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

In summary, the conversation is about studying Tong's chapter on quantizing the Dirac Lagrangian and understanding how to obtain the corresponding Hamiltonian. The discussion covers the canonical commutation relations and how to represent the Hamiltonian as an operator using plane wave solutions. There are also specific questions about the factors and signs in the equations and how they are derived from the commutation relations. The conversation ends with a request for clarification on how the equations in the conversation relate to each other.
  • #1
JD_PM
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TL;DR 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.
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 :smile:
 
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  • #2
JD_PM said:
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.
 
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  • #3
strangerep said:
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

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

strangerep said:
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.

strangerep said:
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.

strangerep said:
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.

strangerep said:
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.
 
  • #4
JD_PM said:
$$-i \gamma^w \frac{\partial}{\partial x^w} e^{i\sum_j k^j x^j}=\gamma^w k^j \delta_j^w e^{i\sum_j k^j x^j}=\gamma^j k^j e^{i\sum_j k^j x^j}=-\gamma^j k_j e^{i\sum_j k^j x^j} \tag{**}$$
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.
 
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  • #5
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).
 

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

1. What is the Dirac field?

The Dirac field is a theoretical concept in quantum field theory that describes the behavior of fermions, which are particles with half-integer spin. It is named after physicist Paul Dirac, who first proposed its existence in the 1920s.

2. What does it mean to quantize the Dirac field?

Quantization is the process of turning a classical field (described by continuous values) into a quantum field (described by discrete values). In the case of the Dirac field, this involves representing fermions as quantized excitations of the field.

3. Why is understanding the wrong way to quantize the Dirac field important?

Understanding the wrong way to quantize the Dirac field is important because it highlights the limitations of certain approaches to quantum field theory. It also helps us better understand the correct way to quantize the Dirac field, which is crucial for accurately describing the behavior of fermions in particle physics.

4. What are some consequences of quantizing the Dirac field incorrectly?

One consequence of quantizing the Dirac field incorrectly is that it can lead to unphysical predictions, such as negative probabilities or infinite energy. This can also result in inconsistencies with experimental data and theoretical predictions.

5. How can we avoid quantizing the Dirac field incorrectly?

To avoid quantizing the Dirac field incorrectly, we must use mathematically rigorous and physically sound methods, such as canonical quantization or path integral quantization. It is also important to carefully consider the symmetries and constraints of the Dirac field when performing the quantization process.

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