How to understand unitarity in QM?

[vanhees71]

The notion of "reality" is loaden with so much confusion by philosophers that it's pracatically useless to be discussed within the natural sciences. For a natural scientist reality is what can be objectively observed.

Why there should be a problem with the empirical fact that I find my bathroom in more or less the same state in the morning as I left it in the evening is not a problem of QT, but it is the only consistent description of this observed stability of matter, given the empirical fact that matter around us consists of atomic nuclei and electrons, bound to atoms, molecules and condensed-matter many-body state.

Further concerning the collapse: As I said, you cannot say, in which state the measured system is after the measurement without giving a sufficiently precise description of the measurement apparatus.

The Stern-Gerlach experiment is the paradigmatic example of a very good approximation for a von Neumann filter measurement, which is rather a preparation process than a measurement. You let a beam of uncharged particles (orginally silver atoms, nowadays with much more precision e.g., realized with neutrons) with magnetic moment run through an appropriately taylored inhomogeneous magnetic field (with a strong homogeneous component in a given direction, say the $z$ direction). As can be pretty easily shown by solving the time-dependent Schrödinger equation, the beam splits (for spin 1/2 particles) into two partial beams, where one beam consists to practically 100% of particles prepared in the spin-up state $|+1/2 \rangle$ and the other in the spin-down state $|-1/2 \rangle$, i.e., position and spin-$z$ component are (almost) perfectly entangled, and thus by blocking one beam (then the particles are absorbed in the material of the blocker and for sure not adequately described as a certain eigenstates of $\sigma_z$), you have left a beam of particles with (almost exactly) determined spin-$z$ component. In QT there's nothing needed to explain this preparation of $\sigma_z$ eigenstates than quantum-theoretical dynamics. There's no need for some mysterious "collapse" dynamics outside of QT, let alone the assumption of an instantaneous action in the entire space (which is, of course, a problem within relativistic physics and thus should not be used to base QT on).

 

[bhobba]

Whoops ... measurement in quantum mechanics is as obvious as line in geometry? That was not clear to me ...

What is a measurement is clear in practice like what a point and line is. Measurements that do not destroy what is being measured are simply a new preparation procedure. You can't always predict the result of a preparation procedure - my view is - big deal. We will have to disagree on this being an big issue - a discussion is likely not to really go anywhere.

Thanks
Bill

 

[PeroK]

But physics is about reality.

Physics tries to explain reality, usually in terms of mathematical models and the results of experiments. For example, you might model a baseball bat as a continuous mass distribution; whereas, "in reality" it is no such thing. We continue to model gravity as a force; whereas, "in reality" it is not a force. And so on.

That's where I agree with the sentiment that I think @bhobba was expressing. Physics is, in a sense, a mapping of "reality" into a scheme of models and experiments. I don't believe we have anything in physics that is reality in an absolute sense.

 

[RockyMarciano]

- its like point and line in Euclidean Geometry - they are undefined in the theory but when you apply it its rather obvious. Is Euclidean Geometry up the creek as well ? (with all due respect to Hilbert and his axioms)

Euclidean geometry is perhaps the wrong name to use here(it would be more appropriate maybe just geometry or differential geometry) as non-euclidean geomtries are actually what's at the root of relativity and QFT. But if something is undefined, instead of ignoring it, wouldn't redefining it in more clear or physical terms work better?

I don't like 'reality' in physics because there are so many different ideas about it.

Thanks
Bill

I think here the word "reality" is actually a substitute for "measurement", which is the true "elephant in the room" of modern science. But again, instead of ignoring it or despising it "because there are so many different ideas about it", why not try to define it mor erigorously in physical and mathematical terms. Just saying that this is unnecessary(even if the critique is precisely that there are many ideas and confusion about it, which should prompt everyone to correct this state of affairs) when it is so obviously it is not, and insisting that everybody doing measurements knows very well what is doing FAPP(even if no one can define it), and that SUAC is the only possible answer only leads to more confusion and disparate pseudo-interpretations(as Tom stoer correctly points out).

 

[Grinkle]

is doing FAPP

What is FAPP?

edit:

Never mind - For All Practical Purposes

 

[mikeyork]

The notion that Schrodinger's eqn tells us the time evolution of an isolated quantum system is based on the idea that a quantum state depends on some external variable called time. However, in reality, our concept of time comes from observing change -- i.e. quantum transitions -- and what we know is that the transition between observations is governed by a unitary transformation. What we see as a "direction" in time is the result of probability rules increasing disorder in the universe -- it's easier to break an egg than to put it back together. Thus time is a measure of that increasing disorder.

To then describe the unitary transformation as $e^{-i\hat{H}t/\hbar}$ where $\hat{H}$ is Hermitian then has a certain circularity and should be seen as a definition of $\hat{H}$. Modeling $\hat{H}$ is then a purely phenomenological process, however successful it is in predicting transitions.

By contrast, S-matrix theory says let's model the unitary operator in terms of its elements ( time-independent scattering amplitudes) directly.

 

[PeroK]

What we see as a "direction" in time is the result of probability rules increasing disorder in the universe -- it's easier to break an egg than to put it back together. Thus time is a measure of that increasing disorder.

It's strange, therefore, that the egg forms from relative disorder in the first place!

And, how is the motion of the Earth not a measure of time? Where is the increasing disorder there?

 

[PeroK]

It's strange, therefore, that the egg forms from relative disorder in the first place!

And, how is the motion of the Earth not a measure of time? Where is the increasing disorder there?

In fact, none of the clocks I can think of measure increasing disorder: sundial, hourglass, pendulum, quartz crystal, caesium atom.

 

[mikeyork]

The motion of the Earth and all the other clocks you mention are measuring change. Anything that changes can be a clock and its interpretation as time is merely a matter of calibration.

The creation of the egg is a low probability event and does not happen in isolation (second law of thermodynamics) but requires energy/interaction with other parts of the universe (just like a refrigerator).

 

[bhobba]

In fact, none of the clocks I can think of measure increasing disorder: sundial, hourglass, pendulum, quartz crystal, caesium atom.

Feynman wrote an interesting essay on that - if I recall its in the Character of Physical Law - the chapter on the distinction of past as future. Again, if I recall correctly, it would seem it's subtly involved with any way to measure time.

But really a new thread is required for it - its way off topic.

Thanks
Bill

 

[Blue Scallop]

About how to understand unitarity in QM.

They say decoherence occurs while the system is still unitarity meaning prior to any outcome.

So if you can somehow interface with any system and no outcome chosen. Is it still unitarity?

 

[Simon Phoenix]

As I said, you cannot say, in which state the measured system is after the measurement without giving a sufficiently precise description of the measurement apparatus.

Hmm. That's a bit of a cop out. For any given observable a sufficiently ingenious experimenter may be able to devise several technically different ways of realizing one of these ideal measurements. Granted it might take a considerable degree of ingenuity. Is your position that QM can not answer the question at all? So, on the assumption that I can construct an experiment to perform one of these idealized measurements, what does QM say about the state after the measurement?

I disagree that one actually needs the specific experimental details to even discuss this. I think QM tells us very nicely what the state is given a particular result - on the assumption that we can actually do one of these kinds of experiments.

On the other hand, is it even a reasonable assumption that these kinds of measurements even exist? I'm not sure - it's certainly beyond my ingenuity to think of a really clean-cut example (I don't consider the Stern-Gerlach set-up to be such a 'clean-cut' example).

position and spin-z component are (almost) perfectly entangled

Ah OK, I see what you mean by this now - the word 'component' threw me a bit and I thought you were talking about entangling different properties of a single system, which I couldn't make sense of. The magnetic field is acting like a beamsplitter. There's an entanglement between the path and the spin - in the same way we have an entanglement of the spatial mode (path) and polarization in a polarizing beamsplitter - which isn't a measurement either but a process by which we can filter different polarization states.

The real measurement is done when there is a subsequent measurement of the particles in one of the beam arms. So in the Stern-Gerlach case the absorption is acting rather like the technique for the production of heralded photons using parametric downconversion.

In the photon polarizing BS case you'd have a mixed state along the lines of $| 0 \rangle \langle 0| + | 1 \rangle \langle 1|$ in each timeslot in one of the arms before making a measurement on the other arm (an absorptive photodetection). After this 'heralding' measurement you'd have pure states in each timeslot (either a polarized photon or the vacuum).

 

[Blue Scallop]

Hmm. That's a bit of a cop out. For any given observable a sufficiently ingenious experimenter may be able to devise several technically different ways of realizing one of these ideal measurements. Granted it might take a considerable degree of ingenuity. Is your position that QM can not answer the question at all? So, on the assumption that I can construct an experiment to perform one of these idealized measurements, what does QM say about the state after the measurement?

I disagree that one actually needs the specific experimental details to even discuss this. I think QM tells us very nicely what the state is given a particular result - on the assumption that we can actually do one of these kinds of experiments.

On the other hand, is it even a reasonable assumption that these kinds of measurements even exist? I'm not sure - it's certainly beyond my ingenuity to think of a really clean-cut example (I don't consider the Stern-Gerlach set-up to be such a 'clean-cut' example).
Ah OK, I see what you mean by this now - the word 'component' threw me a bit and I thought you were talking about entangling different properties of a single system, which I couldn't make sense of. The magnetic field is acting like a beamsplitter. There's an entanglement between the path and the spin - in the same way we have an entanglement of the spatial mode (path) and polarization in a polarizing beamsplitter - which isn't a measurement either but a process by which we can filter different polarization states.

The real measurement is done when there is a subsequent measurement of the particles in one of the beam arms. So in the Stern-Gerlach case the absorption is acting rather like the technique for the production of heralded photons using parametric downconversion.

In the photon polarizing BS case you'd have a mixed state along the lines of $| 0 \rangle \langle 0| + | 1 \rangle \langle 1|$ in each timeslot in one of the arms before making a measurement on the other arm (an absorptive photodetection). After this 'heralding' measurement you'd have pure states in each timeslot (either a polarized photon or the vacuum).

Simon Phoenix. As an expert in practical quantum mechanics and setups. Can you please help think of a practical experimental setup to test whether it is possible to "scan" a unitary evolution without measurements? It is commonly believed that you need measurements to even tell when a system is unitary or entangled.. I want to convince myself this is an ironclad rule. So maybe you can encode some data in a superposition in an experiment setup and without measurements.. one will test whether it can be decoded while in unitary evolution. Any idea how to do the simplest experimental setup for this test?

 

[Simon Phoenix]

As an expert in practical quantum mechanics and setups

Lol. Vanhees and Bill are much more expert than I - I have a rather 'old fashioned' view of QM and I struggle a bit with the more modern (and logically sound) approaches.

Can you please help think of a practical experimental setup to test whether it is possible to "scan" a unitary evolution without measurements? It is commonly believed that you need measurements to even tell when a system is unitary or entangled..

I'm sorry - I don't really understand what you're asking here. Is there another way of phrasing your question? By 'unitary' here, I think you mean separable. To extract any real information you need to do a measurement of some kind.

 

[Blue Scallop]

Lol. Vanhees and Bill are much more expert than I - I have a rather 'old fashioned' view of QM and I struggle a bit with the more modern (and logically sound) approaches.
I'm sorry - I don't really understand what you're asking here. Is there another way of phrasing your question? By 'unitary' here, I think you mean separable. To extract any real information you need to do a measurement of some kind.

I read this in a paper about quantum information:
"The crucial distinction between quantum and classical information appears when one attempts
to use these other states as alternatives for transmitting information. Suppose we attempt
to encode ten bits of information onto a spin- 1/2 particle, by preparing it so that its spin points in
one of 2^10 possible directions. Then we send the particle to you. Can you read out the ten bits?
Of course not. Quantum theory forbids any measurement to distinguish all 1024 possibilities."

How do you do this actual physical experimental setup where you can encode ten bits of information onto a spin 1/2 particle?

 

[zonde]

Quote lifted from a thread in the cosmology forum.

What does it mean to know the exact state of a QM system? QM predicts probabilities that particles will be in one of multiple states when the particles are observed, and when observed, not all properties of a particle are simultaneously knowable to an exact degree (eg position and momentum).

Does knowing the exact state mean I know the probability functions for each particle in a given system, or is it different than that?

I would say that the statement you quoted is rather incorrect. It would be better if instead of "state" he would talk about "quantum state (vector)" like this:

But quantum mechanics, as far as we know, has a property called unitarity: if I know the exact quantum state (vector) of the system at time T, then I can, given enough computer power, calculate the precise quantum state (vector) of the system at any other time, no matter what.

As I see it there is important distinction between (classical) state and quantum state (vector) as they are very different. Description of quantum state (vector) can be transformed into description of classical state only probabilistically via Born rule.

The way I see the difference can be illustrated with crude analogy - let's say that classical state of particles is like actual configuration of molecules in complex object, then quantum state (vector) would be like chemical composition of complex object.

 

[vanhees71]

Hmm. That's a bit of a cop out. For any given observable a sufficiently ingenious experimenter may be able to devise several technically different ways of realizing one of these ideal measurements. Granted it might take a considerable degree of ingenuity. Is your position that QM can not answer the question at all? So, on the assumption that I can construct an experiment to perform one of these idealized measurements, what does QM say about the state after the measurement?

I disagree that one actually needs the specific experimental details to even discuss this. I think QM tells us very nicely what the state is given a particular result - on the assumption that we can actually do one of these kinds of experiments.

On the other hand, is it even a reasonable assumption that these kinds of measurements even exist? I'm not sure - it's certainly beyond my ingenuity to think of a really clean-cut example (I don't consider the Stern-Gerlach set-up to be such a 'clean-cut' example).
Ah OK, I see what you mean by this now - the word 'component' threw me a bit and I thought you were talking about entangling different properties of a single system, which I couldn't make sense of. The magnetic field is acting like a beamsplitter. There's an entanglement between the path and the spin - in the same way we have an entanglement of the spatial mode (path) and polarization in a polarizing beamsplitter - which isn't a measurement either but a process by which we can filter different polarization states.

The real measurement is done when there is a subsequent measurement of the particles in one of the beam arms. So in the Stern-Gerlach case the absorption is acting rather like the technique for the production of heralded photons using parametric downconversion.

In the photon polarizing BS case you'd have a mixed state along the lines of $| 0 \rangle \langle 0| + | 1 \rangle \langle 1|$ in each timeslot in one of the arms before making a measurement on the other arm (an absorptive photodetection). After this 'heralding' measurement you'd have pure states in each timeslot (either a polarized photon or the vacuum).

A photon is a good example. In almost all photon detectors the photon gets absorbed, and for sure then you cannot describe its state as being the one you get by assuming a collapse. That's all what I meant. It's nothing deep but just looking at what experimentalists really do in their labs when they measure all kinds of (microscopic) observables.

 

[Simon Phoenix]

How do you do this actual physical experimental setup where you can encode ten bits of information onto a spin 1/2 particle?

The first thing is to get clear what is being described here. It's a communication channel. Normally we think of a communication channel as some transmission of information between two (or more) parties - that's just transmission in space. Storage and retrieval is just transmission in time. So there's a coding process, a transmission (either in space or time) and then a measurement. The critical thing we're trying to achieve is that the original information (the message) can be reconstructed from the measurement results.

Rather than talk about trying to code 10 bits on a single spin-1/2 particle it's probably easier to think about just trying to encode 1 bit.

So let's think of the obvious way first. Suppose we use the up and down eigenstates of spin-z for our encoding. That's just choosing $| + \rangle _Z$ or $| - \rangle _Z$, respectively.

But how are we to recover that information? We can make a measurement of spin-z and for each particle transmitted we'll get the right answer - we'll get the bit that has been encoded with unit probability (assuming perfect measurements and no other noise sources).

But suppose we decide to measure spin-x instead? Now each measurement will give us the result + or - with 50% probability. With this measurement we'll recover no information (about the message that has been encoded) whatsoever. It's like having a binary channel with 'perfect' noise. No information can be transmitted by such channels. In classical systems we usually tend to think in terms of the noise source being a characteristic of the transmission medium, rather than the measurement process. In QM, however, using a misaligned coding and measuring set-up will be equivalent to a noise source irrespective of how perfect our coding or measurement process is.

This is the problem with the scheme above - it's not the 'coding' that's the issue - it's the retrieval of the information.

Let's suppose we attempt to send 1 bit of information again, but this time we're going to use the state $| + \rangle _Z$ to be the "1" and the state $| - \rangle _X$ to be the "0". No problem with the coding here at all. We can send our message (real information) by coding according to this state preparation procedure. But what are we going to do to retrieve (read) that message?

One simple measurement we could try is simply to measure spin-z for every incident particle. We recover the "1" symbol ( we get the result +) perfectly every time the state $| + \rangle _Z$ is sent, but with this choice of measurement we're going to get the result + or - with 50% probability whenever we measure the particles prepared in the state $| - \rangle _X$. So there's going to be a 25% error rate overall.

That's not the best we can do - to find the optimum we'd have to delve into the POVM formalism and optimize over all possible quantum measurements that could be done, but whatever we do there's going to be an unavoidable error that leads to less than one bit per particle being transmitted.

It's the fact that in order to code more than 1 bit on to a spin-1/2 particle (using the spin states as our coding basis) we'd have to select eigenstates from non-commuting operators - and therein lies the problem - it's this that prevents us from retrieving more than 1 bit per particle using this kind of coding.

 

[Blue Scallop]

Lol. Vanhees and Bill are much more expert than I - I have a rather 'old fashioned' view of QM and I struggle a bit with the more modern (and logically sound) approaches.

Vanhees and Bill are statistical ensemblers. On first impression they seemed to be shut up and calculate types.. but then what if Many worlds, Bohmians, Objective Collapse, Transactional I., etc. were all wrong. What if there was no processes or physical stuff like wave functions or even objective reality. Have you heard of Qbism, where wave function is only a tool inside our mind. Maybe Vanhees and Bills are Qbists in disguised? There is one other possibility where only statistics are real.. What if our world is just output of some infinite processing that occurs in an AdS/CFT domain.. here in the bulk only statistics or ensembles make sense.. particles or field just output or pixels of the programming? then there is no Many worlds, no Bohmians, No Objective Collapse.. Do you agree with this paragraph analysis?

About your last message:

That's not the best we can do - to find the optimum we'd have to delve into the POVM formalism and optimize over all possible quantum measurements that could be done, but whatever we do there's going to be an unavoidable error that leads to less than one bit per particle being transmitted.

It's the fact that in order to code more than 1 bit on to a spin-1/2 particle (using the spin states as our coding basis) we'd have to select eigenstates from non-commuting operators - and therein lies the problem - it's this that prevents us from retrieving more than 1 bit per particle using this kind of coding.

Is it true that superposition of up and down in the spin 1/2 particles can produce any other angles? What do you mean selecting eigenstates from from non-commuting operators?

Can you give other examples of superposition where you can encode some data in the nonorthogonal states (is nonorthogonal states a standard usage term)?

Thanks.

I'm sorry - I don't really understand what you're asking here. Is there another way of phrasing your question? By 'unitary' here, I think you mean separable. To extract any real information you need to do a measurement of some kind.

 

[Simon Phoenix]

Is it true that superposition of up and down in the spin 1/2 particles can produce any other angles? What do you mean selecting eigenstates from from non-commuting operators?

Can you give other examples of superposition where you can encode some data in the nonorthogonal states (is nonorthogonal states a standard usage term)?

A general (pure) spin state of a spin-1/2 particle can be written as $$ | \psi \rangle = a |- \rangle + b e^{ -i \alpha } | + \rangle $$ with $a$ and $b$ real such that $a^2 + b^2 = 1$. This is equivalent to some state $| - \rangle _{\theta , \phi}$ which would be an eigenstate of the spin operator $\hat {\mathbf \sigma}_{\theta , \phi}$ for some direction specified by the angles $\theta , \phi$.

But all this fancy stuff is really saying is that an eigenstate of spin in a given direction can be expanded in a different basis - just like a 2-dimensional vector (an arrow on a sheet of paper, for example) can be expanded as a superposition of any other two non parallel vectors in the same vector space.

In the quantum case it is often convenient, but not necessary, to restrict this expansion to orthogonal basis states. For a single mode EM field it is sometimes useful, for example, to expand a state in a coherent state basis. This is an overcomplete non-orthogonal basis for the mode.

I was hoping the examples I gave previously would be enough but I think you missed the main point. So let's consider what we would have to do to try to code just 2 bits of information on a single spin-1/2 particle.

To code 2 bits of information we're going to have to choose between 4 possible inputs which we can write as 00, 01, 10, and 11.

So we could choose the states $| - \rangle _Z$ and $| + \rangle _Z$ to represent the input symbols 00 and 11, respectively. But now what are we going to choose to represent the symbols 01 and 10? We've "used up" the eigenstates of spin-z. We're going to have to choose states from another basis - but these represent the eigenstates of spin in another direction. Let's pick the states $| - \rangle _X$ and $| + \rangle _X$ to represent the input symbols 01 and 10, respectively.

Now we have a problem because we've got 2 of our input symbols represented by eigenstates of $\hat {\mathbf \sigma}_Z$ and the other 2 input symbols represented by eigenstates of $\hat {\mathbf \sigma}_X$. These 2 operators don't commute - so there's no single measurement we can do that's going to allow us to perfectly distinguish between the 4 possible inputs 00,11, 01, and 10. For one thing it would violate the uncertainty principle if there was such a measurement.

The spin-z eigenstates are not orthogonal to the spin-x eigenstates - there's an overlap of squared magnitude 1/2.

If we can't perfectly distinguish between our possible input states then we can't recover the full 2 bits of information that has been coded. You can work out what you can recover from the example above (I'll leave that as an exercise for the reader). But really what we need is to work that out for all possible choices of input states and that's a more difficult calculation.

Anyway the upshot is that you can't recover more than 1 bit of information - even though we might have coded more than 1 bit.

 

[Blue Scallop]

A general (pure) spin state of a spin-1/2 particle can be written as $$ | \psi \rangle = a |- \rangle + b e^{ -i \alpha } | + \rangle $$ with $a$ and $b$ real such that $a^2 + b^2 = 1$. This is equivalent to some state $| - \rangle _{\theta , \phi}$ which would be an eigenstate of the spin operator $\hat {\mathbf \sigma}_{\theta , \phi}$ for some direction specified by the angles $\theta , \phi$.

But all this fancy stuff is really saying is that an eigenstate of spin in a given direction can be expanded in a different basis - just like a 2-dimensional vector (an arrow on a sheet of paper, for example) can be expanded as a superposition of any other two non parallel vectors in the same vector space.

In the quantum case it is often convenient, but not necessary, to restrict this expansion to orthogonal basis states. For a single mode EM field it is sometimes useful, for example, to expand a state in a coherent state basis. This is an overcomplete non-orthogonal basis for the mode.

I was hoping the examples I gave previously would be enough but I think you missed the main point. So let's consider what we would have to do to try to code just 2 bits of information on a single spin-1/2 particle.

To code 2 bits of information we're going to have to choose between 4 possible inputs which we can write as 00, 01, 10, and 11.

So we could choose the states $| - \rangle _Z$ and $| + \rangle _Z$ to represent the input symbols 00 and 11, respectively. But now what are we going to choose to represent the symbols 01 and 10? We've "used up" the eigenstates of spin-z. We're going to have to choose states from another basis - but these represent the eigenstates of spin in another direction. Let's pick the states $| - \rangle _X$ and $| + \rangle _X$ to represent the input symbols 01 and 10, respectively.

Now we have a problem because we've got 2 of our input symbols represented by eigenstates of $\hat {\mathbf \sigma}_Z$ and the other 2 input symbols represented by eigenstates of $\hat {\mathbf \sigma}_X$. These 2 operators don't commute - so there's no single measurement we can do that's going to allow us to perfectly distinguish between the 4 possible inputs 00,11, 01, and 10. For one thing it would violate the uncertainty principle if there was such a measurement.

The spin-z eigenstates are not orthogonal to the spin-x eigenstates - there's an overlap of squared magnitude 1/2.

If we can't perfectly distinguish between our possible input states then we can't recover the full 2 bits of information that has been coded. You can work out what you can recover from the example above (I'll leave that as an exercise for the reader). But really what we need is to work that out for all possible choices of input states and that's a more difficult calculation.

Anyway the upshot is that you can't recover more than 1 bit of information - even though we might have coded more than 1 bit.

Thanks for your indepth explanations. But I'd better share the following paper to know the context of my questions:

https://arxiv.org/pdf/quant-ph/9601025.pdf

quoting from page 2
"The quantum realization of a bit is a two-state quantum system—for example, a spin- 1/2
particle. A spin- 1/2 particle can be used to send one bit of classical information—and no more
than one bit—encoded in two orthogonal states, e.g., spin “down” for 0 and spin “up” for 1.
Thus it is convenient to denote the state of spin “down” by |0i and the state of spin “up” by
|1i. The difference between classical and quantum two-state systems is that quantum-mechanical
superposition gives a quantum two-state system other possible states, not available to a classical
two-state system: any linear combination of |0i and |1i is also a possible state. For a spin- 1/2
particle these states are in one-to-one correspondence with directions of the particle’s spin.

The crucial distinction between quantum and classical information appears when one attempts
to use these other states as alternatives for transmitting information. Suppose we attempt
to encode ten bits of information onto a spin- 1/2 particle, by preparing it so that its spin points in
one of 2^10 possible directions. Then we send the particle to you. Can you read out the ten bits?
Of course not. Quantum theory forbids any measurement to distinguish all 1024 possibilities."

Simon. In your messages. you were focusing on the first paragraph or orthogonal states and I understood it. I was simply asking now whether the author meant in the second paragraph as encoding the 10 bits by having 1024 angles in one spin-1/2 and trying to encode the information in one angle so that means when a spin 1/2 particle with one nonorthogonal angle say 43 degrees was send from you to me.. it's like sending 10 bits of information because of the 1024 angles? Is this what the second paragraph was describing? I need confirmation if it is.

But in superposition of spin up and spin down.. can the math really put it in a certain angle like 43 degrees? I know generally it is not possible to read nonorthogonal states... but for now just understanding the second paragraph..

 

[Simon Phoenix]

Take the paragraph above and rewrite it :

"The crucial distinction between quantum and classical information appears when one attempts
to use these other states as alternatives for transmitting information. Suppose we attempt
to encode 2 bits of information onto a spin- 1/2 particle, by preparing it so that its spin points in
one of 2^2 possible directions. Then we send the particle to you. Can you read out the 2 bits?
Of course not. Quantum theory forbids any measurement to distinguish all 4 possibilities"

I've just given you an example of an attempt to encode 2 bits using 4 possible spin directions.

Remember that it makes no sense to talk about the information encoded in a single transmission without knowledge of the associated coding scheme. So in the above case we're only able to send up to 2 bits of information per transmission because we're selecting from one of 4 possible inputs to the channel (the 4 possible spin directions).

The coding isn't the problem - we can indeed code 2 bits per transmission quite successfully per spin-1/2 particle. The problem is that we can never recover the full 2 bits at the receiver end of the channel - and in fact the best we can do is just recover 1 bit - even though we have coded 2 bits into each transmission.

 

[Blue Scallop]

Take the paragraph above and rewrite it :

"The crucial distinction between quantum and classical information appears when one attempts
to use these other states as alternatives for transmitting information. Suppose we attempt
to encode 2 bits of information onto a spin- 1/2 particle, by preparing it so that its spin points in
one of 2^2 possible directions. Then we send the particle to you. Can you read out the 2 bits?
Of course not. Quantum theory forbids any measurement to distinguish all 4 possibilities"

I've just given you an example of an attempt to encode 2 bits using 4 possible spin directions.

Remember that it makes no sense to talk about the information encoded in a single transmission without knowledge of the associated coding scheme. So in the above case we're only able to send up to 2 bits of information per transmission because we're selecting from one of 4 possible inputs to the channel (the 4 possible spin directions).

The coding isn't the problem - we can indeed code 2 bits per transmission quite successfully per spin-1/2 particle. The problem is that we can never recover the full 2 bits at the receiver end of the channel - and in fact the best we can do is just recover 1 bit - even though we have coded 2 bits into each transmission.

Again you were focusing on the first paragraph or orthogonal states and I understood it. Please address the second paragraph about encoding the 10 bits by having 1024 angles in one spin-1/2 and trying to encode the information in one angle so that means when a spin 1/2 particle with one nonorthogonal angle say 43 degrees was send from you to me.. it's like sending 10 bits of information because of the 1024 angles? Is this what the second paragraph was describing? I need confirmation if it is.

Maybe you meant since there was no way to distinguish the 1024 angles in one spin 1/2 particle and no coding scheme for it to be possible. It is not possible to encode it by choosing one angle??

 

[Simon Phoenix]

Anyway - we're going a bit off-topic with discussions of coding schemes. This thread is about the meaning and implications of unitarity in QM.

I'm going to hazard a guess where you're going with this. I get the impression (but may be very wrong on this) that you're thinking of coding 2 bits on a single spin-1/2 particle as somehow 'containing' all 4 spin directions in a single state - and then using one of these 'filter' type measurements to split this into 4 distinct paths. Is this what you had in mind? If so, that's not the right way to look at this at all.

 

[Simon Phoenix]

Please address the second paragraph

I did - you just don't yet understand why my reply does so

 

[Blue Scallop]

Anyway - we're going a bit off-topic with discussions of coding schemes. This thread is about the meaning and implications of unitarity in QM.

I'm going to hazard a guess where you're going with this. I get the impression (but may be very wrong on this) that you're thinking of coding 2 bits on a single spin-1/2 particle as somehow 'containing' all 4 spin directions in a single state - and then using one of these 'filter' type measurements to split this into 4 distinct paths. Is this what you had in mind? If so, that's not the right way to look at this at all.

Yes. Is it not possible to encode 4 spin directions in a single spin 1/2 particle? It's up, down, left and right spin. Up and down being orthogonal state and left and right being superposition nonorthogonal state. I'm not asking about decoding them.. just asking how you can encode it and if there is a way..

This is on topic because at the end of the paper it was mentioned "No matter how involved the demonstrations of
the inability to distinguish nonorthogonal quantum states become, the underlying principle is unitarity."

Don't worry. Let's transfer to another thread if I still can't understand your next message. Lol

 

[Simon Phoenix]

"No matter how involved the demonstrations of
the inability to distinguish nonorthogonal quantum states become, the underlying principle is unitarity."

Yes - that's because unitary transformations preserve overlaps. If I have two states $| \psi \rangle$ and $| \phi \rangle$ then their overlap is $\langle \psi | \phi \rangle$. This overlap is important - a zero overlap means the states are orthogonal and can be perfectly distinguished. Conversely a non-zero overlap means the states are non-orthogonal and cannot be perfectly distinguished - they can only be distinguished with a certain probability that depends on the squared magnitude of this overlap.

So if I have some unitary transformation, $\hat { \mathbf U }$, that is applied to the states $| \psi \rangle$ and $| \phi \rangle$ so that $\hat { \mathbf U } | \psi \rangle = | \psi ' \rangle$ and $\hat { \mathbf U } | \phi \rangle = | \phi ' \rangle$ then $$ \langle \psi ' | \phi ' \rangle = \langle \psi | \hat { \mathbf U } ^\dagger \hat { \mathbf U } | \phi \rangle = \langle \psi | \phi \rangle $$ where we have used the definition of unitarity here : $\hat { \mathbf U } ^\dagger = \hat { \mathbf U } ^{-1}$.

 

[Blue Scallop]

Yes - that's because unitary transformations preserve overlaps. If I have two states $| \psi \rangle$ and $| \phi \rangle$ then their overlap is $\langle \psi | \phi \rangle$. This overlap is important - a zero overlap means the states are orthogonal and can be perfectly distinguished. Conversely a non-zero overlap means the states are non-orthogonal and cannot be perfectly distinguished - they can only be distinguished with a certain probability that depends on the squared magnitude of this overlap.

So if I have some unitary transformation, $\hat { \mathbf U }$, that is applied to the states $| \psi \rangle$ and $| \phi \rangle$ so that $\hat { \mathbf U } | \psi \rangle = | \psi ' \rangle$ and $\hat { \mathbf U } | \phi \rangle = | \phi ' \rangle$ then $$ \langle \psi ' | \phi ' \rangle = \langle \psi | \hat { \mathbf U } ^\dagger \hat { \mathbf U } | \phi \rangle = \langle \psi | \phi \rangle $$ where we have used the definition of unitarity here : $\hat { \mathbf U } ^\dagger = \hat { \mathbf U } ^{-1}$.

In Wikipedia.. Unitarity 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 always equals 1."

Decoherence is supposed to preserved unitarity in quantum system.
Yet measurements can destroy unitarity.

Can you just give a simple example how when a system is not measured, and no matter how complex.. unitarity is preserved such that sum of probabilities of all possible outcomes of any event always equals 1? Let's say your eyeglasses are entangled with the environment and your body.. what does it mean the sum of all possible outcomes of any event always equals 1?

 

[Boing3000]

what does it mean the sum of all possible outcomes of any event always equals 1?

Conservation of likelihood

 

[Blue Scallop]

Conservation of likelihood

Ok. Refer to this graph of wave functions in deterministic unitarity evolution...

3^2= 9%
7^2=49%
8^2=64%
-------------
122%

It's not unity or 100%. Why?

 

[Simon Phoenix]

Can you just give a simple example how when a system is not measured, and no matter how complex.. unitarity is preserved such that sum of probabilities of all possible outcomes of any event always equals 1?

OK - let's consider a 2-level system (like a spin-1/2 particle, or a photon's polarization). We can express any pure state as a superposition of states in some basis so that $$ | \psi \rangle = \alpha |0 \rangle + \beta | 1 \rangle $$ where here $\alpha$ and $\beta$ are two complex numbers such that $| \alpha |^2 + | \beta |^2 = 1$. Now it's not too difficult to show that $| 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | = \hat { \mathbf I}$, the identity operator.

Now let's consider performing a unitary transformation $\hat { \mathbf U }$ on our state $| \psi \rangle$ so that we obtain some new state $| \phi \rangle = \hat { \mathbf U } | \psi \rangle$. This new state can be expanded in any basis and in our original basis we're going to have $$ | \phi \rangle = \gamma |0 \rangle + \delta | 1 \rangle $$ The question is whether $| \gamma |^2 + | \delta |^2 = 1$?

Now we have that $\gamma = \langle 0 | \phi \rangle$ and $\delta = \langle 1 | \phi \rangle$ and after a tiny bit of algebra we can write $$ | \gamma |^2 + | \delta |^2 = \langle \psi | \left\{ \hat { \mathbf U } | 0 \rangle \langle 0 | \hat { \mathbf U } + \hat { \mathbf U } | 1 \rangle \langle 1 | \hat { \mathbf U } \right\} | \psi \rangle $$Now we know that unitary transformations preserve these overlaps. That means that if we apply a unitary transformation to an orthonormal basis we'll obtain another orthonormal basis. So if the states $| 0 \rangle$ and $| 1 \rangle$ get transformed to $| 0' \rangle$ and $| 1' \rangle$ under the unitary transformation this means that $| 0' \rangle \langle 0' | + | 1' \rangle \langle 1' | = \hat { \mathbf I}$, so that $$ \hat { \mathbf U } | 0 \rangle \langle 0 | \hat { \mathbf U } + \hat { \mathbf U } | 1 \rangle \langle 1 | \hat { \mathbf U } = | 0' \rangle \langle 0' | + | 1' \rangle \langle 1' | = \hat { \mathbf I } $$which means that $| \gamma |^2 + | \delta |^2 = 1$

And so we see that the sum of the probabilities is preserved under unitary transformation.

 

[Blue Scallop]

OK - let's consider a 2-level system (like a spin-1/2 particle, or a photon's polarization). We can express any pure state as a superposition of states in some basis so that $$ | \psi \rangle = \alpha |0 \rangle + \beta | 1 \rangle $$ where here $\alpha$ and $\beta$ are two complex numbers such that $| \alpha |^2 + | \beta |^2 = 1$. Now it's not too difficult to show that $| 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | = \hat { \mathbf I}$, the identity operator.

Now let's consider performing a unitary transformation $\hat { \mathbf U }$ on our state $| \psi \rangle$ so that we obtain some new state $| \phi \rangle = \hat { \mathbf U } | \psi \rangle$. This new state can be expanded in any basis and in our original basis we're going to have $$ | \phi \rangle = \gamma |0 \rangle + \delta | 1 \rangle $$ The question is whether $| \gamma |^2 + | \delta |^2 = 1$?

Now we have that $\gamma = \langle 0 | \phi \rangle$ and $\delta = \langle 1 | \phi \rangle$ and after a tiny bit of algebra we can write $$ | \gamma |^2 + | \delta |^2 = \langle \psi | \left\{ \hat { \mathbf U } | 0 \rangle \langle 0 | \hat { \mathbf U } + \hat { \mathbf U } | 1 \rangle \langle 1 | \hat { \mathbf U } \right\} | \psi \rangle $$Now we know that unitary transformations preserve these overlaps. That means that if we apply a unitary transformation to an orthonormal basis we'll obtain another orthonormal basis. So if the states $| 0 \rangle$ and $| 1 \rangle$ get transformed to $| 0' \rangle$ and $| 1' \rangle$ under the unitary transformation this means that $| 0' \rangle \langle 0' | + | 1' \rangle \langle 1' | = \hat { \mathbf I}$, so that $$ \hat { \mathbf U } | 0 \rangle \langle 0 | \hat { \mathbf U } + \hat { \mathbf U } | 1 \rangle \langle 1 | \hat { \mathbf U } = | 0' \rangle \langle 0' | + | 1' \rangle \langle 1' | = \hat { \mathbf I } $$which means that $| \gamma |^2 + | \delta |^2 = 1$

And so we see that the sum of the probabilities is preserved under unitary transformation.

Wow, our future science advisor you are.
Well. Let's go conceptual for us without Einstein math abilities.

Many Worlds preserves unitarity because the wave function goes on and goes.
Bohmian mechanics preserves unitarity because the wave function also goes on and on but only one branch is manifested.
Copenhagen doesn't preserve unitarity because measurement collapses the wave function or destabilizes the unitarity.

Now for Many Worlds. If somehow there is a way to perceive all branches at once. Why does this violate unitarity? Can't we say that our probes would entangle with the systems and unitarity is preserved. Or what kind of probing all branches can be allowed conceptually that would still preserve unitarity?

 

[vanhees71]

Another way to put it: Anything defining a system within QT, i.e., the Hilbert space and operator algebra is invariant under unitary and antiunitary transformations, and according to Wigner's theorem any symmetry is realized either by a unitary or antiunitary transformation. Only discrete symmetry groups can be realized by antiunitary transformations. Since time-translation invariance, generated by the Hamiltonian of the system, is a continuous symmetry, it must be realized by unitary transformations.

 

[Simon Phoenix]

Anything defining a system within QT, i.e., the Hilbert space and operator algebra is invariant under unitary and antiunitary transformations, and according to Wigner's theorem any symmetry is realized either by a unitary or antiunitary transformation. Only discrete symmetry groups can be realized by antiunitary transformations. Since time-translation invariance, generated by the Hamiltonian of the system, is a continuous symmetry, it must be realized by unitary transformations.

I wish I could double like this - lovely answer

 

[Simon Phoenix]

Well. Let's go conceptual for us without Einstein math abilities.

I wish I could - but it's very difficult to answer the questions you're asking without some mathematical detail. To go beyond pop science accounts of QM there's really no way to do it (at least it's beyond my capabilities) and you're going to need to get to grips with at least some of the formalism - otherwise everything stays at the pop science level - and that's damnably difficult to get right for QM without being misleading in some way.