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I Time/space symmetry in Dirac Equation

  1. Jun 22, 2016 #1


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    Is [itex]\frac{\partial}{\partial t}[/itex] an operator on Hilbert space? I'm a little confused about the symmetry between spatial coordinates and time in relativistic QM.

    There is a form of the Dirac equation that treats these symmetrically:

    [itex]i \gamma^\mu \partial_\mu \Psi = m \Psi[/itex]

    However, at least in nonrelativistic QM, time is not treated symmetrically with the spatial coordinates, in that the Hilbert space consists of square-integrable functions of space, not space and time. So in NRQM, [itex]\partial_x[/itex] is a Hilbert-space operator, but [itex]\partial_t[/itex] is not, since the elements of the Hilbert space are not functions of time. Another way of putting the distinction is that [itex]t[/itex] is a parameter, while [itex]\hat{x}, \hat{y}, \hat{z}[/itex] are (Hilbert space) operators.

    I'm assuming that the Dirac equation has a similar notion of a Hilbert space that is square-integrable functions of space, so that would mean that despite the symmetry between spatial and time coordinates in the covariant form of the Dirac equation, the meaning of space and time coordinates is different. So I would assume that it's true for the Dirac equation, as well, that [itex]t[/itex] is a parameter, while [itex]\hat{x}[/itex] is an operator? Similarly, [itex]\partial_x[/itex] is a Hilbert space operator, but [itex]\partial_t[/itex] is not?

    Can somebody clarify this for me?
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  3. Jun 22, 2016 #2


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    No, time is not an observable in QT (neither in relativistic nor in non-relativistic). The reason is simply that otherwise the Hamiltonian would always have the entire real axis as a spectrum, which contradicts the very clear observation that there's a stable ground state. If this weren't the case no stable matter, including us, would exist!

    The Dirac equation (as any other relativistic wave equation) has no physically sensible single-particle wave-function interpretation in the sense of "first quantization". Dirac's attempt to interpret it as that automatically lead him to the hole-theoretical formulation of QED, which is equivalent to the now usual QFT formulation but much more cumbersome. The advantage of Fermi over Dirac is that Fermi's sea really exists, while Dirac's is an outdated concept, nowadays substituted by the Feynman-Stueckelberg trick ;-).

    In QFT all operators, representing observables, are built from the field operators.
  4. Jun 22, 2016 #3


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    Time is merely a parameter in the standard formulation of QM. The Hilbert space of the Dirac equation is:
    ## \mathcal{H} = L^2 \left(\mathbb{R}^3, \mathbb{C}^4\right)##. Notice that 3 over there. It should be a 4, but it isn't. When you write the wavefunction as ##\psi (x,y,z,t) ##, the four variables are not really on an equal footing.
  5. Jun 24, 2016 #4

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    This is a terrific question, and one which puzzled me for a long time too! Here's how I stopped losing sleep over it...

    Let's start out with the Lorentz-covariant form of the Dirac Equation as you have given it: $$i\gamma^\mu\partial_\mu\Psi(\mathbf {x},t)=mc\Psi(\mathbf {x},t)$$ For clever choices of ##4 \times 4## matrices ##\mathbf{\alpha}=(\alpha_1,\alpha_2,\alpha_3)## and ##\beta## we can rewrite this in the form: $$i\frac {\partial}{\partial t}\Psi(\mathbf {x},t)=(-ic\mathbf {\alpha} \cdot \nabla + \beta mc^2)\Psi(\mathbf {x},t)$$ By defining $$H_0:=-ic\mathbf {\alpha} \cdot \nabla + \beta mc^2$$ we can write this equation in the form $$i\frac {\partial}{\partial t}\Psi(\mathbf {x},t)=H_0\Psi(\mathbf {x},t)$$Whilst this is getting close to the Hilbert Space formulation, we mustn't jump to conclusions because our solutions ##\Psi(\mathbf{x},t)## are functions of space and time instead of functions of space parametrized by time. To get Schrodinger mechanics, we carefully separate out space and time with a scalpel ##\psi## (lower case psi this time!) defined as follows:$$\psi(t):=\Psi(\cdot,t)$$ It turns out that for each time ##t##, ##\psi(t)## is not only a function of ##x##, but is square-integrable with respect to the following Hilbert Space (depending on which way you like your eggs): $$\mathscr{H} := L^2(\mathbb{R})^4\equiv L^2(\mathbb{R}^3,\mathbb{C}^4) \equiv L^2(\mathbb{R}^3) \otimes \mathbb{C}^4$$ Furthermore, since our operator ##H_0## defined above does not depend either explicitly or implicitly on time, it can safely be interpreted as an operator in ##\mathscr{H}##. (I'll leave out some subtleties about the precise domain on which ##H_0## must be defined as this doesn't impact on your question.)

    Finally, we can now write $$i\frac {d}{dt}\psi(t)=H_0\psi(t)$$ where, for any instant of time ##t##, ##\psi(t)## is a function on ##\mathbb{R}^3## (into ##\mathbb{C}^4##), and is interpreted as the (spatial) wave function of the electron at time ##t##.

    The reason I went through all that is to convince you that it is indeed true that a Lorentz-covariant differential equation can be re-expressed in a form that looks completely asymmetric in space and time. That's why we often use the modifier 'manifestly' to describe equations that are purposely written in a space/time symmetric way.

    It is (to me at least) extremely surprising to first find out that the Hilbert Space formalism (which feels so non-relativistic) can accommodate both a relativistic and a non-relativistic treatment of a system, but it turns out to have a compelling explanation. Hilbert Spaces are flexible enough to carry representations of both the Galilean group and the Lorentz-Poincare group (and pretty much any other algebraic invention actually!). Classical state spaces don't have that linear structure.

    Another good way to reconcile this state of affairs is to revisit good old Maxwell's Equations. We first learn these equations in a very non-relativistic-looking way by considering E-B fields defined on space but which vary in time. Yet, they can be re-expressed in the form of a simple (coordinate-free) tensor field on Minkowski spacetime. You can't get more manifestly Lorentz-covariant than that; there aren't even any transformations to check!
  6. Jun 24, 2016 #5

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    I realized after posting this response that I left out an important part of the OP's query, which asked whether ##\frac {\partial} {\partial t}## is a Hilbert Space operator.

    The answer is definitely no, because the functions ##\Psi(\mathbf{x},t)## do not constitute a Hilbert Space. Specifically, they are not square-integrable (which you can see if you think about what is going on in the time direction).

    Actually though, even if ##\Psi(\mathbf{x},t)## was square-integrable, it wouldn't make sense for it to live in the Hilbert Space of states, because being a function of space and time it represents the entire life of the system (in the same way a world line does). And it makes no sense at all to ask how a world-line evolves in time!

    But what about the ##\frac {d} {dt}## I introduced in the Schrodinger version of the Dirac Equation? $$i\frac {d}{dt}\psi(t)=H_0\psi(t)$$Nope. Although this time our ##\psi## functions belong to the Hilbert Space, the mathematical object ##\frac {d} {dt}## acts upon is not a state, but rather a parametrized curve of states (and it returns one of these as well): $$\frac {d} {dt} : \psi (t) \mapsto \dot\psi (t)$$ A Hilbert Space observable ##A##, on the other hand, is a (usually linear) map ##A:\mathscr{H}\to\mathscr{H}##.
    Last edited: Jun 24, 2016
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