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Variational Calculus and momentum

  • #1
167
0

Homework Statement



Momentum, involving variational calculus.

[tex]\psi^{\dagger}[/tex]

[tex]\bar{\psi}[/tex]

matrices [tex](\alpha, \beta)[/tex]

Homework Equations



[tex]i \bar{\psi} \gamma^0 = i\psi^{\dagger}[/tex]

The Attempt at a Solution




Am I to understand, that the momentum [tex]P=i\psi^{\dagger}[/tex], concentrating on the right hand side, is really just

[tex]i \beta \bar{\psi}[/tex]

I am led to believe this since

[tex]\bar{\psi} \beta = \psi^{\dagger}[/tex]

Would I be right in saying this? If so, since the [tex]\gamma^{0}[/tex] is really just the beta matrix for the time derivatives [tex]\beta \partial_t[/tex], is this just

[tex]i \bar{\psi} \gamma^0 = \psi^{\dagger}\beta[/tex]
 

Answers and Replies

  • #2
strangerep
Science Advisor
3,123
947

Homework Statement



Momentum, involving variational calculus.

[tex]\psi^{\dagger}[/tex]

[tex]\bar{\psi}[/tex]

matrices [tex](\alpha, \beta)[/tex]

Homework Equations



[tex]i \bar{\psi} \gamma^0 = i\psi^{\dagger}[/tex]

The Attempt at a Solution




Am I to understand, that the momentum [tex]P=i\psi^{\dagger}[/tex], concentrating on the right hand side, is really just

[tex]i \beta \bar{\psi}[/tex]

I am led to believe this since

[tex]\bar{\psi} \beta = \psi^{\dagger}[/tex]

Would I be right in saying this? If so, since the [tex]\gamma^{0}[/tex] is really just the beta matrix for the time derivatives [tex]\beta \partial_t[/tex], is this just

[tex]i \bar{\psi} \gamma^0 = \psi^{\dagger}\beta[/tex]
I'm not surprised you're having trouble getting someone to answer this. You haven't really given a "problem statement" in part 1, and its hard to guess what your problem really is.
You might want to re-read the homework help guidelines more carefully.

Maybe it will help if I tell you that the Dirac adjoint ##\bar{\psi}## is defined as $$\bar{\psi} := \psi^{\dagger} \gamma^0$$ and that relations between the ##\alpha,\beta## matrices and the ##\gamma^\mu## matrices can be found in Wikipedia:

http://en.wikipedia.org/wiki/Dirac_equation
 
  • #3
167
0
Yes, wiki is not clear.

What my problem states is how to get from momentum, the relations described in the thread.
 
  • #4
strangerep
Science Advisor
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947
Yes, wiki is not clear.
What my problem states is how to get from momentum, the relations described in the thread.
(Sigh.) Wiki seemed clear enough to me.

I'll say (only) one more time that you need to express your problem/question more clearly/specifically. I still have no idea where your difficulty lies, which makes it impossible to help you. Look at some of the other threads in this forum where people did receive help, and that might give you a clue about how to phrase your questions better.
 
  • #5
167
0
(Sigh.) Wiki seemed clear enough to me.
Well very seldom is a wiki article clear, and I believed my thread was clear enough. Do you have the capabilities to answer my question?

What part of the OP seems... unreadable?
 
  • #6
167
0
Ok... Let's try this again... step by step. Is this momentum?

[tex]i\bar{\psi} \gamma^0 = i\psi^{\dagger}[/tex]
 
  • #7
167
0
And to get from the left to right, you multiply by beta,

[tex]\bar{\psi}\beta = \psi^{\dagger}[/tex]

so to get from right to left, you also multiply by beta, yes?

[tex]\bar{\psi} = \psi^{\dagger} \beta[/tex]
 
  • #8
167
0
and what does this sign mean you used,

:=
 
  • #9
strangerep
Science Advisor
3,123
947
This is the homework forum. We don't just hand people complete answers on a plate here. Rather, we help people to figure out solutions. See also the recently posted sticky thread in this forum:

https://www.physicsforums.com/showthread.php?t=94380


The original title of your thread was "Variational Calculus and momentum", but you didn't quote a specific Lagrangian in the OP. I can only guess that you're trying to work with a Dirac Lagrangian (cf. http://en.wikipedia.org/wiki/Lagrangian#Dirac_Lagrangian ).

Nor did you indicate whether you already understand the concept of "canonical momentum", and how to obtain it from a given Lagrangian.

The symbol ":=" means "is defined by".
E.g., "A := B" means A is defined by B.
 
  • #10
167
0
This is the homework forum. We don't just hand people complete answers on a plate here. Rather, we help people to figure out solutions. See also the recently posted sticky thread in this forum:

https://www.physicsforums.com/showthread.php?t=94380


The original title of your thread was "Variational Calculus and momentum", but you didn't quote a specific Lagrangian in the OP. I can only guess that you're trying to work with a Dirac Lagrangian (cf. http://en.wikipedia.org/wiki/Lagrangian#Dirac_Lagrangian ).

Nor did you indicate whether you already understand the concept of "canonical momentum", and how to obtain it from a given Lagrangian.

The symbol ":=" means "is defined by".
E.g., "A := B" means A is defined by B.
Yes, but in Canonical form - except all I am wanting to know is if this momentum term is written out correctly. I have done the work, I am just waiting for someone to clarify whether I have made a mistake or it is written right.
 
  • #11
strangerep
Science Advisor
3,123
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[...] I have done the work, [...]
If you've done the work, then show it.

(If you'd done this in your original post, as the homework forums guidelines specify, you would have got a proper answer by now.)
 
  • #12
22,097
3,282
help1please, can you please state the entire problem exactly as it is given to you??
 
  • #13
167
0
If you've done the work, then show it.

(If you'd done this in your original post, as the homework forums guidelines specify, you would have got a proper answer by now.)
The work is in the OP. I haven't randomly stuck variables together for the sake of it. If you are familiar with this covariant language (and my OP in not in Chinese) you should be able to confirm what I have written.

This is all I want!
 
  • #14
167
0
help1please, can you please state the entire problem exactly as it is given to you??
I want my OP clarified by someone who knows their covariant language, I wish to know what I have written is correct.
 
  • #15
Fredrik
Staff Emeritus
Science Advisor
Gold Member
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407
If you want people to help you, you need to make it as easy as possible for them to answer. We shouldn't have to go to Wikipedia to remind ourselves of the definitions of these matrices. They should be included right there in the OP. If you had done your part, you would probably have had a complete answer the same day you started this thread. Instead you're being rude to the people trying to help you.

You seem to be asking if it's true that ##i\psi^\dagger=i\beta\bar\psi##, i.e. if ##\psi^\dagger=\beta\bar\psi##. Since the left-hand side is a 1×4 matrix and the right-hand side is a 4×4 matrix times a 1×4 matrix, I say no. (What does the right-hand side even mean?)

I don't understand your argument for why this equality should hold, so if I had seen this thread earlier, I would have asked for clarification, just like strangerep did.
 
  • #16
167
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Except he didn't ask for clarification exactly, he originally pointed me towards wiki and wiki can hardly be understood half the time. It doesn't explain itself well.

Now, why are you saying

[tex]\psi^{\dagger} = \beta \bar{\psi}[/tex] isn't true?

It's pretty standard that to go from psi dagger to psi bar you simply multiply beta on bothe sides of the equation... but yes, you are helping me more than the last poster. You are probing the equations and questioning them, which is what I wanted.
 
  • #17
Fredrik
Staff Emeritus
Science Advisor
Gold Member
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Now, why are you saying

[tex]\psi^{\dagger} = \beta \bar{\psi}[/tex] isn't true?
What I'm saying is that the product of a 4×4 matrix and a 1×4 matrix is undefined. The equality is neither true nor false, since the right-hand side is undefined.

The product XY of two matrices X and Y is only defined when the number of columns of X is equal to the number of rows of Y.

It's pretty standard that to go from psi dagger to psi bar you simply multiply beta on bothe sides of the equation...
According to the Wikipedia article that strangerep linked to, we have ##\bar\psi=\psi^\dagger\gamma^0 =\psi^\dagger\beta##. And we can solve this for ##\psi^\dagger## by multiplying both sides with ##\beta## from the right. The result is ##\bar\psi\beta=\psi^\dagger##. So yes, you can "go from psi dagger to psi bar" and vice versa using these equations, but if you're going to write down something like ##\beta\bar\psi## (with ##\beta## on the left), you're going to have to define what that product means.
 
Last edited:
  • #18
167
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Oh I see, that's just my clumsiness.
 
  • #19
167
0
Right, so that is ok... What about my momentum terms, have I understood them correctly as well?
 
  • #20
Fredrik
Staff Emeritus
Science Advisor
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What about the momentum terms? I don't understand what you're asking. I also don't understand what you're doing towards the end of post #1.
 
  • #21
Hans de Vries
Science Advisor
Gold Member
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Dear, help1please

If you are talking about momentum then everybody will think about
the standard expression for momentum given by:

[tex]
p^\mu ~\propto~ j^\mu ~=~ \bar{\psi}\gamma^\mu\psi
[/tex]

This is a four momentum. po is the energy component and the other
three form the spatial momentum vector. So your OP doesn't seem to
make much sense.

However it seems that you are really talking about another term which
transforms exactly like the momentum p as in.

[tex]
p ~=~\sqrt{p^2_x+p^2_y+p^2_z}
[/tex]

This term is found in the 0th component of the axial vector current
given by:

[tex]
j^\mu_A ~=~ \bar{\psi}\gamma^\mu\gamma^5\psi
[/tex]

The term [itex]j^o_A[/itex] transforms like.

[tex]
\frac{v^2/c^2}{\sqrt{1-v^2/c^2}}
[/tex]

and so the one you're after must be:

[tex]
p ~=~\sqrt{p^2_x+p^2_y+p^2_z} ~\propto~ \bar{\psi}\gamma^o\gamma^5\psi ~=~ \psi^\dagger\gamma^5\psi ~=~ \psi^\dagger\beta\psi
[/tex]

Where the β is more generally used in relativistic chemistry while γ5 is
used more in Quantum Field Theory.

Hans.
 
Last edited:

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