Relativistic quantum mechanics

[redtree]

TL;DR Summary

Question about its formulation

Given that the Minkowski metric implies the Lorentz transformations and special relativity, why do the equations of relativistic quantum mechanics, i.e., the Dirac and Klein-Gordon equations, require a mass term to unite quantum mechanics and special relativity? Shouldn't their formulation in Minkowski 4-space be sufficient for the unification of QM and SR? Furthermore, why wouldn't formulating the Schrodinger equation in Minkowski 4-space also make that equation relativistic?

 

[PeroK]

Summary:: Question about its formulation

Given that the Minkowski metric implies the Lorentz transformations and special relativity, why do the equations of relativistic quantum mechanics, i.e., the Dirac and Klein-Gordon equations, require a mass term to unite quantum mechanics and special relativity? Shouldn't their formulation in Minkowski 4-space be sufficient for the unification of QM and SR? Furthermore, why wouldn't formulating the Schrodinger equation in Minkowski 4-space also make that equation relativistic?

You can have a massless Klein-Gordon equation. But, if you want to study particles with mass then you need a mass term.

The mass term in the Dirac equation has the effect of coupling the spinor components. The massless Dirac equation is called the Weyl Equation. As electrons have mass, you need a mass term.

The SDE is fundamentally not Lorentz invariant, as it has a second derivative of time and a first derivative with respect to space.

 

[redtree]

By SDE, do you mean the Schrodinger equation?

 

[PeroK]

By SDE, do you mean the Schrodinger equation?

Yes.

 

[redtree]

Lorentz invariance applies to 4-vectors. Energy, where $E = \hbar k_0$ and $k_0$ is the Fourier conjugate of time $x_0$, is not a 4-vector, and therefore is not Lorentz invariant.

 

[PeroK]

Lorentz invariance applies to 4-vectors. Energy, where $E = \hbar k_0$ and $k_0$ is the Fourier conjugate of time $x_0$, is not a 4-vector, and therefore is not Lorentz invariant.

What is the relevance of that statement?

 

[vanhees71]

Concerning the massless Dirac equation one should say that it has the additional feature of chiral symmetry, i.e., the projections to the left- and righthanded parts decouple in the equation of motion
$$\psi_L=\frac{1-\gamma_5}{2} \psi, \quad \psi_{R} = \frac{1+\gamma_5}{2} \psi.$$
You still have also a representation of space reflections as a symmetry, which exchange $\psi_L$ and $\psi_R$.

If you have a massless Weyl equation, there's only a lefthanded (or righhanded) spinor field in the game, and you cannot define space reflections as a symmetry anymore. In such theories parity is violated.

 

[redtree]

The Schrodinger equation, formulated in Minkowski 4-space, already incorporates SR by virtue of its formulation in Minkowski 4-space. The Dirac equation is no more relativistic than the Schrodinger equation. The difference is that the Dirac equation describes a 4-vector $\textbf{k} = [k_0, k_1, k_2, k_3] = [k_0 , \vec{k}]$, where $\vec{p} = \hbar \vec{k}$. The Schrodinger equation describes only a component of $\textbf{k}$, namely $k_0$.

 

[redtree]

The inclusion of mass $M$ in the Dirac equation proceeds on the assumption that mass $M = \textbf{k}$.

 

[PeroK]

The Schrodinger equation, formulated in Minkowski 4-space, already incorporates SR by virtue of its formulation in Minkowski 4-space. The Dirac equation is no more relativistic than the Schrodinger equation. The difference is that the Dirac equation describes a 4-vector $\textbf{k} = [k_0, k_1, k_2, k_3] = [k_0 , \vec{k}]$, where $\vec{p} = \hbar \vec{k}$. The Schrodinger equation describes only a component of $\textbf{k}$, namely $k_0$.

Can you write down the Lorentz Invariant SDE you are thinking of?

 

[redtree]

Lorentz invariance applies only to 4-vectors like the Fourier conjugates $\textbf{k}$ and $\textbf{x}$, where $\textbf{x} = [x_0, x_1, x_2, x_3] = [x_0, \vec{x}]$. In position space, the fact that time $x_0$ is not Lorentz invariant has nothing to do with whether an equation is relativistic or not.

 

[redtree]

The same is true for its Fourier conjugate $k_0$.

 

[redtree]

The Schrodinger equation is just a subset of the Dirac equation, just as $k_0$ is a subset of $\textbf{k}$.

 

[PeroK]

The Schrodinger equation is just a subset of the Dirac equation, just as $k_0$ is a subset of $\textbf{k}$.

This makes no sense. You came here to ask a question about relativistic QM; you got an answer; to which you replied with a number of assertions that are either untrue or make no sense.

What is the purpose of this thread?

 

[redtree]

I am trying to understand the relationship between the Dirac and Schrodinger equations. What did I say that makes no sense?

 

[redtree]

I have done nothing other than state the properties of 4-vectors and the Fourier transform.

 

[PeroK]

I am trying to understand the relationship between the Dirac and Schrodinger equations. What did I say that makes no sense?

There is no relativistic SDE. The SDE can be shown to be the non-relativistic limit of the KG equation and the Dirac equation: for example, the four-component Dirac spinor reduces to the two-component spin state in the non-relativistic limit.

 

[redtree]

By relativistic, I assume you mean Lorentz invariant. I agree that the SDE cannot be Lorentz invariant. It describes $k_0$, and $k_0$ cannot be Lorentz invariant any more than its Fourier conjugate time $x_0$ can be Lorentz invariant.

However, just as time $x_0$ describes a component of the Lorentz invariant 4-vector $\textbf{x}$, $k_0$ describes a component of the Lorentz invariant 4-vector $\textbf{k}$.

 

[PeroK]

By relativistic, I assume you mean Lorentz invariant. I agree that the SDE cannot be Lorentz invariant. It describes $k_0$, and $k_0$ cannot be Lorentz invariant any more than its Fourier conjugate time $x_0$ can be Lorentz invariant.

However, just as time $x_0$ describes a component of the Lorentz invariant 4-vector $\textbf{x}$, $k_0$ describes a component of the Lorentz invariant 4-vector $\textbf{k}$.

That statement is of no relevance or consequence. If you simply declare the SDE to hold in one particular reference frame, then the equation that describes the same system in another frame would look entirely different.

Quantities such as four-vectors are Lorentz covariant. A relativistic theory, by definition, must have Lorentz invariant equations.

 

[redtree]

I agree. My point is that the Dirac equation describes $\textbf{k}$ and the SDE describes $k_0$. Both are formulated in Minkowski 4-space. One could easily formulate a Dirac-type equation for $\textbf{x}$ and a SDE-like equation for $x_0$.

Given that $k_0$ is a component of $\textbf{k}$, the Schrodinger equation is embedded in the Dirac equation.

 

[PeroK]

I agree. My point is that the Dirac equation describes $\textbf{k}$ and the SDE describes $k_0$. Both are formulated in Minkowski 4-space. One could easily formulate a Dirac-type equation for $\textbf{x}$ and a SDE-like equation for $x_0$.

Given that $k_0$ is a component of $\textbf{k}$, the Schrodinger equation is embedded in the Dirac equation.

Do it then! Let's see it.

 

[vanhees71]

I don't know, what you are talking about. The Dirac equation describes the time evolution of a Dirac-spinor valued wave function (or rather a Dirac-spinor field operator in the Heisenberg picture of relativistic QFT). I've no clue what it should mean to claim the Dirac equations describes $\vec{k}$ (I guess you mean a momentum with this).

 

[PeterDonis]

@redtree, you need to start giving some references, since you are making claims that make no sense. We need to see where you are getting your information from.

 

[redtree]

The Dirac equation:

\begin{equation}

\begin{split}

\left(i \hbar \gamma^{\mu} \partial_{\mu} - M \right)\psi &= 0

\end{split}

\end{equation}
For $c=\hbar = 1$ and the metric signature $(+,-,-,-)$

\begin{equation}

\begin{split}

M^2 &= (\gamma M)^2 - (\gamma M v)^2

\\

&= E^2 - p^2

\\

&= k_0^2 - \vec{k}^2

\\

&= \textbf{k}^2

\end{split}

\end{equation}

where $\textbf{k} =[k_0, i k_1,i k_2,i k_3 ] = [k_0, i\vec{k}]$, such that $\textbf{k}^2 = k_0^2 - \vec{k}^2$
Thus,

\begin{equation}

\begin{split}

M &= \pm \textbf{k}

\end{split}

\end{equation}

such that the Dirac equation can be written as follows:

\begin{equation}

\begin{split}

\left(i \hbar \gamma^{\mu} \partial_{\mu} - (\pm \textbf{k}) \right)\psi &= 0

\end{split}

\end{equation}
Furthermore,

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \gamma^0 \partial_0 \psi, i \hbar \gamma^j \partial_j \psi \right]

\end{split}

\end{equation}

where $j=1,2,3$
Given the metric signature $(+,-,-,-)$

\begin{equation}

\begin{split}

\left(i \hbar \gamma^j \partial_j \right)^2 \psi &= \left[ -k_1^2,-k_2^2,-k_3^2 \right] \psi

\\

&= -\vec{k}^2 \psi

\end{split}

\end{equation}

where

\begin{equation}

\begin{split}

\left[-k_1^2,-k_2^2,-k_3^2 \right] &= \left[(i (\pm k_1))^2,(i (\pm k_2))^2,(i (\pm k_3))^2 \right]

\\

&= (i(\pm \vec{k}))^2

\end{split}

\end{equation}
Similarly,

\begin{equation}

\begin{split}

\left(i \hbar \gamma^0 \partial_0\right)^2 \psi &= k_0^2 \psi

\\

&= (\pm k_0)^2 \psi

\end{split}

\end{equation}
Thus,

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ \pm k_0 \psi, i (\pm \vec{k}) \psi \right]

\\

&= \left[\pm k_0, i(\pm \vec{k}) \right] \psi

\\

&= (\pm \textbf{k}) \psi

\end{split}

\end{equation}
Which demonstrates that indeed

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi - (\pm \textbf{k}) \psi &= 0

\end{split}

\end{equation}

And the Dirac equation is correct.

 

[redtree]

Do it then! Let's see it.

I'll answer this once I see the response to my previous post.

 

[strangerep]

[...]
And the Dirac equation is correct.

So what? We all knew the Dirac eqn was correct. Indeed, Dirac originally formulated it such that it would satisfy relativistic invariance. In modern terminology, the space of 4-spinors satisfying the Dirac eqn "carries" a representation of the Poincare group, which happens to correspond to spin 1/2.

You can have both massive and massless spin 1/2 fields that are relativistically covariant.

Focussing on "formulating" something in Minkowski space kinda misses the crucial point. What matters is that the field under consideration must correspond to a representation of the Poincare group, and those are characterized by the 2 Poincare Casimir invariants, namely mass and spin.

Given that $k_0$ is a component of ${\textbf k}$, the Schrodinger equation is embedded in the Dirac equation.

Although true (in a sense), that's a somewhat misleading way of saying it. A better way to say it is that the Schrodinger equation can be derived from the Dirac equation in an approximation where $v \ll c$, i.e., in the nonrelativistic limit.

 

[vanhees71]

It's not the fields themselves that provide a unitary representation of the Poincare group. E.g., the transformation of the Dirac field corresponding to boosts is not a unitary transformation. In fact there are no finite-dimensional unitary representation of the proper orthochonous Lorentz group. That's because it's not a compact Lie group. There are of course infinite-dimensional unitary representations of the proper orthochronous Poincare group, realized in quantum field theory. Together with locality, microcausality and stability of the ground state you get "local representations", i.e., the application of the unitary representation leads to transformation laws for the field operators which look like that for the corresponding classical fields. This is the group-theoretical foundation of the relativistic QFTs describing the Standard model of elementary-particle physics.

 

[redtree]

So what? We all knew the Dirac eqn was correct. Indeed, Dirac originally formulated it such that it would satisfy relativistic invariance.

Indeed, I consider it obvious that the Dirac equation describes the 4-vector $\textbf{k}$ but I was asked to prove it.
Now I will demonstrate how the Schrodinger equation is a component of the Dirac equation just as $k_0$ is a component of $\textbf{k}$.
I note:

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \gamma^0 \partial_0 \psi, i \hbar \gamma^j \partial_j \psi \right]

\\

&= \left[ \hbar(\pm k_0) \psi, i \hbar (\pm \vec{k}) \psi \right]

\end{split}

\end{equation}

where $j=1,2,3$
The notation $\pm$ is merely a statement that an observable can take positive or negative values. In most of mathematics, this is assumed to be true unless specified otherwise. Energy and momentum can take negative values according to the Dirac equation. Of course, negative energy can be considered as positive energy with a negative charge, and negative momentum can represented as positive momentum with a negative spin state. In this context, the spin state and charge are represented by the sign of $\vec{k}$ and $k_0$, respectively. Thus, the $\pm$ notation can be dropped such that:

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \gamma^0 \partial_0 \psi, i \hbar \gamma^j \partial_j \psi \right]

\\

&= \left[ \hbar k_0 \psi, i \hbar \vec{k} \psi \right]

\end{split}

\end{equation}

where $j=1,2,3$
The recognition that an observable can be positive or negative (or instead of negative, at least positive with a negative spin state or charge) also means that the Dirac matrix is unnecessary in the Dirac equation when writing the Dirac equation utilizing $k_0$ and $\vec{k}$ form. Without the Dirac matrix, the charge and spin state become part of the observable $k_0$ and $\vec{k}$ as their sign, such that:

\begin{equation}

\begin{split}

\gamma^0 \partial_0 &= \partial_t

\end{split}

\end{equation}

and

\begin{equation}

\begin{split}

\gamma^j \partial_j &= \partial_{\vec{x}}

\end{split}

\end{equation}
Thus,

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \partial_t \psi, i \hbar \partial_{\vec{x}} \psi \right]

\\

&= \left[ \hbar k_0 \psi, i \hbar \vec{k} \psi \right]

\end{split}

\end{equation}
Given

\begin{equation}

\begin{split}

\hat{\mathcal{H}} &= i \hbar \partial_t

\end{split}

\end{equation}

and

\begin{equation}

\begin{split}

E &= \hbar k_0

\end{split}

\end{equation}
I note:

\begin{equation}

\begin{split}

i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ \hat{\mathcal{H}} \psi, i \hbar \partial_{\vec{x}} \psi \right]

\\

&= \left[ E \psi, i \hbar \vec{k} \psi \right]

\end{split}

\end{equation}
Thus, the Schrodinger equation is a component of the Dirac equation, where the Dirac equation demonstrates that energy in the Schrodinger equation can take negative values (or positive values with negative charge).

 

[Tendex]

Energy and momentum can take negative values according to the Dirac equation. Of course, negative energy can be considered as positive energy with a negative charge, and negative momentum can represented as positive momentum with a negative spin state. In this context, the spin state and charge are represented by the sign of and , respectively. [...]The recognition that an observable can be positive or negative (or instead of negative, at least positive with a negative spin state or charge) also means that the Dirac matrix is unnecessary in the Dirac equation when writing the Dirac equation utilizing and form

The math doesn't work like that, you can't simply get rid of the Dirac matrices, they are part of the formalism that includes complexification of the Lie algebras and then applying the reality conditions in real Minkowski spacetime. The time derivative being first or second order makes a difference that cannot be disposed of so easily.

 

[vanhees71]

The Dirac equation is just a possible equation of motion for a wave function corresponding to the two inequavilent fundamental representations of $\text{SL}(2,\mathbb{C})$, which is the covering group of the proper orthochronous Lorentz group $\text{SO}(1,3)^{\uparrow}$. For a very thorough and complete treatment of this group-theoretical approach to relativistic wave equations, see

Weinberg, Quantum Theory of Fields, Vol. 1 or
Sexl, Urbandtke, Relativity, Groups, Particles.

 

[PeroK]

For example, the Dirac equations should be of the form:
$$(\partial_0 - \partial_1 + i\partial_2)\psi_1 + \partial_3 \psi_2 = - im\psi_4$$
Plus three more similar equations. There's no immediate way to decouple these: they are fundamentally equations for a four-component spinor. Certainly the SDE is nowhere to be seen.

 

[vanhees71]

Sure you get the non-relativistic limit by a formal expansion with respect to powers of $1/c$, and also choosing a convenient representations for the Dirac matrices to do that. E.g., if you minimally couple the em. field you are lead to the Pauli equation with a Pauli spinor (throughing away the anti particles). All this can be found in Bjorken&Drell.

 

[PeterDonis]

the Schrodinger equation is a component of the Dirac equation

No, it isn't, it's a non-relativistic approximation to it, as has already been pointed out.

The OP question has been answered. Thread closed.