Several different things (all pleasant and work-related, no disasters) have been keeping me from being a good blogger as of late. Last week, for example, we hosted a visit by Andy Albrecht from UC Davis. Andy is one of the pioneers of inflation, and these days has been thinking about the foundations of cosmology, which brings you smack up against other foundational issues in fields like statistical mechanics and quantum mechanics. We spent a lot of time talking about the nature of probability in QM, sparked in part by a somewhat-recent paper by our erstwhile guest blogger Don Page.
But that’s not what I want to talk about right now. Rather, our conversations nudged me into investigating some work that I have long known about but never really looked into: David Deutsch’s argument that probability in quantum mechanics doesn’t arise as part of a separate ad hoc assumption, but can be justified using decision theory. (Which led me to this weekend’s provocative quote.) Deutsch’s work (and subsequent refinements by another former guest blogger, David Wallace) is known to everyone who thinks about the foundations of quantum mechanics, but for some reason I had never sat down and read his paper. Now I have, and I think the basic idea is simple enough to put in a blog post — at least, a blog post aimed at people who are already familiar with the basics of quantum mechanics. (I don’t have the energy in me for a true popularization at the moment.) I’m going to try to get to the essence of the argument rather than being completely careful, so please see the original paper for the details.
The origin of probability in QM is obviously a crucial issue, but becomes even more pressing for those of us who are swayed by the Everett or Many-Worlds Interpretation. The MWI holds that we have a Hilbert space, and a wave function, and a rule (Schrödinger’s equation) for how the wave function evolves with time, and that’s it. No extra assumptions about “measurements” are allowed. Your measuring device is a quantum object that is described by the wave function, as are you, and all you ever do is obey the Schrödinger equation. If MWI is to have some chance of being right, we must be able to derive the Born Rule — the statement that the probability of obtaining a certain result from a quantum measurement is the square of the amplitude — from the underlying dynamics, not just postulate it.
Deutsch doesn’t actually spend time talking about decoherence or specific interpretations of QM. He takes for granted that when we have some observable X with some eigenstates |xi>, and we have a system described by a state
then a measurement of X is going to return either x1 or x2. But we don’t know which, and at this stage of the game we certainly don’t know that the probability of x1 is |a|2 or the probability of x2 is |b|2; that’s what we’d like to prove.
In fact let’s just focus on a simple special case, where
If we can prove that in this case, the probability of either outcome is 50%, we’ve done the hard part of the work — showing how probabilistic conclusions can arise at all from non-probabilistic assumptions. Then there’s a bit of mathematical lifting one must do to generalize to other possible amplitudes, but that part is conceptually straightforward. Deutsch refers to this crucial step as deriving “tends to from does,” in a mischievous parallel with attempts to derive ought from is. (Except I think in this case one has a chance of succeeding.)
The technique used will be decision theory, which is a way of formalizing how we make rational choices. In decision theory we think of everything we do as a “game,” and playing a game results in a “value” or “payoff” or “utility” — what we expect to gain by playing the game. If we have the choice between two different (mutually exclusive) actions, we always choose the one with higher value; if the values are equal, we are indifferent. We are also indifferent if we are given the choice between playing two games with values V1 and V2 or a single game with value V3 = V1 + V2; that is, games can be broken into sub-games, and the values just add. Note that these properties make “value” something more subtle than “money.” To a non-wealthy person, the value of two million dollars is not equal to twice the value of one million dollars. The first million is more valuable, because the second million has a smaller marginal value than the first — the lifestyle change that it brings about is much less. But in the world of abstract “value points” this is taken into consideration, and our value is strictly linear; the value of an individual dollar will therefore depend on how many dollars we already have.
There are various axioms assumed by decision theory, but for the purposes of this blog post I’ll treat them as largely intuitive. Let’s imagine that the game we’re playing takes the form of a quantum measurement, and we have a quantum operator X whose eigenvalues are equal to the value we obtain by measuring them. That is, the value of an eigenstate |x> of X is given by
![V[|x\rangle] = x .](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-5e36b3a788425fb906bc9ab0c239b73e_l3.png "Rendered by QuickLaTeX.com")
The tricky thing we would like to prove amounts to the statement that the value of a superposition is given by the Born Rule probabilities. That is, for our one simple case of interest, we want to show that
![V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] = \frac{1}{2}(x_1 + x_2) . \qquad\qquad(1)](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-9f6332d9ef7bb79816cf8f4dd66972ef_l3.png "Rendered by QuickLaTeX.com")
After that it would just be a matter of grinding. If we can prove this result, maximizing our value in the game of quantum mechanics is precisely the same as maximizing our expected value in a probabilistic world governed by the Born Rule.
To get there we need two simple propositions that can be justified within the framework of decision theory. The first is:
Given a game with a certain set of possible payoffs, the value of playing a game with precisely minus that set of payoffs is minus the value of the original game.
Note that payoffs need not be positive! This principle explains what it’s like to play a two-person zero-sum game. Whatever one person wins, the other loses. In that case, the value of the game to the two participants are equal in magnitude and opposite in sign. In our quantum-mechanics language, we have:
![V\left[\frac{1}{\sqrt{2}}(|-x_1\rangle + |-x_2\rangle)\right] = - V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] . \qquad\qquad (2)](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-4a0c547d3e57dc5c1665b5397af96dd0_l3.png "Rendered by QuickLaTeX.com")
Keep that in mind. Here’s the other principle we need:
If we take a game and increase every possible payoff by a fixed amount k, the value is equivalent to playing the original game, then receiving value k.
If I want to change the value of a playing a game by k, it doesn’t matter whether I simply add k to each possible outcome, or just let you play the game and then give you k. I don’t think we can argue with that. In our quantum notation we would have
![V\left[\frac{1}{\sqrt{2}}(|x_1+k\rangle + |x_2+k\rangle)\right] = V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] +k . \qquad\qquad (3)](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-64983ac5f96f7e3c63fb82e3b1356a48_l3.png "Rendered by QuickLaTeX.com")
Okay, if we buy that, from now on it’s simple algebra. Let’s consider the specific choice
and plug this into (3). We get
![V\left[\frac{1}{\sqrt{2}}(|-x_2\rangle + |-x_1\rangle)\right] = V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] -x_1 - x_2.](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-8fc282fd0e39e7cb41cfe9b90fd65dad_l3.png "Rendered by QuickLaTeX.com")
You can probably see where this is going (if you’ve managed to make it this far). Use our other rule (2) to make this
![-2 V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] = -x_1 - x_2 ,](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-98f5baeb15378b76bd1d6dba2570d5ad_l3.png "Rendered by QuickLaTeX.com")
which simplifies straightaway to
![V\left[\frac{1}{\sqrt{2}}(|x_1\rangle + |x_2\rangle)\right] = \frac{1}{2}(x_1 + x_2) ,](https://www.preposterousuniverse.com/blog/wp-content/ql-cache/quicklatex.com-b1f54750220fb61d560e1a56fbec2f9c_l3.png "Rendered by QuickLaTeX.com")
which is our sought-after result (1).
Now, notice this result by itself doesn’t contain the word “probability.” It’s simply a fairly formal manipulation, taking advantage of the additivity of values in decision theory and the linearity of quantum mechanics. But Deutsch argues — and on this I think he’s correct — that this result implies we should act as if the Born Rule is true if we are rational decision-makers. We’ve shown that the value of a game described by an equal quantum superposition of states |x1> and |x2> is equal to the value of a game where we have a 50% chance of gaining value x1 and a 50% chance of gaining x2. (In other words, if we acted as if the Born Rule were not true, someone else could make money off us by challenging us to such games, and that would be bad.) As someone who is sympathetic to pragmatism, I think that “we should always act as if A is true” is the same as “A is true.” So the Born Rule emerges from the MWI plus some seemingly-innocent axioms of decision theory.
While I certainly haven’t followed the considerable literature that has grown up around this proposal over the years, I’ll confess that it smells basically right to me. If anyone knows of any strong objections to the idea, I’d love to hear them. But reading about it has added a teensy bit to my confidence that the MWI is on the right track.
Just a quick semi-coherent placeholder note, since I have to run now. As you say, the issue of P in MW is much trickier than if you have some sort of extra collapse in which to insert special new rules. The traditional argument justifying Born is the one that Ben refers to, reproduced by Arkani-Hamed, but that’s long since been known to be invalid, since the limiting procedure is irrelevant.
On Deutsch and decision theory: “Given a game with a certain set of possible payoffs, the value of playing a game with precisely minus that set of payoffs is minus the value of the original game.” What does a “precisely minus payoff” even mean, except in the context of little financial games, where the statement is well-known to be false?
The question is not so much what a rational actor would bet, but how the existence of rational actors can be reconciled with the unitary structure+decoherence. The problem becomes one of why the probabilities for sequential observations factorize, i.e. why the chance of Schroedinger’s cat having survived the Tuesday experiment doesn’t change on Wednesday due to quantum fleas. As has repeatedly been shown, only the standard quantum measure gives the conserved flow needed to allow that factorization and hence allow the existence of rational actors.
So that’s a requirement but not an explanation. The best (only) explanation I’ve seen is by Jaques Mallah. If the state consists of the usual part we think about plus some maximal entropy white noise, a physical definition of a thought as a robust quantum computation, together with ordinary signal-to-noise constraints on robustness (square-root averaging), gives the Born rule from ratios of counts of thoughts!
Why that particular (mixture of low S +high S parts) starting state? Mallah doesn’t like this idea but I suggest the old cheat: anthropic selection. If that type of state is needed to allow the existence of rational actors, nobody will be arguing about why they find themselves part of some other type of state.
I’ll try to get back to fill this in more coherently in 24 hours.
p.s. Zurek’s paper sneaks in context-independent probabilities, and thus doesn’t really address the core question.
How do the coefficients enter into the story at all? It looks like assumptions (2) and (3) make just as much sense if the coefficients for the two states are different, but if that’s true, the we can derive (1) for the case when the coefficients are different as well… in other words, taken at face value, the argument seems to prove that V[a|x_1>+b|x_2>]=1/2(x_1+x+2) no matter what ‘a’ and ‘b’ are.
I revoke my previous question (after actually trying to carry though the math).
I should make at least one small correction to my hasty and over-compact note. The background entropy in Mallah’s picture is high, not maximal.
Since this article doesn’t explain where the absolute square of the amplitude comes in with Deutsch’s argument (48), I have read his paper which introduces it in equations 16 – 21. However I don’t understand the argument. It would be great if someone could explain why the value of eq. 18 equals the LHS of eq. 16, i.e. why is
V[|x1>|y1> + … + |x1>|ym> + |x2>|y m+1> + … + |x2>|yn>] =
V[sqrt(m) |x1> + sqrt(n – m) |x2>]
when y1 + … + ym = y_{m+1} + … + yn = 0? Can this actually be derived or is it an axiom? If the former, it does seem to rely on the state vectors being normalized, which would also need to be postulated?
.
@47
Got a copy of Pauli’s book. Good stuff.
I like this on the first page, written in 1933
“The solution is obtained at the cost of abandoning the possibility of treating physical phenomena objectively, i.e. abandoning the classical space-time and causal description of nature which essentially rests upon our ability to separate uniquely the observer and the observed.”
Combined with the fact that any bound state can be represented in a quantum field theory, it appears we are getting closer to completely abandoning any notion that general relatively is even needed.
Daryl, Thank you for responding (@36&38) to my question. Unfortunately I have been away for a few days and so have been slow to respond, but I hope you are still around and following this discussion as I remain puzzled.
I have no problem with agreeing that your conditions (1)(2)(3) imply the Born rule (and similarly for Sean’s and David’s similarly numbered equations) but I still don’t see how these are implied by decision theory without essentially assuming the Born rule to start with.
Yes, the “states” in question are all eigenfunctions for the same observable, but on the two sides of each value equation (other than (1)) they correspond to different eigenvalues so they are not actually the same states.
In fact the decision theoretic increment of value that is expected from replacing X by X+k and the reversal that comes from replacing X with -X seem to me to be obvious only if we work with the same state and consider the observable to be what is changing.
To ask for these to also apply when the operator stays the same but the states are changed seems to involve an implicit assumption that V(a1|x1>+a2|x2>) is a linear combination of x1 and x2 with coefficients p1(a1,a2) and p2(a1,a2) which are independent of |x1> and |x2>. And to me that looks very much like begging the question.
Is there a way to show (without assuming the usual expectation formula) that V(X+k,|Psi>)=V(X, T(k)|Psi>?
What seems odd about this business of starting with the Hilbert space and inferring a probabilistic interpretation after the fact, is that the Hilbert space itself arises naturally as a way of representing the possible families of probability distributions for observables. In that approach, pioneered by von Neumann and Mackey, and nicely developed and summarized in the books by Varadarajan, the starting point is a lattice of questions (observables with values in {0,1}) and the notion of probability for these seems to be no less elementary than that of decision theoretic “value” since the expected value of a proposition in any state is just the same as the probability that it is observed to be true.
Here’s an example where (2) and (3) is consistent with a different probability rule than Born’s.
Just before we’re measuring the observable X (or “playing the game” in Deutsch’ terminology), we will scale |psi> such that the sum of the expansion coefficients is 1.
1. |psi> = a1 |x1> + a2 |x2>
2. a1 + a2 = 1
This scaling is not as physically illogical as you might think, for instance the collapse of the wavefunction can also be viewed to contain a rescaling of the observed eigenvector immediately after/during the observation.
Let |base> be the sum of all eigenvectors of X.
3. |base> = |x1> + |x2> + …
I’m going to show that the following definition of the expected value of the measurement (“payoff”) satisfies (2) and (3) in this article.
4. V[|psi>] = <base| X |psi>
Here’s how (2) is satisfied:
5. V[a1 |x1 + k> + a2 |x2 + k>]
= a1 (x1 + k) <base|x1 + k> + a2 (x2 + k) <base|x2 + k>
= a1 x1 + a2 x2 + k
= V[a1 |x1> + a2 |x2>] + k
Above, <base|x1 + k> is 1 since |base> is the sum of the eigenvectors, and |x1 + k> is an eigenvector. Similarily, (3) is satisfied because:
6. V[a1 |-x1> + a2 |-x2>]
= -a1 x1 <base|-x1> – a2 x2 <base|-x2>
= -(a1 x1 + a2 x2)
= -V[a1 |x1> + a2 |x2>]
The main result in the article is then reproduced easily:
7. V[(|x1> + |x2>) / 2] = (x1 + x2) / 2
Let me follow the same arguments in this blog article and Deutsch’s, to prove something other than the Born rule:
0. V[a1 |x1> + a2 |x2> + …] = |a1| x1 + |a2| x2 + …
To be able to make the argument we will need to postulate that the state vector should be scaled just prior to the measurement, such that the sum of the absolute values of the probability amplitudes is 1, instead of being normalized.
1. If |psi> = a1 |x1> + a2 |x2> + …, then |a1| + |a2| + … = 1
Because of this postulate, instead of (|x1> + |x2>) / sqrt(2) we will use (|x1> + |x2>) / 2, etc. If we now assume the equivalent of equation (2) and (3) from this blog article:
2’. V[(|-x1> + |-x2>) / 2] = -V[(|x1> + |x2>) / 2]
3’. V[(|x1 + k> + |x2 + k>) / 2] = V[(|x1> + |x2>) / 2] + k
we will end up with the equivalent equation for exactly the same reasons made in this article, since the sqrt(2) never comes into the derivation.
4’. V[(|x1> + |x2>) / 2] = (x1 + x2) / 2
Let’s move over to Deutsch’s article, and the chapter “The general case”. We first want to prove the equivalent of equation (Deutsch.12):
12’. V[(|x1> + |x2> + … + |xn>) / n] = (x1 + x2 + … + xn) / n
The proof is made by induction in two stages. Now I must admit that I don’t understand the first stage, but it doesn’t sound like that will be a problem for my argument (the sqrt(2) is again not used). For the second stage we can use the same arguments of “substitutibility”. Let V[|psi1>] = V[|psi2>] = v, then:
13’. V[(a |psi1> + b |psi2>) / (|a| + |b|)] = v
If we now set
14’. |psi1> = (|x1> + … + |x n-1>) / (n – 1), |psi2> = | V[|psi1>] >, a = n – 1, b = 1
Then (13’) implies:
15’. (x1 + x2 + … + x_{n-1} + V[|psi1>]) / n = V[|psi1>]
Note, (15’) is identical to (Deutsch.15).
Now to the crucial part of Deutsch’s argument. What we want to show and the equivalent of (Deutsch.16) is:
16’. V[ (m |x1> + (n – m) |x2>) / n] = (m x1 + (n – m) x2) / n
and (Deutsch.17), (Deutsch.18), and (Deutsch.20) are:
17’. sum_{a = i}^m |ya> / m or sum_{a = m + 1}^n |ya> / (n – m)
18’. (sum_{a = i}^m |x1>|ya> + sum_{a = m + 1}^n |x2> |ya>) / n
20’. (sum_{a = i}^m |x1 + ya> + sum_{a = m + 1}^n |x2 + ya>) / n
Again, we’re allowed to do this according to the postulate, because we’re just about to do a measurement, and then we need to scale such that the sum of the absolute values of the probability amplitudes is 1.
Equations (Deutsch.19) and (Deutsch.21) are not changed. (Deutsch.22) obviously reads:
22’. sum_a p_a |x_a>, sum_a p_a = 1
The next arguments may pose a problem. They’re supposed to show that even though the above results are valid for p_a being rational numbers, they should also apply if p_a is a real number.
For instance the unitary transformation is imagined to transforms eigenvectors into a higher eigenvalued eigenvectors. The value of the game is then guaranteed to increase. Not so with our postulate, since we need to scale our state vector just prior to a measurement, and in general the scale factor would be different before a unitary transformation and after.
I’m guessing there is an argument for proving how to extend it to real numbers, but I just don’t see it yet. So for now, we will have to be content with the probability amplitudes being rational numbers.
The conclusion of all of this is that the normalization of the state vector is crucial for Deutsch’s derivation.
@Hakon Hallingstad:
your reasoning here is, I’m afraid, totally bogus. I’m not trying to be nasty and I’m sure you won’t take it as such since you seem in other posts to be pretty keen on learning properly how to do these things. One particular error I can spot:
“For instance the unitary transformation is imagined to transforms eigenvectors into a higher eigenvalued eigenvectors. ”
In what follows, m* = “hermitian conjugate of m” not, “times by” :-).
Unitary operators have eigenvalues of magnitude 1. to see this, consider that the definition of a unitary operator is that its inverse is equal to its hermitian conjugate.
UU* = 1 by the definition of unitarity.
If U|a> = m |a> , this implies <a| U* = <a| m*,
but = by unitarity definition.
From above, = m m* , hence m m* = 1.
This means that the magnitude of m is 1. So you can’t have a “unitary transformation” that makes “the eigenvalues higher”. It is a contradiction in terms.
cheers
@DanW
> your reasoning here is, I’m afraid, totally bogus. […]
Right, assumption (61.1) does not hold in Quantum Mechanics proper. I’m interested in knowing about other flaws you can point out, and to see whether those flaws can also be applied to Deutsch’s original arguments.
> One particular error I can spot. […]
I was too careless with my choice of words, so you misunderstood me. I was only trying to refer to Deutsch’s argument on page 12, for instance he says “Now, if U transforms each eigenstate |xa> of X appearing in the expansion of |psi> to an eigenstate |xa’> with higher eigenvalue.” See there for details.