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JD_PM

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- 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

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

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

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##

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

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

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

Any help is appreciated.

Thanks

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

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

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