Psi | \rightarrow |\Psi \rangle |R\rangle \langle R | \langle \Psi|$$ $$=|\alpha|^2 |0 \rangle |R_0\rangle \langle R_0 | \langle 0|+\alpha\beta^* |0 \rangle |R_0\rangle \langle R_1 | \langle 1|+\\ +\alpha^*\beta|1 \rangle |R_1\rangle \langle R_0 | \langle 0| +|\beta|^2 |1 \rangle |R_1\rangle \langle R_1 | \langle 1|$$ I think that the only extra step, key step of decoherence, you want to be shown is the partial trace of $\rho$ over the $R_0/R_1$ degree of freedom. We have $$\mathop{\rm Tr}^{R_0,R_1} \rho = |\alpha|^2 |0\rangle \langle 0| +|\beta|^2 |1\rangle \langle 1 | = \pmatrix{ |\alpha|^2&0\\ 0&|\beta|^2}$$ That's it. Note that the mixed terms didn't contribute anything to the partial trace because $\langle R_0|R_1\rangle=0$ and similarly for its complex conjugate. The main result is that this reduced density matrix, a partial trace, after decoherence has vanishing off-diagonal entries, unlike the density matrix for the pure state $\rho_{\rm pure}$ that we started with.

In more detailed calculations of decoherence, we usually take into account the fact that $|R_0\rangle$ and $|R_1\rangle$ that the environmental degrees of freedom evolve into just a moment later aren't exactly orthogonal to one another. That means that the off-diagonal elements get reduced but they don't quite vanish instantly. However, the inner product is decreasing faster than exponentially as the same information is being imprinted and copied into additional degrees of freedom of the environment. By an exponentially growing avalanche, $\exp(CT)$ of qubits get modified according to the initial decohering bits. Each of these environmental degrees of freedom or qubits contributes a factor of order $I\ll 1$ from the inner product to the off-diagonal entries so the off-diagonal entries go like $$\rho_{12}\sim I^{\exp(CT)}=\exp(-B\exp(CT))$$ where $B=\ln(1/I)$.

To make the off-diagonal entries of the density matrix vanish or almost vanish after some time, the entanglement with the environmental degrees of freedom is essential. If the two subsystems were not entangled, in other words, if the total pure state were a tensor product $|a\rangle\otimes |R\rangle$, the tracing over the $R$ degrees of freedom would give you the pure density matrix $|a\rangle\langle a|$ back: the system $R$ would have no effect on the system $a$ because of the lack of entanglement (lack of correlation).

No, you're wrong. Decoherence makes the off-diagonal entries decrease expo-exponentially in time and I have explained why. It is not a reasonable approximation to assume that the number of degrees of freedom in the environment is bounded. The environment is a sort of an infinite heat bath. At any rate, when the off-diagonal entries begin to substantially drop, e.g. to one millionth, it's well before all the degrees of freedom in the environment are "depleted" which is why the finiteness of their number, even if the number is finite, doesn't affect the functional character of $\rho_{12}(T)$. –  Luboš Motl Mar 2 '13 at 17:52
Your first paper is not doubly exponential, it is $e^{-ct^2}$, and the authors are explicitly directly coupling the electron to infinitely many degrees of freedom (of different frequencies). –  Peter Shor Mar 2 '13 at 18:42