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The transition amplitude for a particle currently in one spacetime point to appear up in another point doesn't respect causality which becomes one of the main reasons to abandon non-relativistic quantum mechanics. We impose the relativistic Hamiltonian $H=\sqrt{c^2p^2+m^2c^4}$ to get the Klein–Gordon equation or more correctly "add" special relativity after 2nd quantizing to fields, which shows how antiparticles crop up and help in preserving causality in this case. Apart from that, the equation is not even Lorentz covariant, which proves it to be non-relativistic.

But why does this occur? I mean, the Schrödinger equation is consistent with the de Broglie hypothesis and the latter is so much consistent with relativity, that some books even offer a "derivation" of the same by equating $E=h\nu$ and $E=mc^2$ probably resulting from a misinterpretation of de Broglie's Ph.D. paper. (A derivation isn't exactly possible though). So, the Schrödinger equation should include relativity in it, right? But it doesn't... How does relativity vanish from the Schrödinger equation or did the de-Broglie hypothesis ever not "include" relativity in any way?

My suspicion—The "derivation" is not possible, so the common $\lambda=h/mv $ with m as the rest mass, doesn't include relativity in any way. End of story. Is this the reason or is there something else?

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  • $\begingroup$ Related: Does relativistic quantum mechanics really violate causality? $\endgroup$
    – Qmechanic
    Jun 21, 2020 at 18:32
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Jun 21, 2020 at 20:40
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    $\begingroup$ This question reads very confused to me. What you are asking is not why the Schroedinger equation is non-relativistic, what you are asking is whether the de Broglie relation can be derived from the relativistic dispersion relation. I suspect you watched a video such as this one: youtube.com/watch?v=xbD_yWgHMVA (and others). The "derivation" in the video is nonsense. $\endgroup$
    – Zorawar
    Jun 22, 2020 at 21:40
  • $\begingroup$ possible duplicate: physics.stackexchange.com/q/257787/84967 $\endgroup$ Jun 23, 2020 at 11:17
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    $\begingroup$ As I recall, Schrodinger tried making his equation compatible with relativity from the start. He kept getting these positively charged electrons and resulting negative energy states and other "nonsense", so he published a non-relativistic treatment of his equation as it applies to the spectra of hydrogen. $\endgroup$
    – R. Romero
    Oct 25, 2021 at 21:52

7 Answers 7

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In non-relativistic Quantum Mechanics (NRQM), the dynamics of a particle is described by the time-evolution of its associated wave-function $\psi(t, \vec{x})$ with respect to the non-relativistic Schrödinger equation (SE) $$ \begin{equation} i \hbar \frac{\partial}{\partial t} \psi(t, \vec{x})=H \psi(t, \vec{x}) \end{equation} $$ with the Hamilitonian given by $H=\frac{\hat{p}^{2}}{2 m}+V(\hat{x}) .$ In order to achieve a Lorentz invariant framework (the SE is only Galilei NOT Lorentz invariant), a naive approach would start by replacing this non-relativistic form of the Hamiltonian by a relativistic expression such as $$ H=\sqrt{c^{2} \hat{p}^{2}+m^{2} c^{4}} $$ or, even better, by modifying the SE altogether such as to make it symmetric in $\frac{\partial}{\partial t}$ and the spatial derivative $\vec{\nabla} .$

However, the central insight underlying the formulation of Quantum Field Theory is that this is not sufficient. Rather, combining the principles of Lorentz invariance and Quantum Theory requires abandoning the single-particle approach of Quantum Mechanics.

  • In any relativistic Quantum Theory, particle number need not be conserved, since the relativistic dispersion relation $E^{2}=c^{2} \vec{p}^{2}+m^{2} c^{4}$ implies that energy can be converted into particles and vice versa. This requires a multi-particle framework.
  • This point is often a little bit hidden in books or lectures. Unitarity and causality cannot be combined in a single-particle approach: In Quantum Mechanics, the probability amplitude for a particle to propagate from position $\vec{x}$ to $\vec{y}$ is $$ G(\vec{x}, \vec{y})=\left\langle\vec{y}\left|e^{-\frac{i}{\hbar} H t}\right| \vec{x}\right\rangle $$ One can show that e.g. for the free non-relativistic Hamiltonian $H=\frac{\hat{p}^{2}}{2 m}$ this is non-zero even if $x^{\mu}=\left(x^{0}, \vec{x}\right)$ and $y^{\mu}=\left(y^{0}, \vec{y}\right)$ are at a spacelike distance. The problem persists if we replace $H$ by a relativistic expression in the SE.

Quantum Field Theory (QFT) solves both these problems by a radical change of perspective.

Remark 1: There are still some cases (however there are a lot subtleties), where one can use RQM in the single-particle approach. Then the SE is replaced by the e.g. Klein-Gordon equation. $$ (\Box+m^2)\;\psi(x)=0 $$ where $\psi(x)$ is still a wave-function.

Remark 2: The Schrödinger equation holds for SR. It's not the SE that fails, it's the non-relativistic Hamiltonian that fails. The Dirac equation is the SE, but with the Dirac Hamiltonian. The Schrodinger equation is valid. $$ i \hbar \frac{\partial \psi(x, t)}{\partial t}=\left(\beta m c^{2}+c \sum_{n=1}^{3} \alpha_{n} p_{n}\right) \psi(x, t)=H_\text{Dirac}\;\psi(x, t) $$

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  • $\begingroup$ Most of what you said was known to me.(I studied QFT already without quite understanding it's necessity).Unfortunately this doesn't answer my question,please notice the quotes in why and where my doubt is--deBroglie is consistent with relativity,but SE is not-why?I am not looking for why SE is not consistent in general(The first few pages of Peskin Schroeder prove them)...Maybe I am not being clear enough in my question.I edited twice already. $\endgroup$ Jun 21, 2020 at 17:39
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    $\begingroup$ @ManasDogra You're using a lot quotes and words like why and magic. No it's not very clear. Please look at my last remark. The SE is perfectly fine. It's the Hamiltonian that fails. My answer was also referring to your title "Why is Schroedinger's equation non-relativistic?" which is a quit general title and should help people searching for an answer to exact that question. $\endgroup$ Jun 21, 2020 at 18:00
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    $\begingroup$ Basically what I am asking is why is de-Broglie includes relativity but SE doesn't---how can this be possible? $\endgroup$ Jun 21, 2020 at 18:08
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    $\begingroup$ I think you have misunderstood the relationship between the Dirac equation and the Schrodinger equation in the full theory. $\endgroup$ Jun 21, 2020 at 22:52
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    $\begingroup$ Minor remarks: 1. SE is not invariant under the Galilean group either, it is only invariant under a central extension of the Galilean group. 2. The KG equation doesn't show that one can use RQM with a single particle, it shows rather the exact opposite. $\endgroup$
    – ACat
    Jun 22, 2020 at 0:55
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To do relativistic quantum mechanics you have to abandon single-particle quantum mechanics and take up quantum field theory.

The Schrödinger equation is an essential ingredient in quantum field theory. It asserts $$ \hat{H} {\psi} = i \hbar \frac{d}{dt} {\psi} $$ as you might guess, but there is a lot of subtlety hiding in this equation when ${\psi}$ refers to a quantum field. If you try to write it using numbers then $\psi$ would be a function of every state of a field $\phi$ which is itself configured over space and time. In $\psi$ you would then have a functional not a function.

In correct terminology, the Schrödinger equation here is covariant, but not manifestly covariant. That is, it would take the same form in some other inertial reference frame, but this is not made obvious in the way the equation has been written down.

But we have here a very different 'beast' to the Schrödinger equation you meet when you first do quantum mechanics. That would now be called single-particle quantum mechanics. $That$ Schrödinger equation is certainly not covariant, and nor is the whole structure of the theory of single-particle quantum mechanics.

The reason for confusion here may be to do with the history of science. Particle physicists started working with the Klein-Gordon (KG) equation under the illusion that it was some sort of relativistic replacement for the Schrödinger equation, and then the Dirac equation was thought of that way too. This way of thinking can help one do some basic calculations for the hydrogen atom for example, but ultimately you have to give it up. For clear thinking you have to learn how to quantise fields, and then you learn that for spin zero, for example, both the Klein-Gordon and the Schrödinger equation have roles to play. Different roles. Neither replaces the other. One asserts what kind of field one is dealing with; the other asserts the dynamics of the field amplitude.$^1$

I have never seen this clearly and squarely written down in the introductory section of a textbook however. Has anyone else? I would be interested to know.

Postscript on de Broglie waves

de Broglie proposed his relation between wave and particle properties with special relativity very much in mind, so his relation is relativistic (the background is that $(E, {\bf p})$ forms a 4-vector and so does $(\omega, {\bf k})$.) Schrödinger and others, in their work to get to grips with the de Broglie wave idea in more general contexts, realised that an equation which was first order in time was needed. As I understand it, the Schrödinger equation came from a deliberate strategy to look at the low-velocity limit. So from this point of view it does seem a remarkable coincidence that that same equation then shows up again in a fully relativistic theory. But perhaps we should not be so surprised. After all, Newton's second law, ${\bf f} = d{\bf p}/dt$ remains exactly correct in relativistic classical dynamics.

$^1$ For example, for the free KG field, the KG equation gives the dispersion relation for plane wave solutions. The Schrödinger equation then tells you the dynamics of the field amplitude for each such plane wave solution, which behaves like a quantum harmonic oscillator.

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  • $\begingroup$ "the Klein-Gordan equation under the illusion that it was some sort of relativistic replacement for the Schrödinger equation" The KG equation is the relativistic form of the Schrödinger equation. $\endgroup$
    – my2cts
    Jun 21, 2020 at 22:54
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    $\begingroup$ @my2cts No, it really isn't. See for example the footnote I have added. $\endgroup$ Jun 21, 2020 at 22:57
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    $\begingroup$ @my2cts If you treat SE as a classical field equation then you can see it as the non-relativistic limit of the KG equation which is a relativistic (classical) field equation. However, as the equation governing the dynamics of a quantum system, it simply doesn't make sense to see SE as the limit of the KG equation because the KG equation doesn't describe the dynamics of any quantum system (it only describes the on-shell condition). The dynamics of both relativistic and non-relativistic systems are described by the SE, of course, the dof involved are different in the relativistic case. $\endgroup$
    – ACat
    Jun 22, 2020 at 1:16
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    $\begingroup$ Just some nit-picking: it's Klein-Gordon (after Walter Gordon), not Klein-Gordan. The name "Gordan" is appropriate in "Clebsch-Gordan coefficients" (after Paul Gordan). $\endgroup$
    – akhmeteli
    Jun 22, 2020 at 3:08
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    $\begingroup$ @akhmeteli Thanks! I corrected it. $\endgroup$ Jun 22, 2020 at 3:12
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An attempt to share the historical development of non-relativistic wave mechanics discovery by E. Schrödinger as related to following query by OP.

"So, the Schrödinger equation should include relativity in it, right? But it doesn't... How does relativity vanish from the Schrödinger equation or did it ever not "include" relativity in any way?"

The course lectures given by Hermann Weyl at ETH, Zurich, 1917 were the starting point of this wave equation journey. Its central idea was, what later came to be known as the gauge transformation. Schrödinger had studied the compiled notes very devotedly in 1921 (Influence on thinking) and often used the central idea in his subsequent work.

He applied the Weyl's measure theory (metric spaces) to the orbits of the electrons in the Bohr-Sommerfeld atomic models. He considered the path of an electron in a single complete orbit and enforced the Weyl condition of the geodetic path, thereby, implying the existence of the quantized orbits. He later realized that, this work already contained the de Broglie's ideas of the Bohr orbit in terms of the electron waves.

In the year 1922, Erwin Schrödinger was suffering the torments of respiratory illness and had moved to Alpine resort of Arosa to recuperate. He had vague ideas about the implications of his formulation about the properties of the electron orbits. It is quite possible, that had he been in a better health, the wave properties of electron could have been clear to him even before de Broglie, from his own work.

Einstein had in-fact cited de Broglie's work in making a connection between the quantum statistics and the wave properties of matter and this was known to Schrödinger, who read most of his papers (Influence on thinking). Schrödinger had later said that "wave mechanics was born in statistics" refering to his work in statsitical mechancs of ideal gases. He said that - his approach was nothing more that taking seriously de Broglie-Einstein wave theory of a moving particle, according to which the particle nature is just like an appendage to the basic wave nature.

In order to think about what kind of waves would satisfy closed obrits and the relavent equations, he was already thinking in relativistic terms (energy -momentum relations) and was thus natural that his attempt to formualate the wave equation would rest upon the foundation of relativistic equations. His first derivation for the wave equation for particles, before his celebrated Quantisierung als Eigenwertproblem (Quantization as an eigenvalue problem) 1926, was left unpublished and was based entirely upon the relativistic theory as given by de Broglie.

The crucial test of any theory at that time was the Hydrogen atom. It was required for any new theory to atleast reproduce some features of the Bohr's work on the H -atom energy levels and the quantum numbers. Further, a relativistic theory must be capable of explainng the fine structure provided by Sommerfeld equation. His relativistic theory did not agree with the experiments because it lacked a key ingredient -the electron spin.

The original manusript of his relativistic wave mechanics formualtion is at best lost and only a notebook of calculations is available in the archives. However, his non-relativistic formulation did indeed go to the print and has become a standard textbook material for undergraduate quantum mechanics course.

References:

  1. A Life of Erwin Schrödinger (Canto original series) by Walter J. Moore.

  2. The Historical Development of Quantum Theory By Jagdish Mehra, Erwin Schrödinger, Helmut Rechenberg.

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  • $\begingroup$ Great info here.Thanks a lot.But still not answering the 'why' thing. $\endgroup$ Jun 22, 2020 at 11:15
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    $\begingroup$ Yeah, that's why i said that is information is from the historical perspective. There will be some good answers discussing the physics part as well. $\endgroup$ Jun 22, 2020 at 11:49
  • $\begingroup$ I just wanted to let you know that the answer to your question "So, the Schrödinger equation should include relativity in it, right? But it doesn't... How does relativity vanish from the Schrödinger equation or did it ever not "include" relativity in any way?" is actually something that he tried to do to at first but could not make matching predictions with the experiments at that time. $\endgroup$ Jun 22, 2020 at 11:59
  • $\begingroup$ The sense in which I said that is because of a seemingly fallatical derivation of debroglie's relation.You are from India right?Then you probably know how much we people are taught in class 11 that debroglie's relation follows from E=mc^2.The facts were known to me but not the details or the reference and it's also beneficial for future users.Thank you. $\endgroup$ Jun 22, 2020 at 12:04
  • $\begingroup$ I know Manas. However, not everything can be made clear in a semester coursework. There is so much to process! $\endgroup$ Jun 22, 2020 at 12:11
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First of all, the terminology is messy. The original Schrödinger equation is nonrelativistic, however, people often call "Schrödinger equation" whatever they want, no matter what Hamiltonian they use, so, "in their book", Schrödinger equation can be relativistic.

So Schrödinger clearly built on de Broglie's relativistic ideas, why did he write a nonrelativistic equation? Actually, he started with a relativistic equation (which we now call the Klein-Gordon equation), however, it did not describe the hydrogen spectra correctly (because it did not take spin into account), so Schrödinger did not dare to publish it. Later Schrödinger noted that the nonrelativistic version (which we now know as the (original) Schrödinger equation) described the hydrogen spectra correctly (up to relativistic corrections:-) ), so he published his nonrelativistic equation.

If you are interested, I'll try to look for the references to the above historical facts.

EDIT (6/21/2020): Actually, I have found the reference: Dirac, Recollections of an Exciting Era//History of Twentieth Century Physics: Proceedings of the International School of Physics "Enrico Fermi". Course LVII. - New York; London: Academic Press, 1977. -P.109-146. Dirac recollects his conversation with Schrödinger that took place in (approximately) 1940.

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  • $\begingroup$ Yes,I came to know about the historical fact from Weinberg's QFT book volume-1.Thanks for clearing the fact about terminologies. $\endgroup$ Jun 22, 2020 at 11:13
  • $\begingroup$ "so, 'in their book', Schrödinger equation can be relativistic" I am surprised that such misnomers occur. Can you give any names (references) of the sinners? $\endgroup$
    – my2cts
    Jun 22, 2020 at 11:25
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    $\begingroup$ @my2cts journals.aps.org/pr/abstract/10.1103/PhysRev.143.978 Actually one can find a lot of such examples, so I would say one cannot call such authors "sinners", it's too late for that:-) $\endgroup$
    – akhmeteli
    Jun 23, 2020 at 0:54
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The Schrödinger equation is non-relativistic by construction. It follows from the nonrelativistic classical energy expression by applying De Broglie's idea to replace $(E,\vec p)$ by $-i\hbar (\partial_t, \vec \nabla)$.

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  • $\begingroup$ I upvoted because you are atleast on the right track to understand my qn,$\lambda=h/mv$ is non-relativistic? But,what of the "derivation" which does $E=mc^2$ and $E=h/neu$ and $ /neu=c/ \lambda$,for matter waves the velocity is v therefore $\lambda=h/mv$.This uses $E=mc^2$,and yet the relativistic nature of $\lambda=h/mv$ vanishes....WHY does THIS occur?..Is it because this "derivation" is incorrect and doesn't make any sense.I say this because many local high schools of our country give this "derivation",and many people believes in this.(I don't know whether to believe or not,I am confused). $\endgroup$ Jun 21, 2020 at 19:23
  • $\begingroup$ @ManasDogra please give a reference to an example of such a derivation. $\endgroup$
    – Ruslan
    Jun 21, 2020 at 20:32
  • $\begingroup$ chem.libretexts.org/Bookshelves/… $\endgroup$ Jun 21, 2020 at 20:40
  • $\begingroup$ Infact,Search google for "derivation of de broglie equation",you will get many sites giving this derivation--It doesn't make quite a sense to me,but literally everyone in our country is taught this at high school,and this makes everyone think that de-Broglie's relations are relativistic--this is supported by a lot of sites on the internet and possibly false interpretations from deBroglie's original paper. I am partially of the opinion that it can't be "derived",such derivtions are not also present in good books like tht by Eisberg,Resnick,Beiser.etc. $\endgroup$ Jun 21, 2020 at 20:41
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    $\begingroup$ Had De Broglie really followed this line of reasoning then he would not have deserved his Nobel prize. A better web article is this: en.wikipedia.org/wiki , complete with a link to the English translation of De Broglie thesis. $\endgroup$
    – my2cts
    Jun 21, 2020 at 23:23
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The de Broglie relationships relate energy $E$ and momentum $p$, with frequency $\nu$ and wavelength $\lambda$ \begin{equation} E = h \nu, \ \ p = \frac{h}{\lambda} \end{equation} There is nothing relativistic about this. In fact this is not even really a full theory. To get a full dynamical theory, you need to relate $E$ and $p$ in some way.

The Schrodinger equation (for one particle) is built on the non-relativistic relationship \begin{equation} E = \frac{p^2}{2m} \end{equation} When I say "built on", I mean if you compute the energy and momentum for an energy eigenstate of a free particle obeying the Schrodinger equation, the energy and momentum will obey the above relationship.

If you want a relativistic theory, you would want to find a wave equation that reproduced the relativistic relationship \begin{equation} E^2 = m^2 c^4 + p^2 c^2 \end{equation} The Klein-Gordon equation is an example of such a wave equation (and indeed Schrodinger tried it first). But, there are problems interpreting the solutions of the Klein-Gordon equation as a wavefunction. We now understand (as others have pointed out on this page) that the problem is that it's not consistent to have an interacting relativistic quantum theory with a fixed number of particles. This leads to the development of quantum field theory.


In the above I restricted myself to the case of writing the Schrodinger equation for one particle. As has also been pointed out on this page, one way to generalize quantum mechanics to quantum field theory is to promote the wavefunction to a wavefunctional (a map from field configurations to complex numbers). The wavefunctional obeys the Schrodinger equation (except now where the Hamiltonian is an operator on a much larger Hilbert space). This Schrodinger equation is relativistic, if the quantum field theory it describes is relativistic. However, this is at a much higher level of sophistication than what I believe was being asked.

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  • $\begingroup$ I believe this is the answer OP was looking for. $\endgroup$
    – Andrea
    Oct 26, 2021 at 9:12
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As it is known Schrödinger's equation diffuses an initially localized particle as far as we want in a time as small as we want (but with small probability).

In formulas the problem is parabolic : $$\partial^2_x u(x,t)=K\partial_t u(x,t)$$ .

However one could use relativistic boundary conditions $u(\pm ct,t)=0$

When solving by spectral methods we diagonalize the lhs, obtaining a time dependent base of eigenfunctions.

Plugging back in the problem, the coefficient of the development in the base are obtained by integration. I was told this were unsolvable, but the solution I could write is as an infinite system of coupled odes for the coefficients of expansion.

The eigenfunctions are $$v_n(x,t)=\cos((2n+1)\frac{\pi x}{2ct})$$, the expansion is written as $$u(x,t)=\sum_{n=0}^\infty a_n(t)v_n(x,t)$$

The solution is formally obtained as an infinite matrix exponentiated.

In this case the particle cannot fly ftl to violate (relativistic) causality.

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