# Understanding the wrong way to quantize the Dirac Field | Part 1

• I

## Summary:

I want to understand how to quantize the Dirac Field in the wrong way in detail. The reasons why this is not the right approach will be discussed in the following part.

## Main Question or Discussion Point

I've been studying Tong's beautiful chapter (pages 106-109; See also Peskin and Schroeder pages 52-58), together with his great lectures at Perimeter Institute, on how to quantize the following Dirac Lagrangian in the wrong way

$$\mathscr{L}=\bar{\psi}(x)(i\not{\!\partial}-m)\psi(x) \tag{5.1}$$

And I want to understand how to get the following Hamiltonian (the idea was to ask everything in one post but I realized it was too lengthy and I decided to split it).

$$H = \int \frac{d^3 k}{(2\pi)^3} \omega_{\vec k} \Big( b_{\vec k}^{s \dagger} b_{\vec k}^s-c_{\vec k}^{s \dagger}c_{\vec k}^s+(2\pi)^3\delta^{(3)}(0) \Big) \tag{5.12}$$

Let's go step by step. We start out by imposing the canonical commutation relations

$$[\psi_{\alpha} (\vec x), \psi_{\beta} (\vec y)]=[\psi_{\alpha}^{\dagger} (\vec x), \psi_{\beta}^{\dagger} (\vec y)]=0$$

$$[\psi_{\alpha} (\vec x), \psi_{\beta}^{\dagger} (\vec y)]= \delta_{\alpha \beta} \delta^{(3)}(\vec x - \vec y) \tag{5.3}$$

Since we're dealing with a free-Lagrangian, any solution to the equations of motion coming out of the Dirac-Lagrangian can be written as a sum of plane waves

$$\psi(\vec x) = \sum_{s=1}^2 \int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x}+c_{\vec k}^{s \dagger}v^s(\vec k) e^{-i\vec k \cdot \vec x}\Big]$$

$$\psi^{\dagger}(\vec x) = \sum_{s=1}^2 \int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^{s \dagger} u^{s \dagger} (\vec k) e^{-i \vec k \cdot \vec x}+c_{\vec k}^s v^{s \dagger}(\vec k) e^{i\vec k \cdot \vec x}\Big] \tag{5.4}$$

By the way, ##(5.3)## commutation relations are equivalent to

$$[b_{\vec k}^r, b_{\vec q}^{s \dagger}] = (2 \pi)^3 \delta^{rs} \delta^{(3)} (\vec k - \vec q)$$

$$[c_{\vec k}^r, c_{\vec q}^{s \dagger}] = -(2 \pi)^3 \delta^{rs} \delta^{(3)} (\vec k - \vec q) \tag{5.5}$$

Where the negative sign avoids getting a negative norm.

OK so far; next we discuss how to get the ##(5.12)## Hamiltonian; I have specific questions.

As we know, the Hamiltonian is defined as

$$H(q,p) := p^a \dot q_a - L \tag{1.7}$$

Here ##\dot q = \dot \psi## and ##p = i \psi^{\dagger}## so

$$\mathscr{H}=i \psi^{\dagger} \dot \psi - \mathscr{L}$$

Question 1: This is a naive one; Tong gets

$$\mathscr{H}=\bar \psi(-i\gamma^i \partial_i + m)\psi \tag{5.8}$$

Which means that ##\mathscr{H}=- \mathscr{L}##. This equality implies that ##i \psi^{\dagger} \dot \psi=0##. This can only be possible (in this case) if ##\dot \psi=0##. But how do we know that? or am I missing something?

OK let's assume ##(5.8)##. We have ##H=\int d^3 x \mathscr{H}##

Now it's time to turn the Hamiltonian into an operator (i.e. get ##(5.12)##).

We start by writing down a sum of plane wave solutions for ##(-i\gamma^i \partial_i + m)\psi##

$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^s(-\gamma^i k_i +m) u^s (\vec k) e^{i \vec k \cdot \vec x}+c_{\vec k}^{s \dagger}(\gamma^i k_i +m)v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{*}$$

Question 2: this is also a naive one; How do ##(-\gamma^i k_i +m)## and ##(\gamma^i k_i +m)## factors show up in equation ##(*)##? Specifically, I do not see why ##k_i## shows up and why the signs are as presented. I know that we have to use the index rule ##\vec k \cdot \vec x := \sum_i x^ik^i = -x^ik_i## to justify the signs but I do not see how to use it.

OK let's assume ##(*)##. Now given the following equations (let's assume them to be true)

$$(\gamma^{\mu} k_{\mu} - m)u(\vec k)= \begin{pmatrix} -m & k_{\mu}\sigma^{\mu} \\ k_{\mu} \bar \sigma^{\mu} & -m \\ \end{pmatrix} u(\vec k)=0 \tag{4.105}$$

$$(\gamma^{\mu} k_{\mu} + m)v(\vec k)= \begin{pmatrix} m & k_{\mu}\sigma^{\mu} \\ k_{\mu} \bar \sigma^{\mu} & m \\ \end{pmatrix} v(\vec k)=0 \tag{4.111}$$

Then Tong asserts that using ##(4.105)## and ##(4.111)## we get

$$(-\gamma^i k_i +m) u^s (\vec k)=\gamma^0 k_0 u^s (\vec k), \ \ \ \ (\gamma^i k_i +m) v^s (\vec k)=-\gamma^0 k_0 v^s (\vec k) \tag{5.9}$$

Thanks to ##(5.9)## we get

$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \sqrt{\frac{\omega_{\vec k}}{2}} \gamma^0 \Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x} - c_{\vec k}^{s \dagger} v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{5.10}$$

Question 3: how to get ##(5.19)## out of ##(4.105)## and ##(4.111)##? Here I am completely lost, so a hint would be appreciated. Besides, I do not see why ##k_0## doesn't show up in ##(5.10)##

Any help is appreciated.

Thanks

Last edited:

Related Quantum Physics News on Phys.org
strangerep
As we know, the Hamiltonian is defined as $$H(q,p) := p^a \dot q_a - L \tag{1.7}$$ Here ##\dot q = \dot \psi## and ##p = i \psi^{\dagger}## so $$\mathscr{H}=i \psi^{\dagger} \dot \psi - \mathscr{L}$$
Question 1: This is a naive one; Tong gets $$\mathscr{H}=\bar \psi(-i\gamma^i \partial_i + m)\psi \tag{5.8}$$ Which means that ##\mathscr{H}=- \mathscr{L}##.
How did you get that? (Tong's eqn(5.1) for ##\mathscr{L}## contains a 4D slash-partial.)

Question 2: this is also a naive one; How do ##(-\gamma^i k_i +m)## and ##(\gamma^i k_i +m)## factors show up in equation ##(*)##? Specifically, I do not see why ##k_i## shows up and why the signs are as presented.
The ##\partial_i## on the LHS means ##\partial/\partial x^i##. So just differentiate by ##x^i## inside the integrand. The only function of ##x## therein is the exponential which involves ##\vec k \cdot \vec x##, which can be written as ##\sum_j k^j x^j## . (Maybe you got hung up by failing to use a different summation symbol from the one in ##\partial_i##? )

Question 3: how to get ##(5.19)## out of ##(4.105)## and ##(4.111)##? Here I am completely lost, so a hint would be appreciated.
You get (5.19) by just using the expression (5.11) for ##H##, with the anticommutation relations (5.15) and (5.16). You don't need the equations of motion (4.105)/(4.111). Just crunch it through.

Besides, I do not see why ##k_0## doesn't show up in ##(5.10)##
Tong's (5.4) skips some steps. In fact, the Fourier transform of ##\psi(x)## starts off as a 4D Fourier transform, but because these fields are "on-shell", we can use the special relativity relation ##m^2 = E^2 - |\vec p|^2## to reduce the 4D transform to the 3D transform shown in (5.10). This is done by expressing ##E## in terms of ##p## and ##m##, and performing the ##E## integral by contour integration and the Cauchy residue theorem. Many textbooks show how to do this. It's not easy, but definitely a great exercise to slog through if you've never confronted something like that before.

JD_PM
How did you get that? (Tong's eqn(5.1) for ##\mathscr{L}## contains a 4D slash-partial.)
You're absolutely right, I missed that eqn(5.1) contains a 4D slash-partial (so that ##\dot \psi=\partial_0 \psi##). We indeed get

$$\mathscr{H}=i \bar \psi \gamma^0 \partial_0 \psi - \mathscr{L}= \bar \psi(-i\gamma^i \partial_i+m)\psi \tag{5.8}$$

The ##\partial_i## on the LHS means ##\partial/\partial x^i##. So just differentiate by ##x^i## inside the integrand. The only function of ##x## therein is the exponential which involves ##\vec k \cdot \vec x##, which can be written as ##\sum_j k^j x^j## . (Maybe you got hung up by failing to use a different summation symbol from the one in ##\partial_i##? )
Ahhh I think I see how it works. Let me first be mathematically precise with the key rule here (please let me know if you do not agree).

$$-i \gamma^w \frac{\partial}{\partial x^w} e^{i\sum_j k^j x^j}=\gamma^w k^j \delta_j^w e^{i\sum_j k^j x^j}=\gamma^j k^j e^{i\sum_j k^j x^j}=-\gamma^j k_j e^{i\sum_j k^j x^j} \tag{**}$$

(At least looks neat! ).

Using ##(**)## I indeed get ##(*)## but that does not necessarily mean that ##(**)## is OK.

You get (5.19) by just using the expression (5.11) for ##H##, with the anticommutation relations (5.15) and (5.16). You don't need the equations of motion (4.105)/(4.111). Just crunch it through.
Btw I meant (5.9), (5.19) was a typo; sorry about that.

Actually it was really easy to get those two (once I saw how of course); more details here.

Tong's (5.4) skips some steps. In fact, the Fourier transform of ##\psi(x)## starts off as a 4D Fourier transform, but because these fields are "on-shell", we can use the special relativity relation ##m^2 = E^2 - |\vec p|^2## to reduce the 4D transform to the 3D transform shown in (5.10).
That's really interesting. Actually I think Tong chose to work in the Schrödinger picture (i.e. time independent operators) to get directly the mode expansion in 3D.

This is done by expressing ##E## in terms of ##p## and ##m##, and performing the ##E## integral by contour integration and the Cauchy residue theorem. Many textbooks show how to do this. It's not easy, but definitely a great exercise to slog through if you've never confronted something like that before.
I like how that 'it's not easy' sounds... would you like to propose an exercise which demands going through such calculation? (samalkhaiat style! ).

I've never done that kind of calculation before so I think it'd be useful.

strangerep
$$-i \gamma^w \frac{\partial}{\partial x^w} e^{i\sum_j k^j x^j}=\gamma^w k^j \delta_j^w e^{i\sum_j k^j x^j}=\gamma^j k^j e^{i\sum_j k^j x^j}=-\gamma^j k_j e^{i\sum_j k^j x^j} \tag{**}$$
I'd write ##k_j x^j## (with implicit summation) in the exponent right from the start. That way you don't need to think about whether your 4D metric is ##(-+++)## or ##(+---)##. So I get
$$-i \gamma^w \partial_w e^{i\, k_j x^j} ~=~ \gamma^w \delta^j_w k_j e^{i\, k_\ell x^\ell} ~=~ \gamma^w k_w e^{i\, k_\ell x^\ell} ~.$$ (My sign is different because Tong seems to be using the convention that a spatial dot product involves a minus sign.)

(Btw, it sometimes helps to remember that coordinates have "naturally" upstairs indices whereas momenta and x-derivatives have downstairs indices.)

I like how that 'it's not easy' sounds... would you like to propose an exercise which demands going through such calculation? (samalkhaiat style! ).
"Samalkhaiat style"?? I would never be so presumptuous.

But, hmm, well,... as a first toe in the water,... you could maybe try to prove eqn(2.40) in P+S. I.e., starting from $$\left. \int\!\! \frac{d^4p}{(2\pi)^4} (2\pi) \,\delta(p^2-m^2) \; \right|_{p^0>0} ~,$$ show that it reduces to $$\int\!\! \frac{d^3p}{(2\pi)^3} \, \frac{1}{2E_p}$$
It should probably be in a new (homework) thread.

Last edited:
JD_PM
I was thinking of posting Part 2 but I finally resolved my doubts while elaborating this question.

Thus a second part will not be posted (at least for now).