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In summary: If the SE doesn't work with SR, what specifically, in terms of the math, is the reason for this? I'm specifically interested in how this relates to the creation and annihilation operators of QFT, as I previously was under the impression that those operators didn't work with the Schrödinger equation, but my professor recently told me they're the same as the ladder operators used in non-relativistic quantum mechanics.

For context:

I posted a few questions a couple months ago, about the relationship between QM and QFT, such as What is the mathematical relationship between the wave functions of QM and the fields in QFT? and I got some very helpful explanations about how QFT requires the creation and annihilation operators, and I've also read (can't remember where) that the problem with the Schrödinger equation formulation of QM (as opposed to, e.g., the Dirac formulation) is that it assumes a single static reference frame, which obviously doesn't work with special relativity, and hence it's replaced by the Dirac equation in QFT. I also saw Schrödinger Equation and Special Relativity where the OP refers to a relativistic version of the Schrödinger equation and is told there is no such version, but no explanation is given for why.

However, there are two things I'm not understanding:

  1. The Schrödinger equation is perfectly compatible with changes of basis, such as converting between position and momentum space via the Fourier transform. The way I understand it from linear algebra and SR, a change of basis is mathematically equivalent to changing reference frames/coordinate systems. And, I mean, obviously changing between position and momentum space isn't the same thing as applying the Lorentz transform to change coordinates in spacetime, both the Lorentz transform and the quantum mechanical operators are linear operators over vector spaces, and the wavefunctions that are solutions to the Schrödinger equation are vectors in the vector space the operators act on. So if we can apply, e.g., the momentum operator to them, shouldn't we also be able to apply the Lorentz transform to them? If not, why?

  2. A couple months ago, when I was reading those responses (such as the second answer to the above linked question) about how what makes QFT fundamentally different from QM is the creation and annihilation operators allowing us to switch reference frames (because, since energy and hence mass is relative, that means particle number is also relative), I got the impression that these operators were something specific to QFT that wouldn't work with the Schrödinger equation. However, in my QM course, we just finished the unit on the quantum harmonic oscillator, including using the ladder operators to find a recursive formula for the wavefunction at an arbitrarily high energy level. Griffiths' covers this in detail in chapter 2 of his QM text, and, at that point in the text at least, he's still starting every derivation with some expression he's previously derived from the Schrödinger equation. So clearly the ladder operators work fine with the Schrödinger equation and I thought nothing of it, until I read the Wikipedia article on those operators, which calls them creation and annihilation operators. I asked my QM professor in office hours and he said that, yes, they're the same creation and annihilation operators used in QFT, and then he started talking about how QED in particular is basically built-up from the quantum harmonic oscillator. I then asked a friend who previously did research in QFT if that was true what my professor said about QED being built up from the harmonic oscillator and he said yes, absolutely. So my question then is this: if QED, at least, is built up from the quantum harmonic oscillator, and QFT more generally uses the creation annihilation operators to change reference frames (which, if I understand correctly, is the main thing the Schrödinger formulation of QM lacks), but Griffiths' shows both that we can model the quantum harmonic oscillator using a wavefunction derived from the Schrödinger equation and that this wavefunction works perfectly well with the aforementioned operators, then how can the Schrödinger equation be incompatible with SR?

On a side note, I did see 'Why' is the Schrödinger equation non-relativistic?, however, while that sounds like the same question as mine, reading the question details, (and a later comment by the OP) what the OP is actually asking about is why the De Broglie equations apparently are compatible with SR while the SE seems not to be. They also appear to already be familiar with the technical reasons the Schrödinger equation is apparently non-relativistic (I say "apparently" and still have my question worded as "is it" because one of the answers to that question says the SE itself IS compatible with relativity but that the Hamiltonian is the problem, and the OP comments saying they already knew that, which has me even more confused because of their initially saying they knew SE wasn't relativistic), whereas those reasons are what I'd like to know in the first place. Hence, please don't merge these questions together, as they're only superficially similar and definitely not duplicates, for the reasons stated above.

Edit: As explained above, this question is not a duplicate of the linked question and is only superficially similar to it. Please actually read the entire question before voting to close, rather than merely going by the title.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Apr 3, 2023 at 6:48

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You ask many questions here and they conflate many things. The answers you were given are partially correct, but not very precise. For example, things being similar does not mean that they are the same. So, I can totally understand why you are confused. I will try to answer at least part of your questions.

The Schrodinger equation and relativistic invariance.

SE describes how the wave function $\psi(x,t)$ evolves with time. Already this means that it attributes a special role to time. Time and coordinates are not treated on the same footing by SE. Thus, one says that SE and the whole approach of quantum mechanics is not manifestly relativistically invariant.

Still, SE is a completely universal equation for evolution in quantum systems. It can be applied -- an is frequently used -- to deal with relativistic quantum systems, including QFT (the Schrodinger picture in canonical quantization). Thus, despite the approach of SE is inherently not relativistic, it can still be used to describe relativistic systems. Due to this drawback of the SE manifest relativistic invariance gets lost and one should check if it is there manually.

For example, one can consider a free non-relativistic particle with the Hamiltonian $\frac{p^2}{2m}$. Using the standard quantization $p\to \hat p = -i\partial_x $ one gets a quantum Hamiltonian, which can be used to write SE in a given case. In this example, neither the system nor its description are relativistically invariant.

In contrast, one can consider a relativistic particle with the Hamiltonian $\sqrt{p^2+m^2}$. Replacing $p\to -i\partial_x $ one gets the respective quantum Hamiltonian. The associated SE reads \begin{equation} i\frac{\partial}{\partial t}\psi = \sqrt{-\partial^i \partial_i +m^2} \;\psi. \qquad (1) \end{equation} This equation is relativistically invariant. This means that when we act with translations \begin{equation} P_\mu \psi \equiv \partial_\mu \psi \end{equation} and Lorentz transformations \begin{equation} J_{\mu\nu} \psi \equiv (x_\mu\partial_\nu-x_\nu \partial_\mu )\psi \end{equation} on the wave function, then solutions of (1) map into solutions of (1). Still, obviously, time derivatives and spatial derivatives enter (1) differently, so this equation is not manifestly relativistically invariant -- the consequence of using SE.

(Let me note that up to factors, $P_i$ that I defined above are just the operators of momenta. Then $P_0$ is the Hamiltonian, once (1) is imposed. Analogously, $J_{\mu\nu}$ are just quantum operators of rotations and boosts)

The system (1) describes a free relativistic particle of mass $m$ and spin 0. It is a completely legitimate quantum-mechanical theory as is. Let me also note, that this is not the Dirac equation. Historically, Dirac derived his equation trying to make the Schrodinger equation Lorentz-covariant. What he got in the end was SE for a free spin-$\frac{1}{2}$ particle (the Dirac equation is often regarded as a classical field equation, which is not the same as a single-particle SE, but let's not get into this), not for a spin-$0$ particle. There are some reasons why he did not get (1), but let us not go into this. This is more on a history side than about physics. The current state of affairs is that you can have consistent relativistic quantum mechanics of free particles of any spin and mass, see e.g. section 2 "Relativistic quantum mechanics" of the book by Weinberg.

Creation and annihilation operators in QM and QFT

Again, there are many things conflated here. In quantum mechanics, creation and annihilation operators are primarily used is only one context: it is a simple trick to solve one particular problem, namely, finding the eigenstates and the eigenvalues of the Hamiltonian for the harmonic oscillator. This is it. In this context, the creation operator applied to an eigenstate maps it to the next eigenstate, while the annihilation operator acts in the opposite direction. In QFT one also uses creation annihilation operators, but their meaning is very different.

Before explaining what they do in QFT, let me first recall that there are two ways to get to QFT: by quantizing a classical field theory or by adding interaction in relativistic quantum mechanics, the latter inevitably requires to create and annihilate particles.

The first approach is more standard. The classical field theory is first of all a classical theory and to get to QFT it needs to be quantized. The classical field theory may be regarded as a limit of a classical mechanics, in which one has infinitely many point particles. With this amendment, classical field theory can be quantizing along the same lines as a system of classical particles. For example, one uses the canonical quantization: finds coordinates and conjugate momenta and then postulates that these no longer commute, but satisfy the canonical commutation relations. In doing so, it is convenient to make the Fourier transform of the field with respect to spatial directions. Then, one finds that different Fourier modes decouple from each other, while each one has the Hamiltonian of a harmonic oscillator with the frequency related to the momentum as $\omega_p = \sqrt{p^2+m^2}$. This is how the harmonic oscillator from quantum mechanics becomes relevant. You also end up dealing with creation and annihilation operators. But their meaning in free QFT is that the creation operator $a^\dagger_p$ creates a particle with momentum $p$, while $a_p$ annihilates it. Note, that there was no notion of a particle in field theory before it was quantized. In fact, this is the whole reason why quntization of field theory was needed: we do see that, say, energy from E-M waves is absorbed in chunks, as if it consists of particles, while classically E-M waves do not have them.

The second approach is followed, e.g. by Weinberg. One starts from a relativistic quantum mechanics, which describes a free single particle. What is missing here is that the theory describes only a single particle, while we want to describe also multi-particle states. We also know that when energies are high enough, processes of creation and of annihilation of particles become unavoidable, so one needs a language to deal with this. What one does is that instead of a single-particle Hilbert space (of quantum mechanics), considers the multi-particle Hilbert space. Creation and annihilation operators are introduces as operators that create or annihilate a particle of a given momentum. One can show that any Hamiltonian can be written in terms of creation and annihilation operators and thus they are used in the following.

To summarize, one does have creation and annihilation operators in QFT. However, unlike in quantum mechanics, in which they raise/lower energy of a single particle in one particular model of the quantum oscillator, in QFT $a$ and $a^\dagger$ annihilate and create particles. So, $a$ and $a^\dagger$ from QFT are completely out of the physical realm of quantum mechanics, as the latter deals only with a single particle, while in QFT $a$ and $a^\dagger$ create and annihilate particles. Still, as I explained above, the Hamiltonian of the free field theory is that of an infinite family of decoupled oscillators. This is how QFT creation and annihilation operators are relevant to the QM ones for the harmonic oscillator.

PS. I believe, it should follow from what I wrote above, $a$ and $a^\dagger$ are introduced in QFT not to be able to change reference frames. Similarly, the usage of $a$ and $a^\dagger$ for the harmonic oscillator does not say anything about its relativistic invariance. Harmonic oscillator in QM is not relativistically invariant.

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    $\begingroup$ Here is someting I'm wondering about: there is a connection between the mathematics that describes propagating waves and the Lorentz transformations. As we know: Maxwell's equations admit a solution that describes propagating electromagnetic waves. The type of symmetry that is expressed by the Lorentz transformations is latent in Maxwell's equations, by virtue of wave propagation being a solution to Maxwell's equations. The Schrödinger equation being a wave equation, is that significant? $\endgroup$
    – Cleonis
    Apr 2, 2023 at 14:51
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    $\begingroup$ I am not sure that I understand what you are asking. I am not even sure what you mean by the mathematics of propagating waves. If we talk about free equations of motion, the simplest way to solve them is by making the Fourier transform. It usually gives you a relation between the frequency $\omega$ and the wave vector $k$. Then, the general solution is expressed as a superposition of exponents $e^{-i(\omega t-kx)}$. Is this propagating waves for you? This applies to any linear differential equation with constant coefficients. Some of them are Lorentz invariant $\endgroup$
    – Dr.Yoma
    Apr 2, 2023 at 17:38
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    $\begingroup$ About wave equation and Lorentz transformations. A discussion of that is present on the website mathpages.com In the section 'the relativity of light'. Scroll to the sentence: "Voigt considered the question of whether there is any linear transformation that leaves the wave equation unchanged." The discussion then proceeds to show that that 'leaves unchanged' requirement is sufficient to arrive at the Lorentz transformations. $\endgroup$
    – Cleonis
    Apr 2, 2023 at 19:22
  • $\begingroup$ ok, I understand what you mean. If you write a particular wave equation, of course, you can study its symmetries. If the equation in question is $(\partial_t^2-\partial^i\partial_i)\phi=0$ then its symmetry is the Poincare algebra. Actually, since there is no mass term there, the symmetry is the conformal algebra. It also depends on what symmetries you consider. If you allow symmetries that act on fields with higher-derivative linear operators, then the symmetry algebra is even larger. It is the so-called higher-spin algebra. $\endgroup$
    – Dr.Yoma
    Apr 2, 2023 at 23:41
  • $\begingroup$ this discussion can be found e g in arxiv.org/abs/hep-th/0206233 $\endgroup$
    – Dr.Yoma
    Apr 3, 2023 at 7:50
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Here's the Schrödinger equation for a free body of mass $m$: $$iħ\frac{∂ψ}{∂t} = -\frac{ħ^2}{2m} ∇^2ψ.$$ The time derivative is first order, the spatial derivatives are second order. It looks like space and time are being treated differently. But, appearances can be deceiving. It's time for some prestidigitation.

Ok. So, watch what happens. Here's the equation now. $$2m×iħ\frac{∂ψ}{∂t} = -ħ^2 ∇^2ψ.$$ Better. Less clunky. That's a start. Now, here's the equation: $$2m×iħ\frac{∂ψ}{∂t} = |-iħ∇|^2ψ.$$ Even better. Things are starting to look the same on both sides. Now, we do a little magic and make $m$ an eigenvalue by changing $ψ(𝐫,t)$ to $ψ(𝐫,t,u)$ with: $$\left(+iħ\frac{∂}{∂t}\right)\left(-iħ\frac{∂}{∂u}\right)ψ = |-iħ∇|^2ψ, \hspace 1em -iħ\frac{∂ψ}{∂u} = mψ.$$ What's $u$? Don't care. But the solution will always be of the form: $$ψ(𝐫,t,u) = e^{imu/ħ} ψ(𝐫,t), \hspace 1em ψ(𝐫,t) ≡ ψ(𝐫,t,0),$$ where the last equation will be our new definition for the original $ψ(𝐫,t)$.

We'll get rid of those pesky little $iħ$'s. Now, here's the Schrödinger equation: $$\left(∇^2 + 2\frac{∂}{∂t}\frac{∂}{∂u}\right)ψ = 0, \hspace 1em \frac{∂ψ}{∂u} = \frac{im}{ħ}ψ.$$

Now, a make a fix: $$\left(∇^2 + 2\frac{∂}{∂t}\frac{∂}{∂u} - α\left(\frac{∂}{∂t}\right)^2\right)ψ = 0, \hspace 1em \frac{∂ψ}{∂u} = \frac{im}{ħ}ψ.$$ What's $α$? Don't care. We'll figure that out later. The non-relativistic Schrödinger equation will now, by definition, be the one where $α = 0$.

What about when $α ≠ 0$? Define $$\hat{ψ}(𝐫,t,s) ≡ ψ\left(𝐫,t,\frac{s-t}{α}\right) \hspace 1em (α ≠ 0).$$ Then the solution and definition become: $$\hat{ψ}(𝐫,t,s) = e^{im(s-t)/(αħ)} ψ(𝐫,t), \hspace 1em ψ(𝐫,t) ≡ \hat{ψ}(𝐫,t,t).$$ The Langoliers Transform. The $s$ time is the Steven King time in Langoliers and the $t$ time is the real world time. The actual solution happens where $s$ and $t$ coincide and everyone's back in normal time. More accurately: it's The Premonition Transform, after the name of The Outer Limits episode that the famed story author clearly ripped off. But that's another discussion for another time.

Then, with the following substitutions: $$ψ(𝐫,t,u) = \hat{ψ}(𝐫,t,t+αu), \hspace 1em \frac{∂ψ}{∂t} = \frac{∂\hat{ψ}}{∂t}+\frac{∂\hat{ψ}}{∂s}, \hspace 1em \frac{∂ψ}{∂u} = α\frac{∂\hat{ψ}}{∂s},$$ we have: $$\left(∇^2 - α\left(\frac{∂}{∂t}\right)^2 + α\left(\frac{∂}{∂s}\right)^2\right)\hat{ψ} = 0, \hspace 1em \frac{∂\hat{ψ}}{∂s} = \frac{im}{ħα}\hat{ψ} \hspace 1em (α ≠ 0).$$ Clearly, this is equivalent to: $$\left(∇^2 - α\left(\frac{∂}{∂t}\right)^2\right)\hat{ψ} = \frac{m^2}{αħ^2}\hat{ψ}, \hspace 1em \frac{∂\hat{ψ}}{∂s} = \frac{im}{ħα}\hat{ψ} \hspace 1em (α ≠ 0).$$ That's the Klein-Gordon equation, when $α = (1/c)^2 > 0$.

More precisely, the general solution is $$\hat{ψ}(𝐫,t,s) = e^{ims/(ħα)}\hat{ψ}(𝐫,t), \hspace 1em \hat{ψ}(𝐫,t) ≡ \hat{ψ}(𝐫,t,0),$$ and it is $\hat{ψ}(𝐫,t)$ that is the solution to the Klein-Gordon equation. In terms of the original $ψ$: $$ψ(𝐫,t) = \hat{ψ}(𝐫,t,t) = e^{imt/(ħα)}\hat{ψ}(𝐫,t) \hspace 1em⇒\hspace 1em \hat{ψ}(𝐫,t) = e^{-imt/(ħα)}ψ(𝐫,t),$$ which is what is the solution to the Klein-Gordon equation.

So, what's $α$? In the non-relativistic case, $α = 0$. In the relativistic case $α > 0$, with $c = \sqrt{1/α}$.

Thus, the non-relativistic Schrödinger equation is not equivalent to the relativistic version and the difference between the two has been made clear for all to see.

So, is the Schrödinger equation compatible with Special Relativity? We'll leave that for you to decide. But before anyone says "no, you need to use the Dirac equation" (or something like that): we could go the other way and turn the question upside down to "is the Dirac equation compatible with non-relativistic theory?"

Wanna see what happens if we start with the Dirac equation and go in the opposite direction, from the relativistic form at $α = (1/c)^2$ to the non-relativistic form at $α = 0$? Maybe a future edit.

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The answer is no

The Oxford Dictionary defines reality as the world or the state of things as they actually exist, as opposed to an idealistic or notional idea of them.

Currently there are two primary ways science attempts to explain and define the “reality” of our universe. The first is Quantum Mechanics or the branch of physics that defines its evolution in terms of the probabilities associated with the wave function. The other is the deterministic environment of Relativity which defines it in terms of a physical interaction between space and time.

Specifically, Relativity would define the observable positions of particles in terms of where the point defining their center of mass is located.

While Quantum Mechanics uses the mathematical interpretation of the wave function to define the most probable position of a particle when observed in terms of one-dimensional point.

Since we all live in the same universe you would expect the probabilistic approach of Quantum Mechanics to be compatible with the deterministic one of Einstein.

Unfortunately, they define two different worlds which appear to be incompatible. One defines existence in terms of the probabilities while the other defines it in terms of the deterministic of properties of space and time.

However, to show why those probabilities appear to be incompatible with Relativity's determinism even though they are not it will be necessary to explain the evolution of quantum environment in terms of a deterministic interaction between the components of a space-time environment.

For example, when we role dice in a casino most of us realize the probability of a six appearing is related to or is caused by its physical interaction with properties of the table in the casino where it is rolled. Putting it another way what defines the fact that six appears is not the probability of getting one but the interaction of the dice with the table and the casino it occupies.

This suggests to show the “reality” behind the wave function in terms of space time one must explain how its environment evolves in terms of how the physical components of space-time interact to define a particles position.

The fact that Relativity defines evolution of space-time in terms of the energy propagated by electromagnetic wave while Quantum Mechanics defines it in terms of the mathematical evolution of the wave function give us a starting point. This is because it suggests the evolution in both is defined in define by a wave.

One can use that commonality to define the position of a particle in terms of the deterministic properties of Relativity because the science of wave mechanics along with the fact Relativity tells us an electromagnetic wave moves continuously through space unless it is prevented from doing so by someone observing or something interacting with it. This would result in its energy being confined to three-dimensional space. The science of wave mechanic also tells us the three-dimensional "walls" of this confinement will result in its energy being reflected back on itself thereby creating a resonant or standing wave in three-dimensional space. This would cause its wave energy to “collapse” and become concentrated at the point in space were a particle would be found.

Additionally, wave mechanics also tells us the energy of a resonant system, such as a standing wave can only take on the discrete or quantized values associated with its fundamental or a harmonic of its fundamental frequency. This means a particle would occupy an extended volume of space defined by the wavelength of its standing wave.

This means what defines the fact that a particle appears where it does may not be determined by the probabilities associated with the wave function but a deterministic interaction of an electromagnetic wave with the physical properties of space-time.

(Note we will use a particles position to make the connection between the probabilities of Quantum Mechanics and the determinism of Relativity but the same logic will apply to all conjugate pairs.)

However, the probabilistic interpretation of the wave function is necessary to define a particle position because in both Quantum mechanics uses which it randomly places with respect to the its center. Therefore, the randomness of where that point is with respect to it will result in its position, when it is observed to be randomly distributed in space. This means one must define its position in terms of probabilities to average the deviations that are caused by that random placement.

Yet Relativity also defines the position of particles in terms of where a point defining their center of mass is located which cannot be precisely define on a quantum level. Therefore, because similar to Quantum Mechanics Relativity cannot precisely determine where that point is located with respect the extended volume of an object it would also have to define their position in terms of probabilities.

However, the large number of particles in objects such as a moon or planet would result in averaging out the deviation of the position of each their individual particles it appears to be deterministic.

This shows why the probabilistic environment of Quantum Mechanics is not incompatible with Relativity's determinism is because each define the position of a particle or object in terms of a one-dimensional point which neither can precisely define with respect to their center.

As was mentioned earlier one can define reality as the world or the state of things as they actually exist, as opposed to an idealistic or notional idea of them.

Therefore, as was shown above one can understand a connection between the probabilistic world of Quantum Mechanics and the deterministic one of Relativity by the fact, they both define position in terms of one-dimensional point and by assuming actual existence of an electromagnetic wave whose evolution can defined the idealistic idea of the wave function.

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    $\begingroup$ Determinism is not postulated or required by relativity—why do you suggest this is the case? More importantly, how does this answer (which doesn't even mention the Schrödinger equation) address the question of the compatibility between the Schrödinger equation and, not reality, but special relativity? Also, you seem to suggest that QM and SR are more broadly irreconcilable, seemingly ignoring relativistic quantum field theory. $\endgroup$
    – Sandejo
    Apr 3, 2023 at 0:36

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