Index Notation & Dirac Notation

[Jilang]

They are only labels for the state. I am only guessing, but what else could they mean? What would be a normal functional form - I'm not sure what you mean by this as wouldn't any notation require some sort of labelling?

 

[TheAustrian]

They are only labels for the state. I am only guessing, but what else could they mean? What would be a normal functional form - I'm not sure what you mean by this as wouldn't any notation require some sort of labelling?

I think you have the meanings right, but like I said, I don't know it for sure because I don't understand this notation.
Hmm... I mean if you wanted to write it with not bra-ket notation, but just how you otherwise mathematics?

 

[micromass]

I think you have the meanings right, but like I said, I don't know it for sure because I don't understand this notation.
Hmm... I mean if you wanted to write it with not bra-ket notation, but just how you otherwise mathematics?

Bra-ket notation is certainly not necessary here. The mathematics behind QM is essentially functional analysis. And most functional analysis texts do not use the bra-ket notation. They use a notation that I think is much simpler and isn't prone to errors. You can try out the excellent book by Kreyszig: https://www.amazon.com/dp/0471504599/?tag=pfamazon01-20

However, most physics texts do use bra-ket notation. I think this is a sad thing, but it's the way it is. It has become standard in physics. So you need to learn it anyway. It will really harm you later down the road not to learn and get used to bra-ket notation. Only a very few books don't use the notation and they are usually written for mathematicians, like https://www.amazon.com/dp/146147115X/?tag=pfamazon01-20 and https://www.amazon.com/dp/0486453278/?tag=pfamazon01-20

 

[Jilang]

I think you have the meanings right, but like I said, I don't know it for sure because I don't understand this notation.
Hmm... I mean if you wanted to write it with not bra-ket notation, but just how you otherwise mathematics?

You would write a ψ with the same subscripts I would think. They may be replaced with i,j,k,l,m etc, but standing for the same physical labels.

 

[TheAustrian]

You would write a ψ with the same subscripts I would think. They may be replaced with i,j,k,l,m etc, but standing for the same physical labels.

Thanks. So basically, that expression is an insanely complicated wavefunction, that which we do not know how it looks like?

 

[Jilang]

Lol, that involves solving the Schroedinger equation. This can be easy or incredibly hard! A good start is Griffiths.

 

[TheAustrian]

I see. I know, it's usually not possible to find analytical solution, I'm aware, I just had problems in understanding what the notation meant, thanks for the clarification :D

 

[Matterwave]

Thanks. So basically, that expression is an insanely complicated wavefunction, that which we do not know how it looks like?

There just happens to be 5 different operators which are diagonalized by the same set of basis states in this problem. The energy, the square of the total orbital angular momentum, the z-component of the orbital angular momentum, the square of the total spin, and the z-component of the spin.

Really, the first 3 variables have to do with the wave function in real space, while the last 2 have to do with the internal degrees of freedom of the particle (spin).

Perhaps a more common notation one might see would be:

$$\left|n,l,m_l\right>\otimes\left|s,m_s\right>$$

This would explicitly separate the internal degrees of freedom from the external ones. Without using bra-ket notation, one might instead construct this state from a wave function $\psi_{n,l,m}$ and a spinor $\chi_{s,m_s}$:

$$\psi_{n,l,m}\otimes\chi_{s,m_s}$$

You might be able to simplify this a little by noting the super selection rule that superpositions do not exist between states of different total spin (e.g. an electron will always have spin 1/2, this number is intrinsic to this particle), so that if we specify the species of particle, we implicitly know the eigenvalue of s, so we may simply write, in either case:

$$\left|n,l,m\right>\otimes\left|m_s\right>$$
or
$$\psi_{n,l,m}\otimes\chi$$

 

[bhobba]

I'm fine with that. So far QM seemed more or less just applied mathematics. I don't really find any Physical interpretation of anything in it so far.

If that's how you found it then Ballentine is most certainly for you.

Here its developed in a fairly rigorous way from just two axioms.

And once you feel comfortable with the stuff in Ballentine do a post and I can explain how it can be reduced to just one axiom. QM from just one axiom? Well of course not - its just each step seems very reasonable and natural - but that is a revelation for the future.

Thanks
Bill

 

[WannabeNewton]

I'm fine with that. So far QM seemed more or less just applied mathematics. I don't really find any Physical interpretation of anything in it so far.

I thought I was the only one who felt this way. Then I asked almost all the people in my QM class and they all felt the same way. Compared to subjects like classical mechanics, EM, statistical mechanics, and GR I would have to agree with you wholeheartedly that QM just seems like applied math and apparently so do the people in my class.

However QM comes to life brilliantly and beautifully in statistical mechanics :)

 

[bhobba]

I thought I was the only one who felt this way. Then I asked almost all the people in my QM class and they all felt the same way. Compared to subjects like classical mechanics, EM, statistical mechanics, and GR I would have to agree with you on that and apparently so do the people in my class.

That's because a good treatment hides the difficult issues of interpretation.

Its how it should be BTW - don't get caught up in that quagmire until you understand its formalism.

Thanks
Bill

 

[atyy]

That's nonsensical. GR is applied math too! QM is a wonderful theory, because it is an "effective" theory, ie. one that is not complete. It has the classical/quantum cut, and its most natural interpretation is "instrumental". (It might be possible that many-worlds makes QM a complete theory, and indeed there are versions that seem very satisfactory, but there is still no consensus on whether many-worlds really works. And if it does, that makes QM even more interesting!)

QM is a theory which is foundationally interesting, in that it (1) has a measurement problem (2) violates the Bell inequalities indicating some sort of nonlocality, yet can be made consistent with relativistic locality (3) together with gravity, and thermodynamics forms the "information loss" problem. It shows the importance of experiments, because I don't think anyone would accept this theory as reasonable, except that it is so successful in describing experiments.

 

[Matterwave]

In my view, any science that deals with probabilities will run into interpretational difficulties (unless all you care about are the average values, like in statistical mechanics). Deterministic theories are much easier to work with conceptually. A leads to B is a very easy logical thing to think about. A might lead to B, C, and D, with probabilities B*, C*, and D* is much more difficult conceptually in my opinion. Even the definition of a probability itself (other than the strict mathematical definition) is mired in conflicting views.

 

[Nugatory]

That's nonsensical. GR is applied math too!

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.

 

[bhobba]

In my view, any science that deals with probabilities will run into interpretational difficulties

Hmmmm. Yes and no.

Probabilities have difficult interpretational issues that are philosophically - how to put it - non trivial - eg the frequentest interpretation in circular unless you base it on something else that isn't such as the Kolmogorov axioms..

But these days with the axiomatic foundation via the Kolmogorov axioms they are manageable in a number of ways eg by the so called Cox axioms or a reasonable implementation of those axioms to applied problems.

In a certain sense Ballentines statistical interpretation is a frequentest view of probabilities in QM, Copenhagen more Bayesian.

At the axiomatic formal level of QM I simply stick to the Kolmogorov axioms to avoid any issues with interpretation.

Thanks
Bill

 

[bhobba]

That's nonsensical. GR is applied math too! QM is a wonderful theory, because it is an "effective" theory, ie. one that is not complete.

I don't think its complete either (specifically I want some mechanism for an improper mixed state to become a proper one), but that is a debatable point.

Personally I think discussions on issues like that don't really go anywhere. They are fine for highlighting the issues involved, but beyond that - its just a talkfest - what we want is experiments.

Thanks
Bill

 

[bhobba]

But I have to say that I'm siding with what I'm hearing from TheAustrian and WbN.

Make that four, and I suspect many more ascribe to it as well.

Thanks
Bill

 

[TheAustrian]

Make that four, and I suspect many more ascribe to it as well.

Thanks
Bill

I'm glad to hear I'm not alone. If I would have said this to my lecturer back then, he would have torn my head off for it.

I thought I was the only one who felt this way. Then I asked almost all the people in my QM class and they all felt the same way. Compared to subjects like classical mechanics, EM, statistical mechanics, and GR I would have to agree with you wholeheartedly that QM just seems like applied math and apparently so do the people in my class.

However QM comes to life brilliantly and beautifully in statistical mechanics :)

I'm okay with interpreting Relativity. It makes sense how it applies to the Physical world. But not QM.

If that's how you found it then Ballentine is most certainly for you.

Here its developed in a fairly rigorous way from just two axioms.

And once you feel comfortable with the stuff in Ballentine do a post and I can explain how it can be reduced to just one axiom. QM from just one axiom? Well of course not - its just each step seems very reasonable and natural - but that is a revelation for the future.

Thanks
Bill

I will give it a read in the upcoming month of June.

 

[atyy]

Query about Ballentine: Does the book even mention that when one does QM, one divides the universe into classical and quantum parts?

Query to TheAustrian: Did your lecturer even mention the need to divide the world into classical and quantum parts? Do you know that this is part of the basic procedure of QM?

 

[TheAustrian]

Query about Ballentine: Does the book even mention that when one does QM, one divides the universe into classical and quantum parts?

Query to TheAustrian: Did your lecturer even mention the need to divide the world into classical and quantum parts? Do you know that this is part of the basic procedure of QM?

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.

 

[stevendaryl]

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.

 

[stevendaryl]

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 |\Psi\rangle \otimes |\Phi \rangle, is the adjoint written as \langle \Phi | \otimes \langle \Psi | or \langle \Psi | \otimes \langle \Phi |. 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:

(\langle \Phi' | \otimes \langle \Psi' |) (|\Psi\rangle \otimes |\Phi \rangle)

do you get

\langle \Phi' | \Psi\rangle \langle \Psi' |\Phi \rangle

or

\langle \Psi' | \Psi\rangle \langle \Phi' |\Phi \rangle

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

 

[TheAustrian]

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 |\Psi\rangle \otimes |\Phi \rangle, is the adjoint written as \langle \Phi | \otimes \langle \Psi | or \langle \Psi | \otimes \langle \Phi |. 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:

(\langle \Phi' | \otimes \langle \Psi' |) (|\Psi\rangle \otimes |\Phi \rangle)

do you get

\langle \Phi' | \Psi\rangle \langle \Psi' |\Phi \rangle

or

\langle \Psi' | \Psi\rangle \langle \Phi' |\Phi \rangle

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?

 

[atyy]

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.

 

[TheAustrian]

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).

 

[atyy]

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.

 

[Jilang]

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.

 

[TheAustrian]

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.

 

[stevendaryl]

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 |\Psi \rangle by its components in an unspecified basis:

\Psi^\alpha

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

\langle \Phi | \Psi \rangle = \Phi_\alpha \Psi^\alpha

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

Composite states would be represented using multiple indices:

|\Phi\rangle \otimes |\Psi \rangle \Rightarrow \Phi^\alpha \Psi^\beta

and the inner product of two such states:

(\langle \Phi' | \otimes \langle \Psi |) (|\Phi\rangle \otimes |\Psi \rangle) \Rightarrow \Phi'_\alpha \Psi'_\beta \Phi^\alpha \Psi^\beta

 

[stevendaryl]

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.

 

[Fredrik]

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 $|x\rangle\otimes|y\rangle$.

Suppose you have a two-particle state |\Psi\rangle \otimes |\Phi \rangle, is the adjoint written as \langle \Phi | \otimes \langle \Psi | or \langle \Psi | \otimes \langle \Phi |. I'm sure there is a convention for that,

I've been wondering about that too. It's a bit embarassing that I don't even know if one of the two obvious conventions is considered standard.

 

[Fredrik]

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.

 

[stevendaryl]

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:

\langle \Psi | \Phi \rangle obviously means the inner product of

|\Phi \rangle with |\Psi \rangle

But for composite states, I think it can be difficult to know what's multiplying what. Abstract indices make this clear.

 

[TheAustrian]

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 |\Psi \rangle by its components in an unspecified basis:

\Psi^\alpha

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

\langle \Phi | \Psi \rangle = \Phi_\alpha \Psi^\alpha

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

Composite states would be represented using multiple indices:

|\Phi\rangle \otimes |\Psi \rangle \Rightarrow \Phi^\alpha \Psi^\beta

and the inner product of two such states:

(\langle \Phi&#039; | \otimes \langle \Psi |) (|\Phi\rangle \otimes |\Psi \rangle) \Rightarrow \Phi&#039;_\alpha \Psi&#039;_\beta \Phi^\alpha \Psi^\beta

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.

 

[atyy]

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.

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.

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.

 

[bhobba]

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

 

[atyy]

Explicitly - no. Implicitly - yes.

Thanks
Bill

Indeed. But implicit to whom, I would ask. Does Ballentine know this, or just bhobba?

 

[bhobba]

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

 

[DeIdeal]

Suppose you have a two-particle state |\Psi\rangle \otimes |\Phi \rangle, is the adjoint written as \langle \Phi | \otimes \langle \Psi | or \langle \Psi | \otimes \langle \Phi |. I'm sure there is a convention for that, but it's pretty arbitrary.

To me it seems natural to write it as \langle\Psi | \otimes \langle \Phi|, since for (finite or infinite-dimensional) vector spaces V,W and their duals V^*,W^*, (V\otimes W)^*\cong V^*\otimes W^*. Just like when you have an operator defined as a tensor products of operators that act separately in V and W, say, S_V\otimes \mathrm{id}_W (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?

 

[bhobba]

Indeed. But implicit to whom, I would ask. Does Ballentine know this, or just bhobba?

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

 

[TheAustrian]

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?

 

[bhobba]

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

 

[atyy]

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.

 

[kith]

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 α_r 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 α_r and α_r 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.

 

[Nugatory]

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").

 

[atyy]

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.

 

[WannabeNewton]

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?

 

[Fredrik]

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.

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.

 

[atyy]

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.

 

[stevendaryl]

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 \langle \Phi|O|\Phi \rangle, it's an assumption that it's equal to the average value of observable O in the state described by |\Phi\rangle.