Index Notation & Dirac Notation

In summary: It simply does not seem very informative and I find myself turned off by it about an otherwise fascinating subject.
  • #71
Nugatory said:
There's a strong element of personal taste here, and... de gustibus non est disputandum.

But I have to say that I'm siding with what I'm hearing from TheAustrian and WbN. Classical mechanics, E&M, SR/GR, I look at the math and map it back into a physical interpretation of how the world works and I think I understand, but QM... Not so much. It's only in QM that "shut up and calculate" is sound advice, and only in QM that all interpretations fall short in one way or another and I'm stuck with what I get when I calculate.

I think that there is a distinction between QM and theories such as GR. In the case of GR, or Newtonian physics, or Maxwell's equations, or just about any theory besides QM, the theory is taken to be a description of what the world is like. Of course, the description could be wrong, or it could be incomplete, or it could be an idealization, but the theories are about what's going on in the world. It says that there are things such as particles and fields and spacetime, and those things have certain properties that evolve according to certain equations.

Quantum mechanics is different, in that it doesn't seem to be making any claims about what the world is like. The entities that you calculate with--wave functions and probability amplitudes--are not claimed to be entities in the real world at all, nor are they claimed to be descriptions of objects in the real world. They are simply calculation tools for making predictions.

Some people say that that's all you need from a theory of physics--a way to make accurate predictions. I guess in some sense that's true. But I think that it's unsatisfying to people who are interested in physics because they want to understand the world.

Let me give an analogy, which might sound unfair, but it seems accurate to me. Suppose that someone came up with a fortune-telling card game, something like Tarot that was ACTUALLY accurate. If you wanted to know whether it would rain tomorrow, you shuffled the deck and dealt out the cards and interpreted the results according to certain rules. If there were such a card game that made repeatable, accurate predictions about weather would people consider that a scientific theory of the weather? Some might, but many would not.

Many people say that the reason we reject as unscientific those superstitious techniques for telling the future such as Tarot, astrology, reading tea-leaves, etc, is because they are either falsified, or else too vague to be falsifiable. But I think people would not consider them scientific theories even if they were accurate, because it does not give us a reasonable understanding of why the predictions are accurate. How does the arrangement of Tarot cards manage to correlate with future events? How does the arrangement of stars and planets manage to correlate with events on Earth? Even if these techniques were accurate, they wouldn't be considered scientific theories, it would just be more data that a scientific theory would need to explain: why those fortune-telling techniques work.

Quantum mechanics isn't quite in the same boat, but emotionally it feels similar to some people. It gives accurate predictions, but it seems to some people that its accurate predictions are more data that needs explaining, rather than a fundamental theory.
 
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  • #72
Getting back to Dirac notation, the point I made a while ago was that it's quite elegant for single-particles, but is ugly when you're dealing with multiple particles.

Suppose you have a two-particle state [itex]|\Psi\rangle \otimes |\Phi \rangle[/itex], is the adjoint written as [itex]\langle \Phi | \otimes \langle \Psi |[/itex] or [itex]\langle \Psi | \otimes \langle \Phi |[/itex]. I'm sure there is a convention for that, but it's pretty arbitrary. Then if you take the inner product of two such composite states:

[itex](\langle \Phi' | \otimes \langle \Psi' |) (|\Psi\rangle \otimes |\Phi \rangle)[/itex]

do you get

[itex]\langle \Phi' | \Psi\rangle \langle \Psi' |\Phi \rangle[/itex]

or

[itex]\langle \Psi' | \Psi\rangle \langle \Phi' |\Phi \rangle[/itex]

Einstein's abstract index notation makes it clear what is being contracted with what.
 
  • #73
stevendaryl said:
Getting back to Dirac notation, the point I made a while ago was that it's quite elegant for single-particles, but is ugly when you're dealing with multiple particles.

Suppose you have a two-particle state [itex]|\Psi\rangle \otimes |\Phi \rangle[/itex], is the adjoint written as [itex]\langle \Phi | \otimes \langle \Psi |[/itex] or [itex]\langle \Psi | \otimes \langle \Phi |[/itex]. I'm sure there is a convention for that, but it's pretty arbitrary. Then if you take the inner product of two such composite states:

[itex](\langle \Phi' | \otimes \langle \Psi' |) (|\Psi\rangle \otimes |\Phi \rangle)[/itex]

do you get

[itex]\langle \Phi' | \Psi\rangle \langle \Psi' |\Phi \rangle[/itex]

or

[itex]\langle \Psi' | \Psi\rangle \langle \Phi' |\Phi \rangle[/itex]

Einstein's abstract index notation makes it clear what is being contracted with what.

very insightful post.
How would these things translate into index notation?
 
  • #74
TheAustrian said:
Our lecturer hated maths. So he just talked a lot about various things. He mentioned that you can learn QM without knowing anything else about Physics, and that when you do QM, you can pretty much forget about everything you've learned so far. We also made some connections to the Lagrangian, and some research paper by Richard Feynman.

To be honest, he was not a good lecturer, and less than 60% of the class passed.

You can approach QM "mathematically", but the physics is not there. To understand QM conceptually, one only needs very simple maths: linear algebra in a finite dimensional vector space with an inner product. The extension to infinite dimensional vector spaces does require finesse and "maths" that you can pick up in Ballentine, but that will not change the physics.

The basic thing about QM (let's talk about Copenhagen or "shut-up-and-calculate"), unlike classical physics, is that it is not a theory of the whole universe. It requires a common sense division of the universe into a macroscopic, non-quantum part in which you and your measuring apparatus are included, and a quantum part. You can see this need below in that quantum mechanics has two time evolution rules - Schroedinger equation and wave function collapse. The wave function collapse occurs when a measurement occurs, ie. when the macroscopic and quantum worlds interact, and the quantum system leaves a macroscopic mark in the macroscopic world.

(1) The state of the quantum part is a ray in a vector space, and can be represented as a unit vector.

(2) The time evolution of the state, as long as no measurement is made, is given by the Schroedinger equation.

(3) To each measurable quantity there is a corresponding operator. When a measurement is made, the outcome of the measurement an eigenvalue of the operator. The probability of a particular outcome is given by the Born Rule, and the state evolution is momentarily not described by the Schroedinger equation. Instead the state collapses into an eigenstate corresponding to the observed eigenvalue. After the measurement, the state evolution is again described by the Schroedinger equation.

Why is this division the most "physical" thing? All our theories are incomplete - Newtonian gravity, E&M in flat spacetime, GR are incomplete. But none of those announce their incompleteness. QM does.
 
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  • #75
atyy said:
You can approach QM "mathematically", but the physics is not there. To understand QM conceptually, one only needs very simple maths: linear algebra in a finite dimensional vector space with an inner product. The extension to infinite dimensional vector spaces does require finesse and "maths" that you can pick up in Ballentine, but that will not change the physics.

The basic thing about QM (let's talk about Copenhagen or "shut-up-and-calculate"), unlike classical physics, is that it is not a theory of the whole universe. It requires a common sense division of the universe into a macroscopic, non-quantum part in which you and your measuring apparatus are included, and a quantum part. You can see this need below in that quantum mechanics has two time evolution rules - Schroedinger equation and wave function collapse. The wave function collapse occurs when a measurement occurs, ie. when the macroscopic and quantum worlds interact, and the quantum system leaves a macroscopic mark in the macroscopic world.

(1) The state of the quantum part is a ray in a vector space, and can be represented as a unit vector.

(2) The time evolution of the state, as long as no measurement is made, is given by the Schroedinger equation.

(3) To each measurable quantity there is a corresponding operator. When a measurement is made, the outcome of the measurement an eigenvalue of the operator. The probability of a particular outcome is given by the Born Rule, and the state evolution is momentarily not described by the Schroedinger equation. Instead the state collapses into an eigenstate corresponding to the observed eigenvalue. After the measurement, the state evolution is again described by the Schroedinger equation.

Why is this division the most "physical" thing? All our theories are incomplete - Newtonian gravity, E&M in flat spacetime, GR are incomplete. But none of those announce their incompleteness. QM does.

This might be a stupid question, but what exactly is a wave-function collapse? Does it have something to do with expectation values? When I calculate an expectation value, is that some probability that I will obtain that value at a wavefunction collapse or its something else? I think I understood parts (1) and (2).
 
  • #76
TheAustrian said:
This might be a stupid question, but what exactly is a wave-function collapse? Does it have something to do with expectation values? When I calculate an expectation value, is that some probability that I will obtain that value at a wavefunction collapse or its something else? I think I understood parts (1) and (2).

Quantum mechanics predicts expectation values. It can predict what the possible outcomes are for a measurement, but it cannot predict the exact outcome on any particular measurement trial. Expectation values are basically averages of a large number (infinite) of outcomes over many repeated measurement trials.

However, on any particular trial, we will get a particular answer. The answer will be one of the eigenvalues of the observable. Quantum mechanics predicts that the state on that single trial after the measurement outcome is observed, is the eigenstate corresponding to the observed eigenvalue. This jump from the state before that particular measurement trial into one of the eigenstates does not obey the Schroedinger equation, and is called wave function collapse.
 
  • #77
You need to bear in mind that the wavefunction is a probability amplitude. You could describe a spinning coin in a similar way. When you make a measurement the function collapses to one of the allowed values. Nothing very sinister.
 
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  • #78
atyy said:
Quantum mechanics predicts expectation values. It can predict what the possible outcomes are for a measurement, but it cannot predict the exact outcome on any particular measurement trial. Expectation values are basically averages of a large number (infinite) of outcomes over many repeated measurement trials.

However, on any particular trial, we will get a particular answer. The answer will be one of the eigenvalues of the observable. Quantum mechanics predicts that the state on that single trial after the measurement outcome is observed, is the eigenstate corresponding to the observed eigenvalue. This jump from the state before that particular measurement trial into one of the eigenstates does not obey the Schroedinger equation, and is called wave function collapse.

So the measurement will be one of the solutions of the Schrodinger Equation?

What are expectation values good for then?

My problem is this, I can do these things, but don't know Why I am doing it.
 
  • #79
TheAustrian said:
very insightful post.
How would these things translate into index notation?

I don't know, because I've never seen QM developed using abstract notation. But I would think it would go something like this:

You represent a state [itex]|\Psi \rangle[/itex] by its components in an unspecified basis:

[itex]\Psi^\alpha[/itex]

Similarly, you represent a dual vector [itex]\langle \Phi |[/itex] by its components, but using lowered indices: [itex]\Phi_\alpha[/itex]. Then the inner product would be written as:

[itex]\langle \Phi | \Psi \rangle = \Phi_\alpha \Psi^\alpha[/itex]

where repeated indices, one lowered and one raised implies a summation or integral.

Composite states would be represented using multiple indices:

[itex]|\Phi\rangle \otimes |\Psi \rangle \Rightarrow \Phi^\alpha \Psi^\beta[/itex]

and the inner product of two such states:

[itex](\langle \Phi' | \otimes \langle \Psi |) (|\Phi\rangle \otimes |\Psi \rangle) \Rightarrow \Phi'_\alpha \Psi'_\beta \Phi^\alpha \Psi^\beta[/itex]
 
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  • #80
Jilang said:
You need to bear in mind that the wavefunction is a probability amplitude. You could describe a spinning coin in a similar way. When you make a measurement the function collapses to one of the allowed values. Nothing very sinister.

Except that in the case of ordinary probability applied to something like a spinning coin, we can make the distinction between the state of the coin, and the state of our knowledge about the coin. Measurement changes the state of our knowledge, but it doesn't change the state of the coin. In quantum mechanics, the wave function is neither clearly about our knowledge, nor is it clearly about the system being studied. It's a strange mix of the two. Attempts to disentangle the two have been unsatisfactory.
 
  • #81
TheAustrian said:
very insightful post.
How would these things translate into index notation?
Why "index" notation, and did you really mean the abstract index notation used for tensors, or are you just talking about vector components and stuff?

The alternative to bra-ket notation (if we're just dealing with a Hilbert space H and its dual space H*) is to write an element of H as x rather than |x>. Then we can use the notation <x,y> for the inner product of x and y. A bra is a map of the form ##y\mapsto \langle x,y\rangle## with domain H. It can be written as ##\langle x,\cdot\rangle##, if we absolutely want to write it out, and not have to invent a new symbol for each bra.

A tensor product would just be written as ##x\otimes y## rather than ##|x\rangle\otimes|y\rangle##.

stevendaryl said:
Suppose you have a two-particle state [itex]|\Psi\rangle \otimes |\Phi \rangle[/itex], is the adjoint written as [itex]\langle \Phi | \otimes \langle \Psi |[/itex] or [itex]\langle \Psi | \otimes \langle \Phi |[/itex]. I'm sure there is a convention for that,
I've been wondering about that too. :smile: It's a bit embarassing that I don't even know if one of the two obvious conventions is considered standard.
 
  • #82
TheAustrian said:
What are expectation values good for then?
The expectation value in state ##\rho## of an observable ##A## is the average result of a long sequence of measurements with devices represented by ##A## on systems represented by ##\rho##.

This is the assumption that turns the mathematics into physics.
 
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  • #83
Fredrik said:
Why "index" notation, and did you really mean the abstract index notation used for tensors, or are you just talking about vector components and stuff?

The point of the abstract index notation for tensors is to make it clear what is getting contracted with what. In the QM case, it's simple enough to figure out if the state is simple:

[itex]\langle \Psi | \Phi \rangle[/itex] obviously means the inner product of

[itex]|\Phi \rangle[/itex] with [itex]|\Psi \rangle[/itex]

But for composite states, I think it can be difficult to know what's multiplying what. Abstract indices make this clear.
 
  • #84
stevendaryl said:
I don't know, because I've never seen QM developed using abstract notation. But I would think it would go something like this:

You represent a state [itex]|\Psi \rangle[/itex] by its components in an unspecified basis:

[itex]\Psi^\alpha[/itex]

Similarly, you represent a dual vector [itex]\langle \Phi |[/itex] by its components, but using lowered indices: [itex]\Phi_\alpha[/itex]. Then the inner product would be written as:

[itex]\langle \Phi | \Psi \rangle = \Phi_\alpha \Psi^\alpha[/itex]

where repeated indices, one lowered and one raised implies a summation or integral.

Composite states would be represented using multiple indices:

[itex]|\Phi\rangle \otimes |\Psi \rangle \Rightarrow \Phi^\alpha \Psi^\beta[/itex]

and the inner product of two such states:

[itex](\langle \Phi' | \otimes \langle \Psi |) (|\Phi\rangle \otimes |\Psi \rangle) \Rightarrow \Phi'_\alpha \Psi'_\beta \Phi^\alpha \Psi^\beta[/itex]

This seems like a sensible way to tackle the problem. Thanks everyone for the help. I've really learned a LOT from this forum discussion. My UG lectures on QM were worthless to be honest.
 
  • #85
TheAustrian said:
So the measurement will be one of the solutions of the Schrodinger Equation?

No. The measurement outcome will be an eigenvalue of the measurement operator. The state after the measurement will be the eigenstate corresponding to the eigenvalue that was the outcome.

The Schroedinger equation tells you how the state evolves between measurements.

TheAustrian said:
What are expectation values good for then?

Quantum mechanics only predicts probabilities or expectation values. An expectation value is simply an "average". Statistical mechanics and Mendelian genetics are two other theories that only give you probabilities or averages. These theories are all useful, even though they only predict expectation values, ie. average quantities.

TheAustrian said:
problem is this, I can do these things, but don't know Why I am doing it.

Think about statistical mechanics. There every physical quantity calculated is an expectation value. It is an average. It tells you the average value if you do the measurement many times. In the same way that statistical mechanics is a useful theory although it only makes predictions about averaged quantities, quantum mechanics is also useful for making predictions about averages. Of course, it is you that has to choose which quantities you are interested in measuring, and quantum mechanics will give you the answer about their average values.
 
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  • #86
atyy said:
Query about Ballentine: Does the book even mention that when one does QM, one divides the universe into classical and quantum parts

Explicitly - no. Implicitly - yes.

Thanks
Bill
 
  • #87
bhobba said:
Explicitly - no. Implicitly - yes.

Thanks
Bill

Indeed. But implicit to whom, I would ask. Does Ballentine know this, or just bhobba? :smile:
 
  • #88
atyy said:
You can approach QM "mathematically", but the physics is not there. To understand QM conceptually, one only needs very simple maths: linear algebra in a finite dimensional vector space

Actually Atyy it can be expressed much more elegantly than that - just one axiom and here it is.

A quantum observation is described by a set of positive operators Ei ∑ Ei = 1 such that the probability of outcome i is determined by the Ei, and only the Ei.

One then invokes a really nice proof of Born from that.

The reason it seems like just math is you can get by with a very vague idea what a measurement/observation is. That's what to a large extent Ballentine does - but he does relate it to an actual archetypical kind of measurement apparatus.

But if you want to pursue it best to start a new thread.

Thanks
Bill
 
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  • #89
stevendaryl said:
Suppose you have a two-particle state [itex]|\Psi\rangle \otimes |\Phi \rangle[/itex], is the adjoint written as [itex]\langle \Phi | \otimes \langle \Psi |[/itex] or [itex]\langle \Psi | \otimes \langle \Phi |[/itex]. I'm sure there is a convention for that, but it's pretty arbitrary.

To me it seems natural to write it as [itex]\langle\Psi | \otimes \langle \Phi|[/itex], since for (finite or infinite-dimensional) vector spaces [itex]V,W[/itex] and their duals [itex]V^*,W^*[/itex], [itex](V\otimes W)^*\cong V^*\otimes W^*[/itex]. Just like when you have an operator defined as a tensor products of operators that act separately in V and W, say, [itex]S_V\otimes \mathrm{id}_W[/itex] (with a "physically relevant" name & form), operate on a product state and you want to have the different operators in the "correct" order, you'd want to have the linear functionals of the dual spaces in the "correct" order when they act on the product state to form a bra-ket.

But I have to say, even though I personally like the notation for the purposes of QM, I had never given it that much thought. Maybe others have different conventions?
 
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  • #90
atyy said:
Indeed. But implicit to whom, I would ask. Does Ballentine know this, or just bhobba? :smile:

It's pretty obvious when you think about the archetypical observation he talks about. Nothing mysterious about it - he simply leaves it up in the air as being utterly obvious the apparatus is classical.

Thanks
Bill
 
  • #91
atyy said:
No. The measurement outcome will be an eigenvalue of the measurement operator. The state after the measurement will be the eigenstate corresponding to the eigenvalue that was the outcome.

The Schroedinger equation tells you how the state evolves between measurements.



Quantum mechanics only predicts probabilities or expectation values. An expectation value is simply an "average". Statistical mechanics and Mendelian genetics are two other theories that only give you probabilities or averages. These theories are all useful, even though they only predict expectation values, ie. average quantities.



Think about statistical mechanics. There every physical quantity calculated is an expectation value. It is an average. It tells you the average value if you do the measurement many times. In the same way that statistical mechanics is a useful theory although it only makes predictions about averaged quantities, quantum mechanics is also useful for making predictions about averages. Of course, it is you that has to choose which quantities you are interested in measuring, and quantum mechanics will give you the answer about their average values.

So expectation values are not particularly important? Wavefunctions, eigenstates and eigenvalues are the important stuff?
 
  • #92
TheAustrian said:
This might be a stupid question, but what exactly is a wave-function collapse? Does it have something to do with expectation values? When I calculate an expectation value, is that some probability that I will obtain that value at a wavefunction collapse or its something else? I think I understood parts (1) and (2).

Ballentine explains it. Enough said.

When you have gone through it we can have a real nifty chat about it and something called Gleasons theroem. It's associated with being able to derive Borns rule from that single axiom I mentioned to Atyy

Thanks
Bill
 
  • #93
TheAustrian said:
So expectation values are not particularly important? Wavefunctions, eigenstates and eigenvalues are the important stuff?

Expectation values, which are just another name for average values, are very important. Quantum mechanics only predicts probabilities. Equivalently, quantum mechanics only predicts expectation values or averages.

Let's say you measure an observable that corresponds to an operator which has only 2 eigenvalues 1 and -1. This means that the outcome of a measurement can only be either 1 or -1. Let's say that quantum mechanics predicts that 60% of the time you get 1, and 40% of the time you get -1. The expectation value or average value of the outcomes for this measurement is (0.6)(1) + (0.4)(-1) = 0.2.
 
  • #94
bhobba said:
It's pretty obvious when you think about the archetypical observation he talks about. Nothing mysterious about it - he simply leaves it up in the air as being utterly obvious the apparatus is classical.
This is not correct. His quantum treatment of the apparatus leads to equation (9.8) about which he says "This final state is a coherent superposition of macroscopically distinct indicator eigenvectors [...]" and "In a typical case, the indicator variable αr might be the position of a needle on a meter or a mark on a chart recorder, and for two adjacent values of the measured variable, r and r , the separation between αr and αr could be several centimeters." and concludes "Each member system of the ensemble consists of an object and a measuring apparatus."

The macroscopic superpositions in the measurement process are the very reason why he takes the ensemble viewpoint.
 
  • #95
atyy said:
Let's say you measure an observable that corresponds to an operator which has only 2 eigenvalues 1 and -1. This means that the outcome of a measurement can only be either 1 or -1. Let's say that quantum mechanics predicts that 60% of the time you get 1, and 40% of the time you get -1. The expectation value or average value of the outcomes for this measurement is (0.6)(1) + (0.4)(-1) = 0.2.

Of course in this example the expectation value is not as interesting as the probability distribution from which it is calculated - the average American house may have 2.8 bedrooms, but I've never seen a house with a fractional bedroom. When the spectrum is continuous (as it is for the position of an unbound particle, for example) the expectation value becomes a much more useful quantity.

(I'm not disagreeing with atyy here, just extending his answer to TheAustrian's question about whether the expectation values or or the eigenvalues are "the important stuff").
 
  • #96
Nugatory said:
Of course in this example the expectation value is not as interesting as the probability distribution from which it is calculated - the average American house may have 2.8 bedrooms, but I've never seen a house with a fractional bedroom. When the spectrum is continuous (as it is for the position of an unbound particle, for example) the expectation value becomes a much more useful quantity.

(I'm not disagreeing with atyy here, just extending his answer to TheAustrian's question about whether the expectation values or or the eigenvalues are "the important stuff").

Yes. I left this point out for simplicity. One could choose to say that the probability distribution of outcomes, ie. the probability distribution of eigenvalues, is more fundamental. It's a matter of taste, because for physically reasonable distributions, the probability distribution and a full set of certain expectation values called cumulants are equivalent. The cumulants are a set of expectation values that can be generated by Taylor series, and whose low order terms are the mean, variance, skew, kurtosis etc.
 
  • #97
TheAustrian said:
So expectation values are not particularly important? Wavefunctions, eigenstates and eigenvalues are the important stuff?

I understand where your state of distraught comes from. In statistical mechanics we take the partition function of a system at equilibrium with an external heat bath and calculate quantities like average pressure, magnetic susceptibility, average energy etc. which all clearly have very important physical applications and are put to use extensively everyday in calculations.

In QM, if I have a pure state, what use are the expectation values of all the observables relevant to the system? What can I actually do with these expectation values? Sure I can calculate them at whim but can I put actually them to use in other calculations of physical applications? Do they hold the same all-important place that they do in statistical thermodynamics? You probably remember them showing up in non-degenerate and degenerate first order time-independent perturbation theory but where else do they show up?

Are these questions more representative of your own?
 
  • #98
TheAustrian said:
So expectation values are not particularly important? Wavefunctions, eigenstates and eigenvalues are the important stuff?
It's at least as important as all the things you mentioned. As I said in post #82, the assumption about the significance of expectation values is essentially what turns the mathematics of QM into physics.


atyy said:
Expectation values, which are just another name for average values,
I have to object to this part. The average value and the expectation value are certainly not defined the same way. That they have the same value is an extremely important assumption, an assumption that's part of the definition of QM.

Same thing with the probability of a possible result and the relative frequency of that result. If you want, you can replace the assumption that expectation values are equal to average values with an assumption that says that probabilities are equal to relative frequencies. These two correspondence rules are essentially equivalent.

There's a similar thing in SR. It always bugs me when people say that proper time is what clocks measure. It's not, at least not by definition. Proper time is the number you get when you integrate a certain function along a timelike curve. That this is equal to what clocks measure is a major assumption. It's the most important correspondence rule in both SR and GR. It's part of the definitions of those theories.
 
  • #99
Fredrik said:
I have to object to this part. The average value and the expectation value are certainly not defined the same way. That they have the same value is an extremely important assumption, an assumption that's part of the definition of QM.

Same thing with the probability of a possible result and the relative frequency of that result. If you want, you can replace the assumption that expectation values are equal to average values with an assumption that says that probabilities are equal to relative frequencies. These two correspondence rules are essentially equivalent.

I don't agree. The expectation and the average value are the same by definition in probability. One is the formal term, the other is the informal term.

I do agree that it is an additional assumption to say that probability is operationally defined as relative frequency.
 
  • #100
atyy said:
I don't agree. The expectation and the average value are the same by definition in probability. One is the formal term, the other is the informal term.

I do agree that it is an additional assumption to say that probability is operationally defined as relative frequency.

Well, whatever it is that you want to call the expression [itex]\langle \Phi|O|\Phi \rangle[/itex], it's an assumption that it's equal to the average value of observable [itex]O[/itex] in the state described by [itex]|\Phi\rangle[/itex].
 
  • #101
stevendaryl said:
Well, whatever it is that you want to call the expression [itex]\langle \Phi|O|\Phi \rangle[/itex], it's an assumption that it's equal to the average value of observable [itex]O[/itex] in the state described by [itex]|\Phi\rangle[/itex].

But it is not an assumption different from the assumption that it is the expectation value of the observable.
 
  • #102
atyy said:
I don't agree. The expectation and the average value are the same by definition in probability. One is the formal term, the other is the informal term.
Nitpicking a bit: Both have formal definitions, and they're different. There are theorems ("laws of large numbers") that prove that the average converges in at least two different ways to the expectation value.

I was ignoring that there's a formal definition of "average". My point was that regardless of what mathematical terms we define, the physics is in the assumptions that tell us how the math is related to the real world. I consider this a supremely important fact; it's the very foundation of the philosophy of mathematics and science. Because of this, I find it rather odd that pretty much everyone but me (definitely not just you) are going out of their way to avoid mentioning these correspondence rules explicitly. Instead they seem to prefer to connect mathematics to reality simply by using the same term for the real-world thing and the corresponding mathematical thing. This is what people do with "proper time" in the relativity forum. I don't like it because it hides what we're really doing.

Edit: Even though "average" and "expectation value" are essentially the same in pure mathematics (because of the laws of large numbers), there's a difference between these concepts and the real-world average, i.e. the average of the measurement results in an actual experiment. The assumption that the real-world average is equal to a number given by a mathematical formula, is a fundamental assumption. This correspondence rule is a hugely important part of the definition of the theory.

atyy said:
I do agree that it is an additional assumption to say that probability is operationally defined as relative frequency.
Probability is typically defined something like this: Let ##X=\{1,2,3,4,5,6\}##. I'll use the notation ##|E|## for the number of elements of ##E##. For each ##E\subseteq X##, define ##P(E)=|E|/|X|##. For each ##E\subseteq X##, the number ##P(E)## is called the probability of E. Note that there's no connection to the real world whatsoever. The probability measure P is just an assignment of numbers in the interval [0,1] to subsets of some set X. Now we can turn this piece of pure mathematics into a falsifiable theory about the real world with a few simple assumptions. Some of them are common to all probability theories. But at least one assumption is part of the definition of the specific probability theory. In our case, that assumption can be that the elements of the set X correspond to the possible results of a roll of a six-sided die.

stevendaryl said:
Well, whatever it is that you want to call the expression [itex]\langle \Phi|O|\Phi \rangle[/itex], it's an assumption that it's equal to the average value of observable [itex]O[/itex] in the state described by [itex]|\Phi\rangle[/itex].
atyy said:
But it is not an assumption different from the assumption that it is the expectation value of the observable.
If "average" refers to the average of the results of actual measurements, then I would say that it's definitely a different assumption. Not only that. It's a very different kind of assumption.

There's a mathematical definition that associates the term "expectation value" with that number. There's a mathematical definition that associates the term "average" with something else. There are theorems that tell us how the average converges to the expectation value. But all of this is pure mathematics. The physics is in the correspondence rules, not in the mathematics, and in this case, the correspondence rule is the assumption that the expectation value is equal to the real-world average (not just the "formal average", i.e. the number that the laws of large numbers are making claims about).
 
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  • #103
Fredrik said:
Nitpicking a bit: Both have formal definitions, and they're different. There are theorems ("laws of large numbers") that prove that the average converges in at least two different ways to the expectation value.

I was ignoring that there's a formal definition of "average". My point was that regardless of what mathematical terms we define, the physics is in the assumptions that tell us how the math is related to the real world. I consider this a supremely important fact; it's the very foundation of the philosophy of mathematics and science. Because of this, I find it rather odd that pretty much everyone but me (definitely not just you) are going out of their way to avoid mentioning these correspondence rules explicitly. Instead they seem to prefer to connect mathematics to reality simply by using the same term for the real-world thing and the corresponding mathematical thing. This is what people do with "proper time" in the relativity forum. I don't like it because it hides what we're really doing.

I agree with your point, but not your nitpicking. The link you give always uses the term "sample average" in the theorems, not the unqualified "average". The main quirk with my terminology is that the unqualified "average" most usually means "mean", which is simply a particular expectation value, ie. the first cumulant. However, I meant it as a synonym for "expectation value". For example, in my colloquial way of using the word "average", the variance is the average of the squared deviation from the mean. Basically, in all my statements I am assuming (1) Kolmogorov axioms (2) relative frequency interpretation of probability, but that is even before quantum mechanics, ie. I assume it for Mendelian genetics and classical statistical mechanics. My assumption (2) is the same as what you are calling the "correspondence" between maths and physics.
 
  • #104
Fredrik said:
Nitpicking a bit: Both have formal definitions, and they're different. There are theorems ("laws of large numbers") that prove that the average converges in at least two different ways to the expectation value.

True. But they are in probability (the weak law) and almost surely (the strong law). To apply them you are going to have to make some assumption such as for all practical purposes if a probability is infinitesimally close to one you can take it as a dead cert. People tend to do that sort of thing unconsciously without even realizing it, such is the very intuitive understanding we all have of probability.

Fredrik said:
My point was that regardless of what mathematical terms we define, the physics is in the assumptions that tell us how the math is related to the real world. I consider this a supremely important fact; it's the very foundation of the philosophy of mathematics and science.

You are correct.

But in applied math, especially with regard to probability, people generally aren't that careful and you make all sorts of unconscious assumptions you generally aren't even aware of when applying it.

That's the reason QM can seem like applied math rather than physics. Its we have this very intuitive understanding of probability that gets us by.

I think most physicists, unlike applied math guys, haven't even done an advanced course on probability and statistical modelling and get by just fine. Pure math guys - that's another story - there are very subtle issues associated with a rigorous approach to probability such as the proof a Wiener process actually exists, is continuous but nowhere differentiable.

In QM as far as an axiomatic treatment is concerned I use Kolmogorov's axioms. Each interpretation, in applying it, has a slightly different take on how its done eg the ensemble interpretation is very frequentest in character, but Copenhagen is rather Bayesian.

Thanks
Bill
 
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  • #105
WannabeNewton said:
I understand where your state of distraught comes from. In statistical mechanics we take the partition function of a system at equilibrium with an external heat bath and calculate quantities like average pressure, magnetic susceptibility, average energy etc. which all clearly have very important physical applications and are put to use extensively everyday in calculations.

In QM, if I have a pure state, what use are the expectation values of all the observables relevant to the system? What can I actually do with these expectation values? Sure I can calculate them at whim but can I put actually them to use in other calculations of physical applications? Do they hold the same all-important place that they do in statistical thermodynamics? You probably remember them showing up in non-degenerate and degenerate first order time-independent perturbation theory but where else do they show up?

Are these questions more representative of your own?

Why do I solve the Schrodinger Equation for? Why are Energy levels important? etc. I know nothing about the theory behind QM. I know only how to calculate questions that come up.
 

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