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Making sense of the canonical anti-commutation relations for Dirac spinors

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When doing scalar QFT one typically imposes the famous 'canonical commutation relations' on the field and canonical momentum: $$[\phi(\vec x),\pi(\vec y)]=i\delta^3 (\vec x-\vec y)$$ at equal times ($x^0=y^0$). It is easy (though tedious) to check that this implies a commutation relation for the creation/annihilation operators $$[a(\vec k),a^\dagger(\vec k')]=(2\pi)^32\omega\delta^3(\vec k-\vec k') $$

When considering the Dirac (spinor) field, it is usual (see e.g. page 107 of Tong's notes or Peskin & Schroeder's book) to proceed analogously (replacing commutators with anticommutators, of course). We postulate \begin{equation} \{\Psi(\vec x),\Psi^\dagger(\vec y)\}=i\delta^3(\vec x -\vec y) \end{equation} and, from them, derive the usual relations for the creation/annihilation operators.

I'd always accepted this and believed the calculations presented in the above-mentioned sources, but I suddenly find myself in doubt: Do these relations even make any sense for the Dirac field? Since $\Psi$ is a 4-component spinor, I don't really see how one can possibly make sense out of the above equation: Isn't $\Psi\Psi^\dagger$ a $4\times 4$ matrix, while $\Psi^\dagger\Psi$ is a number?! Do we have to to the computation (spinor-)component by component? If this is the case, then I think I see some difficulties (in the usual computations one needs an identity which depends on the 4-spinors actually being 4-spinors). Are these avoided somehow? A detailed explanation would be much appreciated.

As a follow-up, consider the following: One usually encounters terms like this in the calculation: $$u^\dagger \dots a a^\dagger\dots u- u\dots a^\dagger a \dots u^\dagger $$ Even if one accepts that an equation like $\{\Psi,\Psi^\dagger\}$ makes sense, most sources simply 'pull the $u,\ u^\dagger$ out of the commutators' to get (anti)commutators of only the creation/annihilation operators. How is this justified?

EDIT: I have just realized that the correct commutation relation perhaps substitutes $\Psi^\dagger$ with $\bar \Psi$ (this may circumvent any issue that arises in a componentwise calculation). Please feel free to use either in an answer.

This post imported from StackExchange Physics at 2014-12-06 00:44 (UTC), posted by SE-user Danu
asked Dec 2, 2014 in Theoretical Physics by Danu (175 points) [ no revision ]
retagged Dec 6, 2014
Most voted comments show all comments
I'm not sure I got your point. In the case of Dirac spinors the anticommutation rules work by components (see for example Peskin & Schroeder p.58): $$\{ \psi_\alpha(\textbf{x}),\psi_\beta^\dagger(\textbf{y}) \} = \delta_{\alpha \beta} \delta^3(\textbf{x}-\textbf{y})$$ so you do have a 4x4 matrix on RHS. Was this your question?

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user glance
@glance The computations are componentwise? This is not obvious to me (the 'standard sources' certainly don't explicitly display these indices), although I already mentioned it as a possibility. If you can carry out the computation this way (avoiding some perhaps minor problems that I foresee) and post it as an answer, I'd be more than happy to accept it!

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
See Peskin and Schroeder eq. 3.86 for verification that the relation is component-wise.

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user joshphysics
@joshphysics damnit, I was looking at 3.89 and didn't see any components, and freaked out. I guess that settles it...

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
Somebody should post that as an answer! (FWIW I came to the same conclusion from Tong's notes)

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user David Z
Most recent comments show all comments
Pretty anticlimactic, duh.

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Daniel
@Daniel lol, yeah... I guess that's how these things are bound to end.

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu

1 Answer

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One usually starts from the CCR for the creation/annihilation operators and derives from there the commutation rules for the fields. However, one can start from either (see for example here about this). Suppose we want then to start from the equal-time anticommutation rules for a Dirac field $\psi_\alpha(x)$: $$ \tag{1} \{ \psi_\alpha(\textbf{x}), \psi_\beta^\dagger(\textbf{y}) \} = \delta_{\alpha \beta} \delta^3(\textbf{x}-\textbf{y}),$$ where $\psi_\alpha(x)$ has an expansion of the form $$ \tag{2} \psi_\alpha(x) = \int \frac{d^3 p} {(2\pi)^3 2E_\textbf{p}} \sum_s\left\{ c_s(p) [u_s(p)]_\alpha e^{-ipx} + d_s^\dagger(p) [v_s(p)]_\alpha e^{ipx} \right\}$$ or more concisely $$ \psi(x) = \int d\tilde{p} \left( c_p u_p e^{-ipx} + d_p^\dagger v_p e^{ipx} \right), $$

and we want to derive the CCR for the creation/annihilation operators: $$ \tag{3} \{ a_s(p), a_{s'}^\dagger(q) \} = (2\pi)^3 (2 E_p) \delta_{s s'}\delta^3(\textbf{p}-\textbf{q}).$$ To do this, we want to express $a_s(p)$ in terms of $\psi(x)$. We have: $$ \tag{4} a_s(\textbf{k}) = i \bar{u}_s(\textbf{k}) \int d^3 x \left[ e^{ikx} \partial_0 \psi(x) - \psi(x) \partial_0 e^{ikx} \right]\\ = i \bar{u}_s(\textbf{k}) \int d^3 x \,\, e^{ikx} \overset{\leftrightarrow}{\partial_0} \psi(x) $$ $$ \tag{5} a_s^\dagger (\textbf{k}) = -i \bar{u}_s(\textbf{k}) \int d^3 x \left[ e^{-ikx} \partial_0 \psi(x) - \psi(x) \partial_0 e^{-ikx} \right] \\ =-i \bar{u}_s(\textbf{k}) \int d^3 x \,\, e^{-ikx} \overset{\leftrightarrow}{\partial_0} \psi(x) $$ which you can verify by pulling the expansion (2) into (4) and (5). Note that these hold for any $x_0$ on the RHS.

Now you just have to insert in the anticommutator on the LHS of (3) these expressions and use (1) (I can expand a little on this calculation if you need it).

most sources simply 'pull the $u, u^\dagger$ out of the commutators' to get (anti)commutators of only the creation/annihilation operators. How is this justified?

There is a big difference between a polarization spinor $u$ and a creation/destruction operator $c,c^\dagger$.

For fixed polarization $s$ and momentum $\textbf{p}$, $u_s(\textbf{p})$ is a four-component spinor, meaning that $u_s(\textbf{p})_\alpha \in \mathbb{C}$ for each $\alpha=1,2,3,4$. Conversely, for fixed polarization $s$ and momentum $\textbf{p}$, $c_s(\textbf{p})$ is an operator in the Fock space. Not just a number, which makes meaningful wondering about (anti)commutators.

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user glance
answered Dec 2, 2014 by glance (65 points) [ no revision ]
By "double arrow" do you mean $\Longleftrightarrow$ or two right arrows ontop of each other?

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Kyle Kanos
I mean the arrow which points on both sides.. the symbol used to indicate a derivative on the right minus a derivative on the left: in $ a \bar{\partial}_\mu b \equiv a \partial_\mu b - (\partial_\mu a) b$ the symbol that would normally be used instead of the bar in $\bar{\partial}$

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user glance
\overset{\leftrightarrow}{\partial} $\overset{\leftrightarrow}{\partial}$

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Robin Ekman
A single lined arrow that points both ways is \leftrightarrow: $\leftrightarrow$. You can put that over by using \overset{up}{down}: $\overset{\leftrightarrow}{\partial_\mu}$.

This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Kyle Kanos

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