# A Weinberg: Quantum Mechanics Without State Vectors

1. Mar 29, 2016

### Agrippa

2. Mar 29, 2016

### ShayanJ

$[\rho,H]=-[H,\rho]$

3. Mar 29, 2016

### vanhees71

The von Neumann equation for the density operator in the Schrödinger picture is
$$\frac{1}{\mathrm{i}\hbar} [\hat{\rho},\hat{H}]+(\partial_t \hat{\rho})_{\text{expl.}}=0.$$

4. Mar 29, 2016

### Zafa Pi

So by reformulating quantum theory the weirdness of the double slit experiment and the violation of Bell's inequality (= the negation of local realism) disappear?
That's pretty weird.

5. Mar 29, 2016

### marcus

Maybe someone else would like to comment. That's how I read what is said in the abstract
"With this definition of a physical state, even in entangled states nothing that is done in one isolated system can instantaneously effect the physical state of a distant isolated system. This change in the description of physical states opens up a large variety of new ..."

6. Mar 29, 2016

### Zafa Pi

I think you're reading too much into that quoted passage. Run of the mill QM doesn't say that "one isolated system can instantaneously effect the physical state of a distant isolated system." That is just a bit of quantum flapdoodle [Gell-Mann]. QM merely says the results of Alice and Bob's measurements are correlated and those correlations will not disappear with a reformulated theory.
How those correlations take place is indeed a mystery, but no more so than how masses make gravitational attraction.

7. Apr 3, 2016

### Agrippa

It would be great to better understand Weinberg's proposal. But I'm not used to density matrix formulation. I only have basic understanding of the state vector formalism he is criticizing. What is the physical significance of the extra terms in his Lindblad equation? First we have $L_n ρ(t) L_m^\dagger$. To me this looks similar to $O_{ij} = <A_i | \hat{O} | A_j>$ which just defines the matrix elements for operator $\hat{O}$ in a basis $|A_1>,...|A_N>$. So is $L_n ρ(t) L_m^\dagger$ doing nothing more than casting the density matrix of the system in the L basis? And then I wonder what the physical significance of subtracting $½(ρ(t) L_m^\dagger L_n + L_m^\dagger L_n ρ(t))$ is. Is the part that's added to the quantum Liouville equation doing nothing more than constructing a matrix out of the unspecified $h_{n,m}$ elements in the $L$ basis? Thanks!

8. Apr 3, 2016

### A. Neumaier

pp.21-24 of my slides http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf might be of help in interpreting the Lindblad equation in a concrete situation. In a microscopic derivation, all three terms arise together; it therefore does not make sense to interpret the three terms separately. The combination is the generator of a completely positive map - that's what counts.

Last edited: Apr 5, 2016
9. Apr 5, 2016

### Agrippa

Thanks for this. A concrete illustration of the Lindblad equation is certainly helpful and your slides are clearly written. But I'm still puzzled.

One prima facie difference between your equation and the general Lindblad equation is that the sum (the sigma) is not present in yours. For an N-dimensional system there is a sum of $N^2-1$ terms. Why is that missing in yours? Why isn't there such a sigma in front of each of your three additional terms (corresponding to a, b, and c)?

Your equation also appears to be missing the $h_{n,m}$ term, i.e. the term for the hermitian matrix elements - the constants that determine the dynamics. What you instead have outside each bracket is just one constant. For example outside the first set of brackets you have 0.002 x 138 - the cavity loss rate. Shouldn't the $h_{n,m}$ term be specified so that it's clear that the matrix elements are continuous functions of the relevant constant (such as cavity loss rate)?

In trying to make sense of your claim that the terms make no sense separately, but together function to guarantee "complete positivity", I stumbled across this. It helped to define $L_n ρ(t) L_m^\dagger$ as a test for whether ρ(t) is positive definite: ρ(t) is positive definite relative to operator $L$ if $L_n ρ(t) L_m^\dagger$ yields a positive number. Is the idea that ($L_n ρ(t) L_m^\dagger - ½(ρ(t) L_m^\dagger L_n + L_m^\dagger L_n ρ(t))$ just functions to ensure all eigenvalues are positive?

10. Apr 5, 2016

### A. Neumaier

https://en.wikipedia.org/wiki/Lindblad_equation gives the most general form. The number $N$ of terms could be arbitrary but one can show that this many are enough. My example just has three nonzero diagonal terms and no nonzero off-diagonal term.

In the initial wikipedia form, only the complete expression generates a completely positive map; thus even the individual expressions whose coefficients are the $h_{n,m}$ don't mean anything. But there is an equivalent diagonal form, derived in the subsection https://en.wikipedia.org/wiki/Lindblad_equation#Diagonalization
which is of the same type as mine, just with more terms, and which can be interpreted physically.

ρ(t) is positive definite iff $x^*ρ(t)x>0$ for all nonzero $x$. But this is unrelated to complete positivity.

Actually my original statement was slightly incorrect - I had meant ''generates a completely positive map'' not ''is a completely positive map''. (I corrected my statement.)

11. Jan 11, 2017

### A. Neumaier

12. Apr 16, 2017

### greypilgrim

This question hasn't been answered yet, and it puzzles me as well. In the second paper, Weinberg says
"Whether in open systems in ordinary quantum mechanics or in closed systems in some modified version of quantum mechanics, in order to avoid instantaneous communication at a distance in entangled states, it is important to require that the density matrix at one time depends on the density matrix at any earlier time, but not otherwise on the state vector at the earlier time."

Since the density operators of both subsystems in a Bell experiment are completely mixed, how can local measurements (be it his suggested Lindblad evolution or any other kind) be correlated?

13. Apr 18, 2017

### kith

The reduced density matrix contains all information which is relevant for experiments which are performed on the subsystem alone.

Weinberg stresses that if the outcomes of such a measurement on subsystem A depended on the reduced density matrix of subsystem B, FTL communication would be possible by altering the reduced density matrix of B.

What Weinberg's statement is silent about is whether the reduced density matrices contain all information about the combined system of A and B. And indeed, they don't contain information about how the outcomes of individual measurements on the subsystems are correlated.

The information about the correlations is only contained in the density matrix of the combined system and not in the reduced density matrices of the subsystems.

14. Aug 15, 2017

### maline

Indeed, treating $\rho$ as "real" does nothing to remove the "weirdness". The crucial point is that the statement
applies only to time-evolution of a given density matrix, but not to the process of the experimenter updating the density matrix in response to measurement results, a.k.a. "collapse", "reduction", etc. - our favorite bugbear.

In this context, the EPR paradox is expressed as follows:
Suppose Alice and Bob, at spacelike separation, share an entangled state described by $\rho$. Alice performs a measurement on her subsystem, views the results, and uses them to describe the post-measurement state with a new density matrix $\rho'$. Bob, on the other hand, although we will assume he knows what measurement Alice planned to perform, cannot know the measurement results. Therefore he must continue to use $\rho$, evolving it with time in accordance with the interactions that take place as part of Alice's experimental protocol. The content of No Signalling Property is that the reduced density matrix $\rho_B$, describing Bob's subsystem, is unaffected by Alice's activities at spacelike separation. But the same cannot be said of $\rho'_B$. In general, since Alice uses new observations to write $\rho'$, $\rho'_B$ will likely contain more information (lower entropy) than $\rho_B$. For the standard EPR case of a singlet pair of spin-1/2 particles, $\rho_B$ is the completely mixed state while $\rho'_B$ is the pure state with spin opposite to Alice's result.

As long as we think of density matrices as describing knowledge- basically the instrumentalist viewpoint- there is nothing strange about this: Alice has more knowledge and so she can write a "better" density matrix. But if density matrices are "real", then which one, $\rho$ or $\rho'$, should we consider to be the "true" state? Alice's new knowledge is certainly correct and "part of reality", so it seems her update must be called a "real change of the state"- good old collapse. And of course, this change is nonlocal. Alice's measurement has instantaneously generated new information constraining Bob's subsystem; information that was not part of the previous "state of reality".

The only way out is MWI, which Weinberg dislikes: $\rho$ indeed remains the true state of the whole system, now a "multiverse", while $\rho'$ is Alice's description of the particular "branch" she now finds herself in. "Probabilites" (whatever that means in MWI) for experiments Bob may perform are described by $\rho_B$, while $\rho'_B$ describes "probabilities" for the branches where those results are eventually compared with Alice's particular result.

All this is well known. I do not understand what Weinberg hopes to gain by assuming the density matrix is real. In particular, in approaches like GRW (objective stochastic collapse) which is the direction he seems to prefer, the Lindblad equation describes the evolution of $\rho$ only from the perspective of one who has not seen the measurement results, so it certainly should not be taken to represent "reality".