# The Choi-Jamiolkowski Isomorphism: You’re Doing It Wrong!

As the dear departed Quantum Pontiff used to say: New Paper Dance! I am pretty happy that this one has finally been posted because it is my first arXiv paper since I returned to work, and also because it has gone through more rewrites than Spiderman: The Musical.

What is the paper about, I hear you ask? Well, mathematically, it is about an extremely simple linear algebra trick called the Choi-Jamiolkwoski isomorphism. This is actually two different results: the Choi isomorphism and the Jamiolkowski isomorphism, but people have a habit of lumping them together. This trick is so extremely well-known to quantum information theorists that it is not even funny. One of the main points of the paper is that you should think about what the isomorphism means physically in a new way. Hence the “you’re doing it wrong” in the post title.

### First Level Isomorphisms

For the uninitiated, here is the simplest way of describing the Choi isomorphism in a single equation:
$\Ket{j}\Bra{k} \qquad \qquad \equiv \qquad \qquad \Ket{j} \otimes \Ket{k},$
i.e. the ismomorphism works by turning a bra into a ket. The thing on the left is an operator on a Hilbert space $\mathcal{H}$ and the thing on the right is a vector in $\mathcal{H} \otimes \mathcal{H}$, so the isomorphism says that $\mathcal{L}(\mathcal{H}) \equiv \mathcal{H} \otimes \mathcal{H}$, where $\mathcal{L}(\mathcal{H})$ is the space of linear operators on $\mathcal{H}$.

Here is how it works in general. If you have an operator $U$ then you can pick a basis for $\mathcal{H}$ and write $U$ in this basis as
$U = \sum_{j,k} U_{j,k} \Ket{j}\Bra{k},$
where $U_{j,k} = \Bra{j}U\Ket{k}$. Then you just extend the above construction by linearity and write down a vector
$\Ket{\Phi_U} = \sum_{j,k} U_{j,k} \Ket{j} \otimes \Ket{k}.$
It is pretty obvious that we can go in the other direction as well, starting with a vector on $\mathcal{H}\otimes\mathcal{H}$, we can write it out in a product basis, turn the second ket into a bra, and then we have an operator.

So far, this is all pretty trivial linear algebra, but when we think about what this means physically it is pretty weird. One of the things that is represented by an operator in quantum theory is dynamics, in particular a unitary operator represents the dynamics of a closed system for a discrete time-step. One of the things that is represented by a vector on a tensor product Hilbert space is a pure state of a bipartite system. It is fairly easy to see that (up to normalization) unitary operators get mapped to maximally entangled states under the isomorphism, so, in some sense, a maximally entangled state is “the same thing” as a unitary operator. This is weird because there are some things that make sense for dynamical operators that don’t seem to make sense for states and vice-versa. For example, dynamics can be composed. If $U$ represents the dynamics from $t_0$ to $t_1$ and $V$ represents the dynamics from $t_1$ to $t_2$, then the dynamics from $t_0$ to $t_2$ is represented by the product $VU$. Using the isomorphism, we can define a composition for states, but what on earth does this mean?

Before getting on to that, let us briefly pause to consider the Jamiolkowski version of the isomorphism. The Choi isomorphism is basis dependent. You get a slightly different state if you write down the operator in a different basis. To make things basis independent, we replace $\mathcal{H}\otimes\mathcal{H}$ by $\mathcal{H}\otimes\mathcal{H}^*$. $\mathcal{H}^*$ denotes the dual space to $\mathcal{H}$, i.e. it is the space of bras instead of the space of kets. In Dirac notation, the Jamiolkwoski isomorphism looks pretty trivial. It says
$\Ket{j}\Bra{k} \qquad \qquad \equiv \qquad \qquad \Ket{j} \otimes \Bra{k}.$
This is axiomatic in Dirac notation, because we always assume that tensor product symbols can be omitted without changing anything. However, this version of the isomorphism is going to become important later.

### Conventional Interpretation: Gate Teleportation

In quantum information, the Choi isomorphism is usually interpreted in terms of “gate teleportation”. To understand this, we first reformulate the isomorphism slightly. Let $\Ket{\Phi^+}_{AA’} = \sum_j \Ket{jj}_{AA’}$, where $A$ and $A’$ are quantum systems with Hilbert spaces of the same dimension. The vectors $\Ket{j}$ form a preferred basis, and this is the basis in which the Choi isomorphism is going to be defined. Note that $\Ket{\Phi^+}_{AA’}$ is an (unnormalized) maximally entangled state. It is easy to check that the isomorphism can now be reformulated as
$\Ket{\Phi_U}_{AA’} = I_A \otimes U_A’ \Ket{\Phi^+}_{AA’},$
where $I_A$ is the identity operator on system $A$. The reverse direction of the isomorphism is given by
$U_A \Ket{\psi}\Bra{\psi}_A U_A^{\dagger} = \Bra{\Phi^+}_{A’A”} \left ( \Ket{\psi}\Bra{\psi}_{A”} \otimes \Ket{\Phi_U}\Bra{\Phi_U}_{A’A} \right )\Ket{\Phi^+}_{A’A”},$
where $A^{\prime\prime}$ is yet another quantum system with the same Hilbert space as $A$.

Now let’s think about the physical interpretation of the reverse direction of the isomorphism. Suppose that $U$ is the identity. In that case, $\Ket{\Phi_U} = \Ket{\Phi^+}$ and the reverse direction of the isomorphism is easily recognized as the expression for the output of the teleportation protocol when the $\Ket{\Phi^+}$ outcome is obtained in the Bell measurement. It says that $\Ket{\psi}$ gets teleported from $A^{\prime\prime}$ to $A$. Of course, this outcome only occurs some of the time, with probability $1/d$, where $d$ is the dimension of the Hilbert space of $A$, a fact that is obscured by our decision to use an unnormalized version of $\Ket{\Phi^+}$.

Now, if we let $U$ be a nontrivial unitary operator then the reverse direction of the isomorphism says something more interesting. If we use the state $\Ket{\Phi_U}$ rather than $\Ket{\Phi^+}$ as our resource state in the teleportation protocol, then, upon obtaining the $\Ket{\Phi^+}$ outcome in the Bell measurement, the output of the protocol will not simply be the input state $\Ket{\psi}$, but it will be that state with the unitary $U$ applied to it. This is called “gate teleportation”. It has many uses in quantum computing. For example, in linear optics implementations, it is impossible to perform every gate in a universal set with 100% probability. To avoid damaging your precious computational state, you can apply the indeterministic gates to half of a maximally entangled state and keep doing so until you get one that succeeds. Then you can teleport your computational state using the resulting state as a resource and end up applying the gate that you wanted. This allows you to use indeterministic gates without having to restart the computation from the beginning every time one of these gates fails.

Using this interpretation of the isomorphism, we can also come up with a physical interpretation of the composition of two states. It is basically a generalization of entanglement swapping. If you take $\Ket{\Phi_U}$ and $\Ket{\Phi_{V}}$ and and perform a Bell measurement across the output system of the first and the input system of the second then, upon obtaining the $\Ket{\Phi^+}$ outcome, you will have the state $\Ket{\Phi_{UV}}$. In this way, you can perform your entire computational circuit in advance, before you have access to the input state, and then just teleport your input state into the output register as the final step.

In this way, the Choi isomorphism leads to a correspondence between a whole host of protocols involving gates and protocols involving entangled states. We can also define interesting properties of operations, such as the entanglement of an operation, in terms of the states that they correspond to. We then use the isomoprhism to give a physical meaning to these properties in terms of gate teleportation. However, one weak point of the correspondence is that it transforms something deterministic; the application of a unitary operation; into something indeterministic; getting the $\Ket{\Phi^+}$ outcome in a Bell measurement. Unlike the teleportation protocol, gate teleportation cannot be made deterministic by applying correction operations for the other outcomes, at least not if we want these corrections to be independent of $U$. The states you get for the other outcomes involve nasty things like $U^*, U^T, U^\dagger$ applied to $\Ket{\psi}$, depending on exactly how you construct the Bell basis, e.g. choice of phases. These can typically not be corrected without applying $U$. In particular, that would screw things up in the linear optics application wherein $U$ can only be implemented non-deterministically.

Before turning to our alternative interpretation of Choi-Jamiolkowski, let’s generalize things a bit.

### Second Level Isomorphisms

In quantum theory we don’t just have pure states, but also mixed states that arise if you have uncertainty about which state was prepared, or if you ignore a subsystem of a larger system that is in a pure state. These are described by positive, trace-one, operators, denoted $\rho$, called density operators. Similarly, dynamics does not have to be unitary. For example, we might bring in an extra system, interact them unitarily, and then trace out the extra system. These are described by Completely-Positive, Trace-Preserving (CPT) maps, denoted $\mathcal{E}$. These are linear maps that act on the space of operators, i.e. they are operators on the space of operators, and are often called superoperators.

Now, the set of operators on a Hilbert space is itself a Hilbert space with inner product $\left \langle N, M \right \rangle = \Tr{N^{\dagger}M}$. Thus, we can apply Choi-Jamiolkowski on this space to define a correspondence between superoperators and operators on the tensor product. We can do this in terms of an orthonormal operator basis with respect to the trace inner product, but it is easier to just give the teleportation version of the isomorphism. We will also generalize slightly to allow for the possibility that the input and output spaces of our CPT map may be different, i.e. it may involve discarding a subsystem of the system we started with, or bringing in extra ancillary systems.

Starting with a CPT map $\mathcal{E}_{B|A}: \mathcal{L}(\mathcal{H}_A) \rightarrow \mathcal{L}(\mathcal{H}_B)$ from system $A$ to system $B$, we can define an operator on $\mathcal{H}_A \otimes \mathcal{H}_B$ via
$\rho_{AB} = \mathcal{E}_{B|A’} \otimes \mathcal{I}_{A} \left ( \Ket{\Phi^+}\Bra{\Phi^+}_{AA’}\right ),$
where $\mathcal{I}_A$ is the identity superoperator. This is a positive operator, but it is not quite a density operator as it satisfies $\PTr{B}{\rho_{AB}} = I_A$, which implies that $\PTr{AB}{\rho_{AB}} = d$ rather than $\PTr{AB}{\rho_{AB}} = 1$. This is analogous to using unnormalized states in the pure-state case. The reverse direction of the isomorphism is then given by
$\mathcal{E}_{B|A} \left ( \sigma_A \right ) = \Bra{\Phi^+}_{A’A}\sigma_{A’} \otimes \rho_{AB}\Ket{\Phi^+}_{A’A}.$
This has the same interpretation in terms of gate teleportation (or rather CPT-map teleportation) as before.

The Jamiolkowski version of this isomorphism is given by
$\varrho_{AB} = \mathcal{E}_{B|A’} \otimes \mathcal{I}_{A} \left ( \Ket{\Phi^+}\Bra{\Phi^+}_{AA’}^{T_A}\right ),$
where $^T_A$ denotes the partial transpose in the basis used to define $\Ket{\Phi^+}$. Although it is not obvious from this formula, this operator is independent of the choice of basis, as $\Ket{\Phi^+}\Bra{\Phi^+}_{AA’}^{T_A}$ is actually the same operator for any choice of basis. I’ll keep the reverse direction of the isomorphism a secret for now, as it would give a strong hint towards the punchline of this blog post.

### Probability Theory

I now want to give an alternative way of thinking about the isomorphism, in particular the Jamiolkowski version, that is in many ways conceptually clearer than the gate teleportation interpretation. The starting point is the idea that quantum theory can be viewed as a noncommutative generalization of classical probability theory. This idea goes back at least to von Neumann, and is at the root of our thinking in quantum information theory, particularly in quantum Shannon theory. The basic idea of the generalization is that that probability distributions $P(X)$ get mapped to density operators $\rho_A$ and sums over variables become partial traces. Therefore, let’s start by thinking about whether there is a classical analog of the isomorphism, and, if so, what its interpretation is.

Suppose we have two random variables, $X$ and $Y$. We can define a conditional probability distribution of $Y$ given $X$, $P(Y|X)$, as a positive function of the two variables that satisfies $\sum_Y P(Y|X) = 1$ independently of the value of $X$. Given a conditional probability distribution and a marginal distribution, $P(X)$, for $X$, we can define a joint distribution via
$P(X,Y) = P(Y|X)P(X).$
Conversely, given a joint distribution $P(X,Y)$, we can find the marginal $P(X) = \sum_Y P(X,Y)$ and then define a conditional distribution
$P(Y|X) = \frac{P(X,Y)}{P(X)}.$
Note, I’m going to ignore the ambiguities in this formula that occur when $P(X)$ is zero for some values of $X$.

Now, suppose that $X$ and $Y$ are the input and output of a classical channel. I now want to think of the probability distribution of $Y$ as being determined by a stochastic map $\Gamma_{Y|X}$ from the space of probability distributions over $X$ to the space of probability distributions over $Y$. Since $P(Y) = \sum_{X} P(X,Y)$, this has to be given by
$P(Y) = \Gamma_{Y|X} \left ( P(X)\right ) = \sum_X P(Y|X) P(X),$
or
$\Gamma_{Y|X} \left ( \cdot \right ) = \sum_{X} P(Y|X) \left ( \cdot \right )$.

What we have here is a correspondence between a positive function of two variables — the conditional proabability distribution — and a linear map that acts on the space of probability distributions — the stochastic map. This looks analogous to the Choi-Jamiolkowski isomorphism, except that, instead of a joint probability distribution, which would be analogous to a quantum state, we have a conditional probability distribution. This suggests that we made a mistake in thinking of the operator in the Choi isomorphism as a state. Maybe it is something more like a conditional state.

### Conditional States

Let’s just plunge in and make a definition of a conditional state, and then see how it makes sense of the Jamiolkowski isomorphism. For two quantum systems, $A$ and $B$, a conditional state of $B$ given $A$ is defined to be a positive operator $\rho_{B|A}$ on $\mathcal{H}_A \otimes \mathcal{H}_B$ that satisfies
$\PTr{B}{\rho_{B|A}} = I_A.$
This is supposed to be analogous to the condition $\sum_Y P(Y|X) = 1$. Notice that this is exactly how the operators that are Choi-isomorphic to CPT maps are normalized.

Given a conditional state, $\rho_{B|A}$, and a reduced state $\rho_A$, I can define a joint state via
$\rho_{AB} = \sqrt{\rho_A} \rho_{B|A} \sqrt{\rho_A},$
where I have suppressed the implicit $\otimes I_B$ required to make the products well defined. The conjugation by the square root ensures that $\rho_{AB}$ is positive, and it is easy to check that $\PTr{AB}{\rho_{AB}} = 1$.

Conversely, given a joint state, I can find its reduced state $\rho_A = \PTr{B}{\rho_{AB}}$ and then define the conditional state
$\rho_{B|A} = \sqrt{\rho_A^{-1}} \rho_{AB} \sqrt{\rho_A^{-1}},$
where I am going to ignore cases in which $\rho_A$ has any zero eigenvalues so that the inverse is well-defined (this is no different from ignoring the division by zero in the classical case).

Now, suppose you are given $\rho_A$ and you want to know what $\rho_B$ should be. Is there a linear map that tells you how to do this, analogous to the stochastic map $\Gamma_{Y|X}$ in the classical case? The answer is obviously yes. We can define a map $\mathfrak{E}_{B|A}: \mathcal{L} \left ( \mathcal{H}_A\right ) \rightarrow \mathcal{L} \left ( \mathcal{H}_B\right )$ via
$\mathfrak{E}_{B|A} \left ( \rho_A \right ) = \PTr{A}{\rho_{B|A} \rho_A},$
where we have used the cyclic property of the trace to combine the $\sqrt{\rho_A}$ terms, or
$\mathfrak{E}_{B|A} \left ( \cdot \right ) = \PTr{A}{\rho_{B|A} (\cdot)}.$
The map $\mathfrak{E}_{B|A}$ so defined is just the Jamiolkowski isomorphic map to $\rho_{B|A}$ and the above equation gives the reverse direction of the Jamiolkowski isomorphism that I was being secretive about earlier.

The punchline is that the Choi-Jamiolkowski isomorphism should not be thought of as a mapping between quantum states and quantum operations, but rather as a mapping between conditional quantum states and quantum operations. It is no more surprising than the fact that classical stochastic maps are determined by conditional probability distributions. If you think of it in this way, then your approach to quantum information will become conceptually simpler a lot of ways. These ways are discussed in detail in the paper.

### Causal Conditional States

There is a subtlety that I have glossed over so far that I’d like to end with. The map $\mathfrak{E}_{B|A}$ is not actually completely positive, which is why I did not denote it $\mathcal{E}_{B|A}$, but when preceeded by a transpose on $A$ it defines a completely positive map. This is because the Jamiolkowski isomorphism is defined in terms of the partial transpose of the maximally entangled state. Also, so far I have been talking about two distinct quantum systems that exist at the same time, whereas in the classical case, I talked about the input and output of a classical channel. A quantum channel is given by a CPT map $\mathcal{E}_{B|A}$ and its Jamiolkowski representation would be
$\mathcal{E}_{B|A} \left (\rho_A \right ) = \PTr{A}{\varrho_{B|A}\rho_A},$
where $\varrho_{B|A}$ is the partial transpose over $A$ of a positive operator and it satisfies $\PTr{B}{\varrho_{B|A}} = I_A$. This is the appropriate notion of a conditional state in the causal scenario, where you are talking about the input and output of a quantum channel rather than two systems at the same time. The two types of conditional state are related by a partial transpose.

Despite this difference, a good deal of unification is achieved between the way in which acausally related (two subsystems) and causally related (input and output of channels) degrees of freedom are described in this framework. For example, we can define a “causal joint state” as
$\varrho_{AB} = \sqrt{\rho_A} \varrho_{B|A} \sqrt{\rho_A},$
where $\rho_A$ is the input state to the channel and $\varrho_{B|A}$ is the Jamiolkowski isomorphic map to the CPT map. This unification is another main theme of the paper, and allows a quantum version of Bayes’ theorem to be defined that is independent of the causal scenario.

### The Wonderful World of Conditional States

To end with, here is a list of some things that become conceptually simpler in the conditional states formalism developed in the paper:

• The Born rule, ensemble averaging, and quantum dynamics are all just instances of a quantum analog of the formula $P(Y) = \sum_X P(Y|X)P(X)$.
• The Heisenberg picture is just a quantum analog of $P(Z|X) = \sum_Y P(Z|Y)P(Y|X)$.
• The relationship between prediction and retrodiction (inferences about the past) in quantum theory is given by the quantum Bayes’ theorem.
• The formula for the set of states that a system can be ‘steered’ to by making measurements on a remote system, as in EPR-type experiments, is just an application of the quantum Bayes’ theorem.

If this has whet your appetite, then this and much more can be found in the paper.

### 21 responses to “The Choi-Jamiolkowski Isomorphism: You’re Doing It Wrong!”

1. Blake Stacey

I was very happy to see this paper appear on the arXiv last night, as after a first read-through, it looks like this might help me cook up a quantum generalization of something I’ve been working on in classical information theory. Nice blog post, too! I’m sure I’ll have questions once I get the chance to work through the algebra in more depth.

2. Cesar Rodriguez-Rosario

This interpretation of the isomorphism in terms of conditional states is great! The intuition you provide is very nice, I’m a fan. I’ll go read the paper now to learn more details.

3. Blake Stacey

I noticed a couple minor cosmetic points in section A of the Introduction. Corrections are in brackets:

Generally, a region will refer to a collection of elementary region[s].

and

e.g., the input and output spaces for a quantum channel are assigned different labels[.]

(I’ve spent so much time this year rewriting manuscripts that I can’t help but copy-edit. Sigh.)

4. Guillaume

Thanks for explaining the difference between Choi isomorphism and Jamiolkowski isomorphism in your paper, that was heplful to me !

5. marozols

Hi Matt. Thanks a lot for taking your time to write this post—it really clarified things to me. I was always puzzled by how the maximally entangled state and the Choi matrix is normalized, but now it all makes sense when I think of them as conditional probability distributions. I’m surprised that these kind of analogies are not widely known and part of standard curriculum, since they make the quantum concepts much more intuitive and presumably easier to learn.

Btw, I wonder if your remark about doing the whole computation before the input is supplied also applies classically.

6. mleifer

Thanks. I’m glad you appreciated it.

The remark about doing the whole computation before the input does apply classically, but it is pretty useless. What it corresponds to is choosing an input at random and then making a copy of it. The correlated state of the two copies then corresponds to the maximally entangled state. Then, you run the computation on one of the copies. When you want to perform the computation on your chosen input you then just check whether it is the same as the randomly chosen input, this being analogous to the Bell measurement. If it is then you accept the output of the computation in the second copy and if it isn’t then you reject it and start the whole process again. Obviously, this is a pretty useless procedure as the probability of correctly guessing the input goes down exponentially with the number of bits, but this is conceptually no different from what is going on in quantum gate teleportation.

7. Andrew Tan

I like the analogies you draw in this paper very much.

Often, one use of the word “epistemic” as in “psi-epistemic” is that an underlying ontic state is compatible with more than one wave function.

However, the analogies in this paper also seem “epistemic” in some sense, because of the analogy to Bayesian inference. Yet the formalism is presumably consistent with Bohmian mechanics, since it is just a reformulation of quantum mechanics.

So is the “epistemic” notion required to use the analogies in this paper less strict than “psi-epistemic”, in the sense that these analogies seem applicable to Bohmian mechanics, which is “psi-ontic”?

8. mleifer

The formalism does not contradict any viable interpretation of quantum theory, since it is just a reformulation of the mathematics of the theory. However, a reformulation can be suggestive, in the sense of casting doubt upon the assumptions used to arrive at a particular interpretation. In this case, the similarities between quantum states, operations, etc. and classical probabilities suggests that those objects in quantum theory should be interpreted in the same sort of way as their counterparts in classical probability theory. This casts doubt on the idea that the wavefunction should be treated as analagous to a real physical wave, as it is in Bohmian mechanics.

Regarding the terminology “epistemic”, it simply means something that refers to knowledge. Therefore, it is appropriate to call Bayesianism an epistemic interpretation of probability theory, and to use the same term for the quantum generalization. The usual distinction between psi-epistemic and psi-ontic models applies to a framework that makes a whole bunch of additional assumptions, e.g. realism, that classical probability operates at the ontic level, etc. That is an interesting framework for various reasons, but most psi-epistemicists would reject one or more of those assumptions, so they would be looking for a different type of evidence that quantum states represent knowledge. One may view the analogies presented in the paper as evidence that we should look for an ontology that supports the idea that a nonclassical probability theory operates at the ontic level. That would be my view, but alternatively, if you like Copenhagenish interpretations, you might view the analogies as further evidence in favour of that type of view.

9. Andrew Tan

Does the idea that the conditional states formalism (1) “suggests that those objects in quantum theory should be interpreted in the same sort of way as their counterparts in classical probability theory” have to conflict with (2) “the idea that the wavefunction should be treated as analogous to a real physical wave, as it is in Bohmian mechanics.”?

Let’s suppose (2) Bohmian mechanics in some version were found to be experimentally true, by finding deviations from quantum mechanics, consistent with quantum non-equilibrium. Would that necessarily be an experimental falsification of (1) the quantum-classical analogies suggested from the quantum conditional states formalism?

Or could I consistently believe in both (1) and (2) simultaneously? After all (2) feels a lot like kinetic theory while (1) seems like the canonical ensemble of classical statistical mechanics.

10. mleifer

Your proposal stikes me as a bit odd. If nonequilibrium Bohmian mechanics is shown to be true then the analogies of the conditional states formalism would only work in the equilibrium state. You want to argue that this is like the canonical state only applying in thermal equilibrium. Picking out a set of states as special is of course what happens in equilibrium. However, here we are dealing with a full-blown generalization of probability theory rather than just some preferred set of states within a given probability theory. It seems strange to me that a theory of Bayesian inference would somehow get “turned on” by the passage to equilibrium and be inapplicable otherwise. This is because, in my view, the theory of probabilistic inference is a “higher level” theory, more akin to logic than physics, so it seems to me that it should be determined prior to considerations of how systems equilibriate into special states. Of course, one can take other views about the nature of probability theory, in which case the proposal may not look so odd.

11. Andrew Tan

I’m not sure this is right, but it looks as if the conditional states formalism doesn’t contain standard probability. If so, then it would seem that neither standard probability nor the conditional states formalism is a necessary logic, before physics enters. Could it can be argued that standard probability and the conditional states formalism are each special cases of a more general structure (say, objects for which a de Finetti representation holds)?

Then the passage to equilibrium would “turn on” the appropriate version of the more general object, eg. conditional states for quantum equilibrium and standard probability for thermal equilibrium.

12. mleifer

The conditional states formalism does contain classical probability as a special case. You just need to make all the operators diagonal in a preferred product basis.

I don’t think your idea that quantum equilibrium leads to the conditional states formalism and then thermal equilibrium leads to classical probability works out. For one thing, in classical statistical mechanics, classical probability is perfectly adequate for handling systems out of thermal equilibrium. Secondly, even in quantum physics, thermal systems need not obey classical probability. For example, at very low temperatures, the system is very close to being in its ground state and for realistic Hamiltonians this state is entangled. Thermal states only look “classical” if you restrict yourself to only making energy measurements on the system, in which case it looks just like a classical Gibbs distribution, but those are not the only measurements we can make. Further, the energy basis is not the way in which classical probability is embedded in the conditional states formalism.

13. Andrew Tan

Yes, my idea doesn’t work for quantum thermal equilibrium. At the back of my mind I was thinking of some hierarchy like dBB -> quantum equilibrium = conditional states -> quantum thermal equilibrium -> classical thermal equilibrium = standard probability, but that leaves the aesthetic problem that from your point of view the conditional states formalism should come before physics.

When you say that standard probability applies to non-equilibrium situations, do you include the completely deterministic case of Newtonian mechanics for a single particle in a potential? (I’m trying to understand whether you mean that in classical mechanics, standard probability always works, and never has to get turned on.)

14. mleifer

“I’m trying to understand whether you mean that in classical mechanics, standard probability always works, and never has to get turned on.”

Yes, that’s what I mean. In Newtonian mechanics you can use whatever probability distribution on phase space you like and compute its evolution via the Liouville equation. It does not have to be the equilibrium case.

Also, the emergence of classicality and the passage to thermal equilibrium are somewhat independent of each other. Although it is the case that many examples of decohering enviroments also thermalize the system and vice versa, this is not always be true. It depends on the relative strengths of the system Hamiltonian and the system-environment interaction, and on whether the environment is itself in a thermal state. Therefore, you can have systems that behave classically without being in thermal equilibrium.

15. Andrew Tan

If standard probability applies to Newtonian mechanics via the Liouville equation, shouldn’t it apply also to nonequilibrium Bohmian mechanics, since it seems to be classical deterministic evolution from an initial condition (different phase space, and some analogue of the Liouville equation)?

If that’s possible, could one say that the conditional states formalism also applies to Bohmian nonequlilibrium, since the conditional states formalism contains standard probability?

16. mleifer

“If standard probability applies to Newtonian mechanics via the Liouville equation, shouldn’t it apply also to nonequilibrium Bohmian mechanics, since it seems to be classical deterministic evolution from an initial condition (different phase space, and some analogue of the Liouville equation)?”

Yes it does. One just puts an ordinary classical probability distribution over the particle configurations and derives the analog of the Liouville equation from the guidance equation. More generally, you could put a probability measure on the space of wavefunctions as well, but I’ve never seen anyone do that.

“If that’s possible, could one say that the conditional states formalism also applies to Bohmian nonequlilibrium, since the conditional states formalism contains standard probability?”

You really seem to want to stretch things so that Bohmian mechanics looks compatible with the conditional states formalism. Yes, in principle you could do this, but it is not a very natural thing to do from either perspective. In the conditional states formalism, classical probability applies if all the quantum states and conditional states involved are diagonal in a common product basis. Therefore, you would have to imagine that there were two quantum states associated with a quantum system. The first is the usual Bohmian wavefunction, but then you also need a state that is diagonal in a product basis to represent the classical probabilities over the particle configurations. Even in equilibrium you would still have two quantum states.

Now, one of the virtues of the Bohmian interpretation is that it does not require a modification of classical probability theory. The probabilities within it are just ordinary classical probabilities and can be interpreted in any of the standard ways (e.g. Bayesian, frequentist, etc.). Therefore, for a Bohmian it is very bizarre to introduce an exotic type of probability where none is needed. On the other hand, from the conditional states point of view, quantum states should be thought of as generalized probabilities. In your reconciliation, you have one quantum state that isn’t to be interpreted in that way, i.e. the usual Bohmian wavefunction that guides the particle trajectories, and another that is interpreted that way, but however is always in the classical special form. Thus, the interpretation of a quantum state as a generalized probability is playing absolutely no role.

Ultimately, if you have two viable accounts of quantum theory then it will always be possible to munge them together in some way in order to make them look consistent because, at the end of the day, both accounts must make the exact same predictions. However, if the two accounts are based on opposite ideas about the ontology underlying quantum theory then the union will likely be a kludge and I fail to see the point in pursuing it.

17. Andrew Tan

“You really seem to want to stretch things so that Bohmian mechanics looks compatible with the conditional states formalism.”

Yes, because the conditional states formalism and Bohmian mechanics both seem very nice and natural to me, so I was wondering if it’s possible to eat my cake and have it! Thanks for the lengthy discussion! I think I understand the issues much better now.

I think I have only one last question for the moment. Does the conditional states formalism apply to infinite dimensional Hilbert spaces?

18. mleifer

In the paper, we only deal with the finite dimensional case, but that is mainly for simplicity and there is no reason in principle not to use it in the infinite dimensional case. In fact, the whole construction could be formulated at the level of operator algebras.

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