# B Eigenstates & Subspaces

1. Nov 13, 2016

### bluecap

Are Everett branches (or relative states) the eigenstates or the Hilbert subspaces (or others?)?

Once in a branch (or world), what law of QM would be broken if you can cut off the branch you are sitting on and revert back to the global state vector (isn't the quantum eraser, etc. about upcollapsing wave function or recoherence of the branches?)?

2. Nov 13, 2016

### Staff: Mentor

When you brought this topic up in a previous thread I asked you to point at some actual math, since you were bringing it up at an "I" level. The same applies here. Please show some actual math behind what you describe as "Everett branches", "eigenstates", "Hilbert subspaces", etc. This topic cannot be discussed productively if all we have is vague ordinary language.

3. Nov 13, 2016

### Staff: Mentor

This is easy: the law of unitary evolution.

No. Once again, if you were looking at the actual math, it would be a lot easier for you to understand what is going on. Your ordinary language descriptions are misleading you.

4. Nov 13, 2016

### bluecap

Sorry I changed the level from "I" to "B" because Im just learning the math and concepts laymen way. You wrote elsewhere:

Many accounts of Many Worlds always refer to the detector hits in the double slits as Worlds or Branches. But these are for simply systems like single-particle system and double slit experiment. In our macroscopic world we are living in one branch of the so called universal wave function. So you can't model it as Eigenstates anymore like you explained before. So for macroscopic world.. the subspace is the so called World or Branch in MWI? I just need a conceptual understanding so just want advisors to confirm yes or no or exactly what is the mathematical equivalents of the famous Worlds or Branches in simple layman terms. About Subspace, you explained thus:

Is it a known thing that the subspace is the MW branches?

You have a point that it's better to just think of the maths and forgets about interpretations. No problem. But I just want to know what is the QM math counterparts for MW branches or worlds.. That's all. Thank you.

5. Nov 13, 2016

### Staff: Mentor

I'm not sure this topic can be productively discussed at the "B" level, but I'll try.

That is not what I said before, but it's not worth trying to disentangle that here. For this discussion I suggest that you forget whatever you think you know about eigenstates. I won't use that term at all in what follows.

Sort of. See below.

If I give you an answer to this, it won't be at the "B" level. To even understand the math at all, you need more background than that. But here is the best answer I can give.

Let's take an even simpler experiment than the double slit: a Stern-Gerlach device, that measures the spin of a single quantum particle. For concreteness, let's say it's an electron. We orient the device in a certain direction, which we'll call the $z$ direction, and we send the electron through the device. The device has two output "slots", which we'll call "up" and "down", and the electron comes out through one or the other of these slots, corresponding to whether its spin in the $z$ direction is up or down. Those are the only two possible results.

Mathematically, we can write the state of the electron as a vector $\vert \psi \rangle$ in a Hilbert space. (I'll use the state vector formalism rather than the wave function formalism since it's easier to understand.) For the case of a single electron when all we're interested in is its spin, the Hilbert space is the simplest possible one that is not trivial. The two states that the electron can be in when it comes out of the Stern-Gerlach device oriented in the $z$ direction are two particular state vectors in this Hilbert space, which we can call $\vert +z \rangle$ and $\vert -z \rangle$. Note that these two states are orthogonal and linearly independent, so they can be thought of as basis states for the Hilbert space we are using--any state vector in that Hilbert space can be written as a linear combination of these two states.

But suppose that the electron enters the Stern-Gerlach device in some state $\vert \psi \rangle$ that is not $\vert +z \rangle$ or $\vert -z \rangle$. What happens then? To figure that out, we use the fact I referred to above, that any state $\vert \psi \rangle$ can be written as a linear combination of $\vert +z \rangle$ and $\vert -z \rangle$, i.e., $\vert \psi \rangle = a \vert +z \rangle + b \vert -z \rangle$, where $a$ and $b$ can be any two complex numbers that satisfy the condition $|a|^2 + |b|^2 = 1$. (This condition ensures that the probabilities of all possible measurement results add up to 1.) The two terms in this linear combination, $a \vert +z \rangle$ and $b \vert -z \rangle$, are what the MWI refers to as "worlds" or "branches". Note that there is one world/branch for each possible measurement result.

6. Nov 14, 2016

### bluecap

Ok that's clear. Thanks. For the double slit experiment.. there are multiple possible measurement results in terms of the hits in the detectors. So the worlds or branches are each detector hit or are there only 2 worlds or branches in the double slits in terms of the left or right slit? But then replacing the word world/branch with "possible measurement result", it doesn't seem to make sense that the double slit experiment has only 2 possible measurement result (left and right slit).

For your example of the Stern-Gerlach experiment above. It's great time to introduce the Eigenstate. So for the observable spin, it is in up and down spin Eigenstate? This is to distinguish how is Eigenstate math related to "each possible measurement result". The term $a \vert +z \rangle$ and $b \vert -z \rangle$ can never be called Eigenstate?

7. Nov 14, 2016

### secur

For this particular measurement |+z> and |-z> would be the eigenstates, which can also be called eigenspinors. See https://en.wikipedia.org/wiki/Eigenspinor.

8. Nov 14, 2016

### bluecap

What cases or measurement results when branches (worlds) are not Eigenstates? And what cases or measurement results branches (worlds) are Eigenstates?

9. Nov 14, 2016

### Staff: Mentor

@bluecap please do not quote an entire post when you respond. Only quote the particular things you are responding to.

10. Nov 14, 2016

### Staff: Mentor

Yes, there is one world or branch for each possible measurement result.

That's because the double slit experiment does not have those two measurement results; in that experiment you don't measure which slit the particle goes through, you only measure where it hits on the detector. So the possible measurement results are all the possible places it could hit on the detector.

If you are talking about a variant of the double slit experiment where we do measure which slit the particle goes through, then the measurement result is more complicated, because for each possible place the particle could hit on the detector--call such a place point P--you have two possible measurement results, not one. Instead of just "particle hit detector at point P", you have "particle went through slit #1 and hit detector at point P", and "particle went through slit #2 and hit detector at point P". And you have such a pair of results for each possible point P, so there are twice as many possible measurement results as for the ordinary double slit (where you don't measure which slit the particle went through). And, of course, in this variant the pattern of hits on the detector that you get when you record the results for a large number of particles is different from the ordinary version, because the latter shows an interference pattern and the former does not.

11. Nov 14, 2016

### Staff: Mentor

I'm still not understanding why you think this is such an important question. But I'll try to answer it.

Go back to the case of a spin measurement in the $z$ direction, and remember we are talking about the MWI. As secur says, the eigenstates of this observable are $\vert +z \rangle$ and $\vert -z \rangle$. A general state $\vert \psi \rangle$ is not an eigenstate of this observable. And under the MWI, there is no collapse, so the system starts in the state $\vert \psi \rangle$ and stays in the state $\vert \psi \rangle$. What changes is the state of the detector, so let's add that to the model.

In the Stern-Gerlach device, the "detector" is actually just the direction of the particle's motion when it comes out--i.e., which slot it comes out of. So let's assume that the particle goes in moving to the right, and comes out moving either up (for $\vert +z \rangle$) or down (for $\vert -z \rangle$). That means the particle's complete state going in is not just $\vert \psi \rangle$, but $\vert R, \psi \rangle$, because it has spin state $\psi$ and is moving to the right. Then the effect of the device can be written as follows:

$$\vert R, \psi \rangle = \vert R, a \left( +z \right) + b \left( -z \right) \rangle$$

turns into

$$a \vert U, +z \rangle + b \vert D, -z \rangle$$

where $U$ and $D$ denote the particle moving up or down. In other words, what the Stern-Gerlach device does is to entangle the momentum of the particle with its spin.

Now, the fact that $\vert +z \rangle$ and $\vert -z \rangle$ are eigenstates of spin in the $z$ direction isn't really relevant any more, because the particle's state is not just its spin. If we want to talk about eigenstates, we have to ask whether $\vert U, +z \rangle$ and $\vert D, -z \rangle$ are eigenstates of any measurement operator, and if so, which one? It turns out that, in this particular case, they are: $\vert U \rangle$ and $\vert D \rangle$ are eigenstates of $z$ momentum, and the operators for $z$ momentum and $z$ spin commute, so they can be jointly measured (basically, after each particle comes out of the Stern-Gerlach device, we detect which slot it came out of, and we can do that without changing its spin). But that is not always true.

For example, let's try to model the double slit experiment (the original version, where we don't detect which slit the particle goes through) along the same lines. The particle comes out of the source in some state $\vert \psi \rangle$, which we don't need to write down explicitly, we just stipulate that the source emits every particle in this same state. Mathematically, to predict the probability of the particle hitting at a particular spot on the detector, we compute an amplitude for it coming through slit 1 and an amplitude for it coming through slit 2, and add the amplitudes, then take the squared norm of the resulting complex number. So, heuristically, if we pick a point P on the detector, we compute a state for the particle at P that looks like this:

$$\vert \Psi \left( P \right) \rangle = \left( H_{1P} H_{S1} + H_{2P} H_{S2} \right) \vert \psi \rangle$$

where the $H$ operators are the Hamiltonians describing the time evolution from the source to slits 1 and 2, and from slits 1 and 2 to the point P on the detector. The squared norm of $\vert \Psi \left( P \right) \rangle$ then gives the probability of the particle hitting at point P.

If we really want to write down the full state of the particle, including all possible points P on the detector, then that would look something like this:

$$\vert \Psi \rangle = \Sigma_P \vert \Psi \left( P \right) \rangle = \left[ \Sigma_P \left( H_{1P} H_{S1} + H_{2P} H_{S2} \right) \right] \vert \psi \rangle$$

i.e., a linear combination of the states for all possible points P. (I have left out normalization here, i.e., the exact computation of the coefficients of each $\vert \Psi \left( P \right) \rangle$ in the above, which is why I say this is only heuristic.)

If we now compare this to the Stern-Gerlach measurement discussed above, the key difference is that in this case, there is no obvious measurement operator that $\vert \Psi \rangle$, the states $\vert \Psi \left( P \right) \rangle$, or any of their components, are eigenstates of. Of course once a particle is detected at a particular point P, we can say it is in a position eigenstate corresponding to point P; but the state we wrote down above is not such a state, or a linear combination of such states. In other words, knowing which states are and are not eigenstates doesn't really tell us what we want to know, namely, what are the probabilities of the different possible measurement results?

12. Nov 14, 2016

### bluecap

Let's say a radioactive material was tracked by a scanner at exactly 1 AM and 2 1/10000+ seconds owned by Comey. If it decays and the scanner clicks. Then he wakes up and unable to sleep and not in mood the next day and he sent the letter to congress reminding Congress of email stuff. If it doesn't decay and the scanner doesn't click. Then he continues to sleep and in good mood and didn't send the letter. In the former case, let's say Trump wins because of that. In the latter case, let's say Clinton wins because of that. We have two branches. Are they still modellable by subspace? But subspace is still in superposition being part of Hilbert space. So in principle can't one cut off the Trump branch you are sitting and get back to the global superposition where you can see the other branch where Clinton won. Earlier you wrote about Unitary Evolution. What is unitary evolution violated when you jump off from the Trump Branch? What is the equivalent in wave function evolution when subspace is not the same as collapse and not in Eigenstates at all.

(edit I send this a minute after you sent the last message)

13. Nov 14, 2016

### Staff: Mentor

This is too involved for a B level thread, and contains too many misconceptions for me to try to disentangle them. You will need to spend some time actually working through the math of QM, and then come back and start an "I" level thread.

The quick B-level answer to "why is unitary evolution violated" is that unitary evolution conserves information, and "cutting off" any branch destroys information, so such an operation manifestly can't be unitary. Similarly, trying to "rejoin" branches while still preserving the unique information in each branch (as opposed to a "quantum eraser" experiment that erases the information that distinguished the branches) creates information, so it manifestly can't be unitary. If you want more than that, then you'll have to do as I just said and get to the point where you can have an "I" level discussion.

14. Nov 15, 2016

### bluecap

I have thought of this by googling "unitary evolution conserves information". But here is a counter argument. Supposed you are watching the double slit experiment and even though the detector hit occurs in certain spots.. you could say communicate to one spot/hit (or branch) and inform it that the next hits were so and so location.. that particular spot/hit (or branch) would know the location of others. How do you describe this effect using the language of "unitary evolution". How could it still conserve information when you inform the hits or branches of the others location (that it couldn't know on its own)?

15. Nov 15, 2016

### secur

16. Nov 15, 2016

### bluecap

That's why we are called laymen because we are not expert in math. I've been reflecting on the words "Unitary Evolution" for quite some hours. It is defined as:

"In quantum physics, unitarity is a restriction on the allowed evolution of quantum systems that ensures the sum of probabilities of all possible outcomes of any event is always 1. [Wiki]"

As the wave function evolves.. what is the context probabilities of all possible outcomes of any event is always 1. Any simple math for this? And what's it connection to the born rule that would probabilities to be less or more than one? But if you have another wave piggybacking on the wave function and can track where the particle collapses.. and it can communicate back to the wave function.. won't this mean unitary evolution is conserved so branch can communicate back to the state vector?

17. Nov 15, 2016

### Staff: Mentor

No, you can't. The fact that this sounds meaningful in ordinary language does not mean it actually is meaningful when you do the math. But the reason why goes beyond a "B" level thread.

I don't have to because it violates unitary evolution and hence is inconsistent with QM. The reason why goes beyond a "B" level thread.

You don't need to "reflect on the words", and you shouldn't try to understand this topic by reading Wikipedia. You need to actually look at a QM textbook and work through the math. Ballentine is a good one to start with.

This is not consistent with QM; there is no such "other wave" in QM. Go look at the math.

18. Nov 15, 2016

### Staff: Mentor

This topic cannot be productively discussed further at the "B" level. Thread closed.