Where does the QM creation operator get its energy?

[joneall]

We are all familiar with ladder operators, such as QM harmonic oscillators or in QFT to produce energy states which are interpreted as particles. But when a creation operator raises the energy level of a system, where does that energy come from?

 

[PeroK]

An operator acting on the Hilbert Space is not a physical process.

 

[Demystifier]

We are all familiar with ladder operators, such as QM harmonic oscillators or in QFT to produce energy states which are interpreted as particles. But when a creation operator raises the energy level of a system, where does that energy come from?

Here is a classical analogy. Let $n$ be the number of apples. Let $A$ be an operator that acts on number $n$ and transforms it into $n+1$. Mathematically,
$$An=n+1$$
Now ask yourself the analog question: Where does organic matter come from needed to turn $n$ apples into $n+1$ apples? The answer is that it doesn't. Can you explain why?

 

[joneall]

This is in answer to PeroK's response.

I am clearly missing something. Sure, the equation is a mathematical construct. But we are using it to understand real ... well, apples, in the other example. Is it just that the ladder-operator equation is incomplete? So how to complete it? In other words, I can't explain why no organic apple matter is involved.

 

[f95toli]

For a closed system. the energy has to be there from the start; meaning the total energy stays constant as the system evolves.

A simple example would be a two-level system such as a spin exchanging energy with a cavity. The latter would be modeled as a harmonic oscillator and the creation operator would be used to construct raising /lowering operators which add/remove a photon from the cavity. If we initialise the system in such as way that the spin is in an excited state the energy can therefore slosh between the spin (which is oscillating between "up" and "down") and the cavity (where photons are created/destroyed)*
If we instead initialise the system with the spin in the ground state and no energy in the cavity nothing happens.

*This is my attempt to describe the Jaynes-Cummings Hamiltonian...

 

[joneall]

Then what is it? This leads into another subject, which I should probably open separately. If the 1st-quantization (Schrôdinger) wave function is a probability amplitude, what is the second-quantization "wave function"?
Clearly, saying An = n+1 is not taking everything into account, which I guess answers my question. Which was really more of an observation.

 

[PeroK]

I am clearly missing something.

Did you not understand that Hilbert space and the mathematical formalism of QM are not the physical processes? If you want to add an electron to a system, then you need to get an electron from somewhere. But, the ladder operator is not that physical process.

In other words, the mathematical formalism of QM is abstract.

 

[PeroK]

PS if we raise an object of mass $m$ a height $h$ above the ground, then we have an increase in the potential energy of $mgh$.

That mathematical operation of multiplication represents some unspecified physical process that is needed to raise the object.

My question to you is: if we write $\Delta PE = mgh$ then where did the energy come from to carry out that mathematical operation?

 

[joneall]

My question to you is: if we write $\Delta PE = mgh$ then where did the energy come from to carry out that mathematical operation?

That's a better way to put it. Thanks.

 

[joneall]

An operator acting on the Hilbert Space is not a physical process.

Excuse me if I'm picking nits or still confused. Since we must be precise, do you really mean an operator acts on a Hilbert space? I would have said it operated on an object such as a state vector in the H space. These are the kind of fine points that QM textbooks, at least what I have read, don't go into much as I recall.

 

[PeroK]

Excuse me if I'm picking nits or still confused. Since we must be precise, do you really mean an operator acts on a Hilbert space? I would have said it operated on an object such as a state vector in the H space. These are the kind of fine points that QM textbooks, at least what I have read, don't go into much as I recall.

It's both. Generally an operator is defined on a set, which implies it acts individually on every member of the set. A Hilbert space is collectively all the relevant state vectors (kets). You would normally write $$L: H \to H$$where $L$ is some linear operator from the Hilbert Space $H$ to itself.

 

[PeterDonis]

An operator acting on the Hilbert Space is not a physical process.

This is not really a good answer, since some operators, namely the self-adjoint ones, model actual physical processes.

A better answer would be that, since the ladder operators are not self-adjoint, they should not be expected to model physical processes.

 

[PeterDonis]

Sure, the equation is a mathematical construct. But we are using it to understand real [things]

See my response to @PeroK in post #12. Only some mathematical objects in the abstract theory correspond to physical objects or processes. The ladder operators in QM are not among them. That's why the question you posed in the OP is not answerable.

 

[PeroK]

This is not really a good answer, since some operators, namely the self-adjoint ones, model actual physical processes.

A better answer would be that, since the ladder operators are not self-adjoint, they should not be expected to model physical processes.

A self adjoint operator represents an observable. But, the state is not changed by a simple action of the operator on the original state. Instead, the state changes probabilistically to an eigebstate of the operator. That's not applying the operator directly.

 

[PeterDonis]

A self adjoint operator represents an observable. But, the state is not changed by a simple action of the operator on the original state. Instead, the state changes probabilistically to an eigebstate of the operator. That's not applying the operator directly.

All this is fine, but I don't think it changes the point that the ladder operators don't represent observables, and indeed don't represent anything "physical", so it's pointless to ask "physical" questions about them. Whereas one could imagine asking "physical" questions about a self-adjoint operator, like "what kind of physical device would realize this operator?"

 

[joneall]

All this is fine, but I don't think it changes the point that the ladder operators don't represent observables, and indeed don't represent anything "physical", so it's pointless to ask "physical" questions about them. Whereas one could imagine asking "physical" questions about a self-adjoint operator, like "what kind of physical device would realize this operator?"

Is this a general result, or is it because the ladder operators are not Cartan generators of the appropriate Lie algebra?

 

[PeroK]

Is this a general result, or is it because the ladder operators are not Cartan generators of the appropriate Lie algebra?

I would say that the operator algebra underpinning QM is used to describe the laws of physics. But, the operators are abstract mathematical tools and not themselves subject to the laws of physics. Instead, they obey the laws of linear algebra.

 

[joneall]

So the laws of physics obey those of linear algebra.

 

[topsquark]

So the laws of physics obey those of linear algebra.

QM is usually written in terms of linear operators because they are simpler to work with. But in reality this is only a good approximation. If you would like to see a non-linear version take a look at the sine-Gordon equation. (No, that's not a typo!)

Sorry, I'm a lousy web searcher. The best I could find was this article and you have to have a subscription to view it. Perhaps you or someone else will have better luck. Here's the link.

-Dan

 

[vanhees71]

QM is usually written in terms of linear operators because they are simpler to work with. But in reality this is only a good approximation. If you would like to see a non-linear version take a look at the sine-Gordon equation. (No, that's not a typo!)

There's not the slightest hint that QM is only an approximation. What has the sine-Gordon equation to do with that anyway?

 

[topsquark]

There's not the slightest hint that QM is only an approximation. What has the sine-Gordon equation to do with that anyway?

There is an application of the SG equation to the Thirring model. I admit that I don't know any details about it.

I recognize that the empirical data supports QFT as being a very good model for the experiments we've been working with. However, as in most Physics, I see no reason to assume that linearity is more than a reasonable approximation to most problems.

Perhaps I'm wrong, but I've seen no treatment that says otherwise. However I recognize that you have much more experience in the field than I do and admit that I may be mistaken.

-Dan

 

[PeroK]

So the laws of physics obey those of linear algebra.

That's one way to put it!

 

[Vanadium 50]

Teachers have to be very careful when teaching QM, because the raising operators in one class get their energy from the lowering operator in another, so the two need to always be perfectly in balance.

 

[strangerep]

Teachers have to be very careful when teaching QM, because the raising operators in one class get their energy from the lowering operator in another, so the two need to always be perfectly in balance.

Oh, very good! Now I understand why redneck migration from Australia to New Zealand increases the total IQ of both countries!

 

[CoolMint]

We are all familiar with ladder operators, such as QM harmonic oscillators or in QFT to produce energy states which are interpreted as particles. But when a creation operator raises the energy level of a system, where does that energy come from?

It is the same underlying quantum field with detections/'particles'/ and conservation laws holding exactly on average. There is no violation of conservation laws in quantum mechanics. Treating the field as something separate from its 'particles' creates unnecessary confusion. The field and its 'particles' are one and the same.

 

[PeterDonis]

Treating the field as something separate from its 'particles' creates unnecessary confusion.

No, it is necessary because there are plenty of quantum field states that do not have particle interpretations.

The field and its 'particles' are one and the same.

This is obviously false given the observation just made above.

 

[CoolMint]

Does this mean that the field states which do not have particle-like observations are separate from the states which have particle-like interpretations? Or are they a configuration of the same overall entity which is the quantum field

 

[Nugatory]

The field and its 'particles' are one and the same.

This is obviously false given the observation just made above.

Or unobviously true - there’s a lot of wiggle room in natural language.

 

[PeterDonis]

Does this mean that the field states which do not have particle-like observations are separate from the states which have particle-like interpretations? Or are they a configuration of the same overall entity which is the quantum field

It seems like you are asking whether states of the quantum field are states of the quantum field. The answer to this should be obvious.

 

[PeterDonis]

Does this mean that the field states which do not have particle-like observations are separate from the states which have particle-like interpretations?

This is really too vague to answer. I suggest taking some time to look at the actual math.