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Shouldn't Lagrangians be real (hermitian)?

by pellman
Tags: hermitian, lagrangians, real
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pellman
#1
Sep17-10, 10:46 AM
P: 582
I find the Lagrangian associated with the Dirac equation given in texts as

[tex]\mathcal{L}=\bar{\psi}\left(i\gamma^\mu \partial_\mu - m\right)\psi[/tex]

or

[tex]\mathcal{L}=i\bar{\psi}\gamma^\mu \partial_\mu \psi- m\bar{\psi}\psi[/tex]

[tex]\mathcal{L}=i \psi^{\dagger}\gamma^0\gamma^\mu \partial_\mu \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

Taking the hermitian conjugate of the expression on the right, we get

[tex](-i) \partial_\mu\psi^{\dagger}\left(-\gamma^\mu\right) \gamma^0 \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

[tex]=-i \partial_\mu\psi^{\dagger}\gamma^0\gamma^\mu \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

[tex]=-i \partial_\mu\bar{\psi}\gamma^\mu \psi- m\bar{\psi}\psi[/tex]

which, as far as I can tell, is not equal to [tex]\mathcal{L}[/tex]

So if it is not hermitian, is that ok?
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ismaili
#2
Sep17-10, 10:55 AM
P: 161
Quote Quote by pellman View Post
I find the Lagrangian associated with the Dirac equation given in texts as

[tex]\mathcal{L}=\bar{\psi}\left(i\gamma^\mu \partial_\mu - m\right)\psi[/tex]

or

[tex]\mathcal{L}=i\bar{\psi}\gamma^\mu \partial_\mu \psi- m\bar{\psi}\psi[/tex]

[tex]\mathcal{L}=i \psi^{\dagger}\gamma^0\gamma^\mu \partial_\mu \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

Taking the hermitian conjugate of the expression on the right, we get

[tex](-i) \partial_\mu\psi^{\dagger}\left(-\gamma^\mu\right) \gamma^0 \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

[tex]=-i \partial_\mu\psi^{\dagger}\gamma^0\gamma^\mu \psi- m\psi^{\dagger}\gamma^0\psi[/tex]

[tex]=-i \partial_\mu\bar{\psi}\gamma^\mu \psi- m\bar{\psi}\psi[/tex]

which, as far as I can tell, is not equal to [tex]\mathcal{L}[/tex]

So if it is not hermitian, is that ok?
You could do integration by part in the kinetic term.
After throwing away the surface term, you could see that the Lagrangian is real.
jostpuur
#3
Sep17-10, 11:30 AM
P: 2,066
IMO the Lagrange's function should be real. It is unfortunate that the mainstream theoretical physicists don't agree with this.

Recall that the Lagrange's function [itex]L[/itex] is obtained by integrating the Lagrange's density [itex]\mathcal{L}[/itex] over spatial coordinates with fixed time variable. So you cannot do integration by parts with the time derivative there. So not only the Lagrange's density is not real, also the Lagrange's function is not real too.

The action is obtained by integrating the Lagrange's function over time, however, this doesn't make the action real either! If you choose some arbitrary field configuration [itex]\psi(x,t)[/itex] in spacetime domain [itex]\mathbb{R}^3\times [T_1,T_2][/itex], and integrate the action for the field to propagate from [itex]T_1[/itex] to [itex]T_2[/itex], the surface terms don't necessarily vanish at these endpoints.

pellman
#4
Sep17-10, 12:08 PM
P: 582
Shouldn't Lagrangians be real (hermitian)?

Thanks, ismaili. I think jostpuur is correct though.

But why do we want the Lagrangian, or at least the action integral, to be real anyway. what is wrong with having a complex action?
jostpuur
#5
Sep17-10, 12:17 PM
P: 2,066
I get headache from trying to comprehend how complex actions fit in the path integral picture. We are interested in a quantity

[tex]
\int \mathcal{D}q\; \exp\Big(\frac{i}{\hbar}S(q)\Big)
[/tex]

How is the "cancellation due to oscillations"-argument supposed to work, if [itex]S[/itex] is not real? If the imaginary component is positive, then it will make the convergence more reliable. But negative imaginary component would cause bad divergence.

Of course... you never really do that with the Dirac Lagrangian, so I guess it's not a problem. But I see a problem from theoretical point of view.
pellman
#6
Sep17-10, 04:07 PM
P: 582
Ok. If it makes your head hurt, I'm not going think about it.

So what should we use for a Dirac Lagrangian? How about ...

[tex]\mathcal{L}=i\bar{\psi}\gamma^\mu \partial_\mu \psi - i\left(\partial_\mu\bar{\psi}\right)\gamma^\mu \psi - 2m\bar{\psi}\psi[/tex]

?
calhoun137
#7
Sep18-10, 12:03 PM
P: 21
It might help you to think about the path integral formalism. We make up a Lagrangian with the Laplacian in it, and then describe the path using the eigen-states of the laplacian. In computing these path integrals, it's really nice to be able to use the power of complex analysis, i.e. the residue theorem. So making the Lagrangian complex is a formal trick that is just used to manually compute some integrals. I don't have a copy of peskin and shroder, but as I recall this was discussed in that book.

It's clear that if the Lagrangian corresponds to a measurable physical quantity, that the Lagrangian operator must be Hermetian. This still doesn't prevent us from applying complex analysis for computing path integrals, fortunately.
pellman
#8
Sep18-10, 02:49 PM
P: 582
Ok. I am going to hold off this question until I get a better grasp of how results are actually calculated in QFT, esp with path integrals. Thanks.
Parlyne
#9
Sep19-10, 01:47 AM
P: 546
Quote Quote by pellman View Post
Ok. If it makes your head hurt, I'm not going think about it.

So what should we use for a Dirac Lagrangian? How about ...

[tex]\mathcal{L}=i\bar{\psi}\gamma^\mu \partial_\mu \psi - i\left(\partial_\mu\bar{\psi}\right)\gamma^\mu \psi - 2m\bar{\psi}\psi[/tex]

?
Divide that by 2 and you've got a perfectly valid version of the free Dirac Lagrangian. In most applications, however, adding [itex]C \partial_\mu\left(i\bar{\psi}\gamma^\mu\psi\right)[/itex], where C is a constant, will have no physical effect. The reason for this is that the integral over spacetime of this term is equivalent to a surface integral of [itex]i\bar{\psi}\gamma^\mu\psi[/itex] over the surface that spans spacial and temporal infinity. Since (pretty much by definition) we can't have any kind of current passing through that surface, we can take this integral to be 0.
pellman
#10
Sep19-10, 05:50 PM
P: 582
why divide by 2? The equations of motion are homogeneous in [tex]\mathcal{L}[/tex], so any multiplicative constant should drop out.

The scale of the Lagrangian matters in the definition of the conjugate momentum, of course, and, consequently, the Hamiltonian also, but that shouldn't effect the physics, should it? At least, it doesn't classically. Is there more to the quantum story?
lucid
#11
Sep23-10, 06:05 AM
P: 5
why does the lagrangian or lagrangian density have to be real?

it's not an observable in general. it's a lorentz scalar and nothing more. it is constructed from observables -- such as the energy and momenta, etc..

you can try to impose it to be real, but then you will get no damping in the path integral.

there is one line of argument however -- pursued in Frankel's geometry in physics -- which argues that it is just relativised energy -- in which you might expect it to be real. however, i think that argument is more philosophical than anything else.
Ben Niehoff
#12
Sep23-10, 11:15 AM
Sci Advisor
P: 1,594
The action indeed must be real. And as you can see by integrating by parts, the Dirac action is real.

Also, the overall scale of the Lagrangian is important in quantum mechanics, so it is a good habit to get the constants right.
pellman
#13
Sep23-10, 01:21 PM
P: 582
Quote Quote by Ben Niehoff View Post
Also, the overall scale of the Lagrangian is important in quantum mechanics
how so? see my last remark above.
lucid
#14
Sep23-10, 02:48 PM
P: 5
in field theory, depending on the interactions, the size of the lagrangian can be scaled up or down by redefining the fields. if the action/lagrangian is scale invariant, then nothing really changes. however, if it isn't the interaction terms change in size.

in the quantum mechanics case, you can similarly scale the lagrangian up or down, and in generic cases (non-scale invariant) this will lead to the resizing of the interactions -- such as the rescaling of the electron charge, etc (for a gauge potential interaction)
samalkhaiat
#15
Sep23-10, 04:51 PM
Sci Advisor
P: 910
[QUOTE]
Quote Quote by lucid View Post
why does the lagrangian or lagrangian density have to be real?
In QFT, the requirement of reality (or more accurately of the Hermiticity) of the Lagrangian leads naturally to a unitary S-matrix. I suppose you know that the S-matrix must be unitary.

sam


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