# Many Worlds and conservation of energy

1. Dec 14, 2013

### tom.stoer

I would like to re-start the discussion regarding conservation of energy in the many worlds interpretation. Kith did not agree with my example, so hopefully it becomes clearer now where I see problems regarding the interpretation.

Let's prepare a particle in a state like

$|\psi\rangle = a_h|h\rangle + a_v|v\rangle$

$a_h^2 + a_v^2 = 1$

Both states with horizontal (h) and vertical (v) polarization are energy eigenstates

$H|h,v\rangle = E|h,v\rangle$

In an experiment with an appropriate polarizator we observe either horizontal or vertical polarization. It seems that the other component has vanished, collapsed, became invisible or something like that. The energy - which was distributed over both component is now contained in the remaining or visible one.

So for the initial preparation we find

$\langle H \rangle = a_h^2 E + a_v^2 E = E$

After the measurement due to decoherence and after tracing out environmental d.o.f. the reduced density matrix can be approximated (!) by

$\rho \simeq a_h^2 |h\rangle\langle h| + a_v^2 |v\rangle\langle v| + \ldots$

where ... represents effectively suppressed interferences terms.

Then again we find for the energy

$\langle H \rangle = a_h^2 E + a_v^2 E = E$

Mathematically this is fine, but the interpretation is rather weird.

1) In the collapse interpretation (which I don't like b/c of unphysical ingredients) the wave function collapses, e.g.

$\rho \to |h\rangle\langle h|$

so the energy which was "contained" the vertical component is contained in the surviving horizontal component after the collapse.

2) in the many worlds interpretation both components remain real, but one component becomes "effectively invisible", so for example I observe the horizontal component only.

In order to calculate the probability whether I will observe branch "h" in future I use

$p_h = \text{tr} (\rho \, P_h) = a_h^2$

After the observation I can't use the same formula to find what I observe (namely "h" with certainty).

So the logical problem is this
1) I have a formula to calculate what I will observe in future (the probability)
2) There is no formula to calculate what I did observe

In order to calculate the energy I always have to use this formula, so before and after the observation I use

$\langle H \rangle = a_h^2 E + a_v^2 E = E$

So the logical problem I see is the following:
1) the polarization "v" became invisible for me due to decoherence and branching
2) the energy contained in branch "v" is still visible to me (even so the polarization "v" is invisible)
3) the formulas to calculate the energy E before and after the observation are identical

So I can't calculate what did observe (!) and I have to use a formula to calculate the energy I measure where the energy which is contained in the invisible branch is still visible.

Of course all this is mathematically consistent, but there's - at least to me - still some weird stuff in the interpretation.

Last edited: Dec 14, 2013
2. Dec 14, 2013

### Staff: Mentor

Same here - consistent but weird.

Thanks
Bill

3. Dec 14, 2013

### tom.stoer

nothing else to say?

4. Dec 14, 2013

### Staff: Mentor

Not really.

Its pretty obvious from a link I gave previously - energy gets 'diluted' with each branch, or as you say - the stuff in the other worlds becomes invisible. That occurs at an accelerating exponential rate and the rate of dilution is really really fast.

Thanks
Bill

5. Dec 14, 2013

### tom.stoer

The strange thing is that what I will observe depends on the quantum state and all its branches, but what I actually do observe does not. Regarding the polarization I can forget about all other branches, only one polarization from one branch is visible. But for other observables like energy the contribution of other branches is still important.

I would call the following a paradox: let's prepare a state

$|\psi\rangle = \sqrt{1-\xi^2}|h\rangle + \xi|v\rangle$

with

$\xi \simeq 0$

Suppose we are in the very unlikely situation to observe "v". That means in some sense we observe a portion of energy

$(1-\xi^2)E$

from other branches, even so we do not observe these branches.

So the statement that other branches become invisible is not correct, not for all observables.

6. Dec 14, 2013

### audioloop

indeed, more and more energy at each moment.

.

7. Dec 14, 2013

### tom.stoer

No, not more and more energy. Total energy is conserved.

8. Dec 14, 2013

### audioloop

i dont think so, the hyper branching creates more and more realities, then you need inifinite energy at the beginning of each branching and at each point.

.

9. Dec 14, 2013

### tom.stoer

That's not what the equation says: energy is not conserved per branch but over all branches; and total energy does not increase during branching. The energy I calculate is not calculated per branch, but as a sum over all branches. Therefore energy per branch is irrelevant, all what counts is total energy over all branches. And this total energy remains visible in each branch, therefore in some sense other branches remain visible in a certain sense - we observe their energy

Last edited: Dec 14, 2013
10. Dec 14, 2013

### audioloop

the problem is, that you can't encapsulate the reality in an equation.

11. Dec 14, 2013

### tom.stoer

I agree. But an interpretation of equations describing reality should not produce too many paradoxa. Otherwise I would not call it an interpretation. And then "should up and calculate" is the best choice

12. Dec 14, 2013

### Staff: Mentor

You do not need infinite energy. Any real number can be subdivided indefinitely.

Thanks
Bill

13. Dec 14, 2013

### audioloop

fully concur.

14. Dec 14, 2013

### The_Duck

The phrases I quoted seem misguided to me. You are talking as if a superposition means that the energy of the system is distributed over the different branches of the superposition. This isn't the right way to think about it. The system has a certain probability to be found in each branch, and each branch has a certain energy. The total energy is not a sum of contributions from each branch; rather, each branch has its own total energy.

For example, suppose I have a system in a superposition of two momentum eigenstates, say $(|p=k\rangle + |p=-k\rangle)/\sqrt{2}$. Can I say that the total momentum is zero, which is a sum of a contribution $k/2$ from the first branch and $-k/2$ from the second branch? No, that statement does not make sense. The system does not have a definite total momentum. The *expectation value* of the momentum is zero, and this *expectation value* is a sum of $k/2$ from the first branch and $-k/2$ from the second branch, but this is a different statement. The expectation value of the momentum is an abstract statistical construct, and should not be confused with the actual momentum, which is not definite in this example.

No. In my example after a measurement of the momentum is performed, the density matrix looks like

$\frac{1}{2}|-k\rangle\langle -k| + \frac{1}{2} |k\rangle\langle k|$

In MWI we interpret this as two different worlds. The reason we do this is that if you find yourself in the first branch, you can without loss of predictive power throw away the second branch and represent the system through the density matrix $| -k \rangle \langle -k |$. The second branch has absolutely no effect on you if you find yourself in the first branch. The total momentum of the system is now $-k$ (according to someone in the first branch) and in no way does it consist of a sum over a "visible" and "invisible" portion.

I have used momentum above, but the same applies to energy in your example.

15. Dec 15, 2013

### tom.stoer

I fully agree; this is nothing else but my example.

I could do that, but then I do not learn anything from MWI. I could throw away the second branch. I could throw away the prefactor of my own branch. I could throw away the prefactor when calculating the energy. But then I would do something the formalism doesn't tell me to do.

The total momentum is always the total momentum in the full Hilbert space. That's the way momentum is constructed and that's the way it is conserved. Of course one can calculate the momentum in one branch, but this momentum need not conserved.

If you define energy as you did it (energy observed per branch) then it is no longer allowed to sum over all these branch-specific energies b/c the sum of them growth during branching and is not conserved. So you introduce two different concepts of energy, namely i) energy in the full Hilbert space which is conserved but partially invisible and ii) branch-specific energy. One should be clear about that.

The problem I see is that the MWI wants to take the formalism seriously and introduce no new ingredients like a collapse. But through the backdoor one has to do exactly this in some cases. That weakens the hole idea.

16. Dec 15, 2013

### stevendaryl

Staff Emeritus
When you observe the energy of a system, you get an energy eigenvalue, not the expected value. To measure the expected value, you have to recreate the initial state and measure the energy again and again. If the system under consideration is the entire universe, then you can't return it to the state before you made the measurement. So I wouldn't call the expectation value for the energy "visible".

17. Dec 15, 2013

### stevendaryl

Staff Emeritus
It actually seems to me that conservation of energy is more problematic in the Copenhagen interpretation. Suppose that we have an observable (say spin in the z-direction) that does not commute with the Hamiltonian. So spin eigenstates are not energy eigenstates. Now, if we prepare an energy eigenstate, then measure the spin, then measure the energy, the final energy can be different from the initial energy. So energy is not conserved for observations.

In terms of the wave function for the entire universe (which is the fundamental object in MWI), energy is always conserved.

18. Dec 15, 2013

### The_Duck

You still sound like you are confusing the expectation value of an observable, which is a sum over branches of a superposition, with the value of the observable, which is not.

You shouldn't be summing energy over branches in the first place, regardless of your interpretation.

What is "energy in the full Hilbert space?" Is it $\langle E \rangle$? Because $\langle E \rangle$ is not the total energy, it is the expectation value of the total energy.

Last edited: Dec 15, 2013
19. Dec 15, 2013

### tom.stoer

this discussion is about what we WILL measure (not what we DID measure) and therefore I need the expectation value, not the eigenvalue.

20. Dec 16, 2013

### kith

I don't think this is correct. If you want to calculate the expectation value after the observation, you should use the formula <H>=Eh=E. This is equal to your <H> only because of the degeneracy of the inital state. If the initial superposition is not an energy eigenstate (Eh≠Ev) your <H> gives a wrong prediction in general.