# How should I think about decoherence?

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Malamala
Hello! Assume we have a 2 level system that is in a superposition of its 2 states:

$$\psi = \frac{|0>+|1>}{\sqrt(2)}$$

If we put it in a chamber at a given temperature (say that the chamber has perfect vacuum) the system will lose its coherence due to the blackbody radiation from the chamber walls. So the density matrix will go (after a long amount of time) as:

$$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

I understand that decoherence happens as the informations leaks from the system in the environment. But I am not sure how do the blackbody photons learn about the superposition state of the system. I can imagine that a photon interacting with the 2 level system would learn about its position or momentum, but I am not sure what physical interaction would lead to the photon learning about the superposition state. Is the position/momentum related to the internal state of the system? Thank you!

Mentor
Moderator's note: Thread moved to quantum forum.

Mentor
Assume we have a 2 level system that is in a superposition of its 2 states
This state only appears to be a superposition because of the basis you chose. You could equally well choose a basis in which ##\psi## is one of the basis states (the other would be ##\left( \ket{0} - \ket{1} \right) / \sqrt{2}##, in which case this state would not be a superposition.

If we put it in a chamber at a given temperature (say that the chamber has perfect vacuum) the system will lose its coherence due to the blackbody radiation from the chamber walls.
This is true, but it has nothing whatever to do with whether or not the original state is a "superposition" (since that, as shown above, is basis dependent anyway, and no actual physics can be basis dependent).

The decoherence has to do with the interaction of the system with the black body radiation. That interaction involves many more degrees of freedom in addition to the single degree of freedom of the quantum state of your two-level system. So writing things in terms of just your two-level system cannot tell you anything about the interaction or decoherence.

the density matrix
After interaction with the black body radiation, the density matrix is no longer pure, it's mixed (because the system is entangled with the radiation and has no definite pure state of its own; you can only derive its density matrix by tracing over the radiation states). So while what you say about how the density matrix is formally correct, it leaves out an essential element, the change from a pure to a mixed density matrix due to interaction with the radiation.

I am not sure how do the blackbody photons learn about the superposition state of the system
They don't. They interact with the system and become entangled with it. Decoherence happens because there are many, many possible black body radiation degrees of freedom that could interact with the two-level system, and it is uncertain which one actually will, so the joint state of system + radiation becomes an entangled state with amplitudes for many different possible interactions, which cannot be kept track of. None of this, as above, has anything to do with whether the original state of the two-level system was a superposition in your chosen basis. (Note that the interaction with the black body radiation would be substantially the same, and would still cause decoherence, if the original state of the two-level system were ##\ket{0}##, or ##\ket{1}##, or any other linear combination of them besides the one you wrote down--in short, any state whatever.)

PeroK, Spinnor and Twigg
Gold Member
I highly, highly recommend checking out John Preskill's lecture notes. It's a bit of a slog if you're not familiar with the concepts of quantum information theorists, but it's very much worthwhile. Specifically, check out chapter 3, section 3.4.2 (page 27). Your two level system corresponds to the ket ##|\psi \rangle_A## in Preskill's notation, which interacts with an "environment" of blackbody photons which have their own quantum state ##|\phi \rangle_E##. The treats the interaction as a quantum channel described by \begin{align*} |0 \rangle_A \otimes |0 \rangle_E &\rightarrow \sqrt{1-p} |0\rangle_A \otimes | 0 \rangle_E + \sqrt{p} |0 \rangle_A \otimes |1 \rangle_E \\ |1 \rangle_A \otimes |0 \rangle_E &\rightarrow \sqrt{1-p} |1\rangle_A \otimes | 0 \rangle_E + \sqrt{p} |1 \rangle_A \otimes |2 \rangle_E \end{align*}
What does this mean? The qubit and BB photon are initially unentangled, and then they interact with probability p. 1-p of the time, they do not interact and they stay in the same state as before. However, with probability p the photon will record what the initial state of the qubit was (without changing the qubit state). Preskill mathematically derives the evolution that you showed, namely that the off-diagonal elements of the density matrix (i.e., the coherences) shrink.

Edit: Sorry, I was missing a piece to this post:

but I am not sure what physical interaction would lead to the photon learning about the superposition state.

This depends on the details of your two-level qubit. Say it's a spin-1/2 particle. Then the BB photons interact with the qubit by a magnetic dipole interaction: $$\cal{H} = -\mathbf{\mu} \cdot \mathbf{B}$$
How does this interaction look like Preskill's quantum channel? We assume the interaction has a weak Rabi rate (low probability of changing the qubit state). In this case, the main effect of the interaction is that the photons pick up a phase that depends on the qubit's spin. You can interpret this as an interaction between the photon's polarization and the electron's spin. If the spin is parallel to the photons' magnetic field, the photons accrue less phase from the interaction (lower energy -> less phase), and vice versa. This phase accrued corresponds to the environment wavefunction ##|\phi\rangle_E## in Preskill's formalism.

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Delta2 and vanhees71
Malamala
@PeterDonis @Twigg thanks a lot! This make more sense now (also I will look into those lecture notes). One other thing that confuses me now, as @PeterDonis said, my notion of superposition in the initial post was related to a choice of basis so it was not physical. However, if I set my qubit in the ##\frac{|0>+|1>}{\sqrt{2}}## state, it will not keep its coherence and will end up being in ##|0>## or ##|1>## with 50% classical probability. However, if I use the ##\frac{|0>\pm|1>}{\sqrt{2}}## as the basis and set my qubit in the ##|0>## state, the qubit won't decohere to the ##\frac{|0>\pm|1>}{\sqrt{2}}## with 50% probability. It will stay in the ##|0>## state i.e. if I start with my qubit in the ground state and let it in a thermal bath for a long time, and measure its state it will still be in the ground state. Why is this the case if the choice of basis doesn't matter (assume that the lifetime of both states are infinite)?

Mentor
Why is this the case if the choice of basis doesn't matter
If ##\ket{0}## is the unique ground state of the system, then that is a basis independent property. It is that property, not anything about the choice of basis, that is actually behind the phenomena you describe.

However, the phenomena you describe are only correct if there is a unique ground state of the system. If there isn't (for example, a qubit with no external fields, so the entire state space is degenerate), then your reasoning about which states will or will not remain unchanged under time evolution (you say "decohere" but that's the wrong term; what you are talking about is unitary time evolution without any measurement and hence without any decoherence) is not correct. For a qubit with no external fields, so the entire state space is degenerate, every qubit state is an eigenstate of the free (non-interacting) Hamiltonian and is unchanged by time evolution.

Also, the term "choice of basis" is ambiguous. Earlier, when I said your choice of basis can't affect the physics, I was using "choice of basis" to mean the basis you choose to write down the wave function. But "choice of basis" can also mean the basis you choose to make an actual measurement (for example, which specific orientation you choose to measure the spin of your qubit). That choice of basis obviously does affect the physics.

(assume that the lifetime of both states are infinite)?
You can't assume the lifetime of both states is infinite if they have different energies. If ##\ket{0}## is a ground state and ##\ket{1}## is an excited state, ##\ket{1}## must have a finite lifetime.

PeroK and vanhees71
Gold Member
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In the Schrödinger picture of time evolution the state ket ##|\psi \rangle## (i.e., a pure state, ##\hat{\rho}=|\psi \rangle \langle \psi|##) is time-independent if and only if it is an energy eigenstate of a Hamiltonian that is not explicitly time-dependent.

Gold Member
Sorry for bringing this thread back from the dead. I've been on a trip and wanted to get back to this.

However, the phenomena you describe are only correct if there is a unique ground state of the system. If there isn't (for example, a qubit with no external fields, so the entire state space is degenerate), then your reasoning about which states will or will not remain unchanged under time evolution (you say "decohere" but that's the wrong term; what you are talking about is unitary time evolution without any measurement and hence without any decoherence) is not correct.

I think this point is valid, but only describes half the story. For the interaction of a qubit with a blackbody photon, there are two loss channels. One is the depolarizing channel (see Preskill's notes Chapter 3, section 3.4.1, on page 24), and the other is the dephasing channel (section 3.4.2, also what I described in post #4). Only in the depolarizing channel do you need the qubit states to be non-degenerate. My understanding is that you can have decoherence of degenerate qubit states by dephasing.

Of course, if you have a qubit where all the lifetimes are zero, then all the transitions must be forbidden and the blackbody photons won't interact at all with the qubit. However, most real-world qubits have meta-stable information-bearing states ("qubit states", which can in principle be degenerate) and have other dipole-allowed transitions that are used for state preparation and readout (and thus also make the qubit states vulnerable to interactions with blackbody photons).

Usually, qubit states aren't degenerate, but this is a technical convenience. It's easier to select a state for readout with high fidelity by its energy than it is to select a state by, for example, its Zeman m-value.

Mentor
Usually, qubit states aren't degenerate
In the absence of external fields, they are. But of course in the absence of external fields you can't really measure anything about a qubit.

Twigg
Gold Member
as:

(1111)→(1001)

I understand that decoherence happens as the informations leaks from the system in the environment. But I am not sure how do the blackbody photons learn about the superposition state of the system. I can imagine that a photon interacting with the 2 level system would learn about its position or momentum, but I am not sure what physical interaction would lead to the photon learning about the superposition state. I
I am reading this thread to learn the issue. It is just a beginner's saying.
LHS matrix with normalization has secular equation of
$$(\lambda-1/2)^2- 1/4 =0$$
and
$$\lambda = 1,0$$
that confirms it is a pure sate.
As for RHS matrix
$$\lambda = 1/2,1/2$$
that confirms it is a mixed state.
Transformation of matrix without conserving eigenvalues is unusual to me and it shows the difficulty for me to grasp the mathematics.

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2022 Award
In the absence of external fields, they are. But of course in the absence of external fields you can't really measure anything about a qubit.
If you have a single spin with spin quantum number ##s##, the eigenspaces of ##\hat{s}_3## are all one-dimenional, i.e., they're are all non-degenerate. That's so by definition, because a space with fixed spin quantum number ##s## is defined by an irreducible representation of SU(2).

Mentor
If you have a single spin with spin quantum number ##s##, the eigenspaces of ##\hat{s}_3## are all one-dimenional, i.e., they're are all non-degenerate.
A qubit, which is what we are talking about, has ##s = 1/2##, so there is only one eigenspace of ##\hat{s}^2##, with two eigenvectors in it, representing the two possible results of a measurement of ##\hat{s}^3##. In the absence of an external field, both of these eigenstates have the same energy, hence they are degenerate.

Gold Member
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Well, yes. But the ##\hat{s}_3## eigenvalues are non-degenerate. Of course the energy eigenstates of a free particle ##\hat{H}=\hat{\vec{p}}^2/(2m)## are degenerate. To make them unique you need to specify 4 compatible observables, e.g.,##\vec{p}## and ##\hat{\sigma}_3## (which are energy eigenvectors with eigenvalues ##E=\vec{p}^2/(2m)##) or ##\hat{H}##, ##\hat{\vec{l}}^2##, ##\hat{l}_3##, and ##\hat{s}_3##.

Einstein44
Mentor
But the ##\hat{s}_3## eigenvalues are non-degenerate.
Are they? If I have a qubit that is not interacting with anything else (no external fields, etc.), then you're saying ##\hat{s}_3## up has a different energy from ##\hat{s}_3## down? Why would it?

Gold Member
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I guess we discuss different things here. If you only look at the spin degrees of freedom, i.e., ##\vec{s}## and you have a fixed spin-quantum number ##s=1/2## (or any other ##s \in \{0,1/2,1,\ldots \}##) then the eigenvalues of ##\hat{s}_3##, ##m_s \in \{-s,-s+1,\ldots,s-1,s \}## are all non-degenerate, i.e., the eigenspaces are one-dimensional.

If you of course consider the full discription of the particle, with the operator algebra spanned by ##\hat{\vec{x}}##, ##\hat{\vec{p}}##, and ##\hat{\vec{s}}##. You need always four independent compatible observables to determine the common eigenvectors uniquely, i.e., making the common eigenspaces for each set of eigenvalues one-dimensional. If you look for energy eigenvalues you need besides ##\hat{H}## three other compatible observables. For the free particle you can choose ##\hat{\vec{p}}## and ##\hat{s}_3## (which are a CONS of energy eigenvectors with eigenvalues ##E_{\vec{p}}=\vec{p}^2/(2m)##) or ##\hat{H}##, ##\hat{\vec{L}}^2##, ##\hat{L}_3##, and ##\hat{s}_3##.

Mentor
I guess we discuss different things here.
Perhaps you are using "degenerate" to mean something different than "has different energy" (or more precisely "different eigenvalues of the Hamiltonian"). Can you answer the question I asked in post #14?

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I tried to answer this in #15. We have discussed simply different things. You considered the full description of a single particle in non-relativistic QM. I thought we were discussing only the spin, ignoring all other degrees of freedom.

Of course you are right. If you consider energy eigenvalues of a free particle with spin ##s \neq 0## the energy eigenvalues are all degenerate: For a particle with ##s=0##, all are degenerate except the ground state, which is uniquely determined by the common eigenvector of the three momentum components ##\hat{\vec{p}}##, ##p_1=p_2=p_3=0##; for ##s\neq 0## this ground state is ##2s+1##-fold degenerate corresponding to the possible eigenvalues of ##\hat{s}_3##, which can be diagonalized in addition to the ##\hat{\vec{p}}##. All other energy eigenstates are degenerate since for each ##E>0## the momentum eigenvectors with ##|\vec{p}|=\sqrt{2mE}## are eigenvalues of ##\hat{H}## with the same eigenvalue.

In general, "degneracy" of eigenstates is not restricted to energy. If you have an operator, a eigenvalue is called "degenerate" if the eigenspace related to this eigenvalue is not one-dimensional.

Mentor
I thought we were discussing only the spin, ignoring all other degrees of freedom.
The spin degrees of freedom were the only ones I asked about. The question I asked in post #14 remains the same whether you treat the qubit as having other degrees of freedom (such as the configuration space position-momentum ones) or not, since in the absence of any interactions the spin degrees of freedom are completely uncoupled from any others.

In general, "degneracy" of eigenstates is not restricted to energy. If you have an operator, a eigenvalue is called "degenerate" if the eigenspace related to this eigenvalue is not one-dimensional.
Yes, agreed. For example, the energy eigenvalues of the (non-interacting) qubit are degenerate, but the spin eigenvalues are not ("up" and "down" are two different eigenvalues for ##\hat{s}_3##).

vanhees71