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Recently in my class we studied quantization of fields and I'm brooding over an argument/ motivation on the construction of the quantization of the Klein-Gordon field. Recall the "classical" Klein-Gordon field is a solution of Klein Gordon-equation Klein Gordon-equation and looks like

$$\phi(\vec{x},t) = \int c \cdot d^3p\left[a(\vec{p})\mathrm{e}^{+i(\vec{p}\cdot\vec{x}-E_pt)} + b(\vec{p})\mathrm{e}^{-i(\vec{p}\cdot\vec{x}-E_pt))}\right] $$

where $c$ is an appropriate normalization constant and $a(\vec{p})$ and $b(\vec{p})$ are coefficients with respect the expansion with respect the eigen vector basis of the hamiltonian. When we quantize the $a(\vec{p})$ and $b(\vec{p})$ become operators $\hat{a}(\vec{p})$ and $\hat{b}(\vec{p})$ in

$$\hat{\phi}(\vec{x},t) = \int c \cdot d^3p\left[\hat{a}(\vec{p})\mathrm{e}^{+i(\vec{p}\cdot\vec{x}-E_pt)} + \hat{b}^(\vec{p})\mathrm{e}^{-i(\vec{p}\cdot\vec{x}-E_pt))}\right] $$

and in the lecture we called $\hat{a}(\vec{p})$ the "creation" operator and $\hat{b}(\vec{p})$ the "annihilation" operator. But why not reversed? I not understand why $\hat{a}(\vec{p})$ is now the creation and $\hat{b}(\vec{p})$ annihilation. Therefore why the creation corresponds to exponention with negative sign and annihilation with positive and not vice versa?

As a "reason" or let say a motivation my lecturer explained it as follows:

If we consider a process with initial state described by wave function $\phi_i e^{-iE_it}$ and final state described by wave function $\phi_f e^{-iE_ft}$ and we want to calculate the probability amplitude then when we integrate over $\int_{-\infty}^{+\infty} dt \int d^3 \vec{x}$ the integrand is given by

$$(\phi_f e^{-iE_if})^* \hat{\phi}(\vec{x},t) \phi_i e^{-iE_it} = (\phi_f)^* e^{+iE_if}) \hat{\phi}(\vec{x},t) \phi_i e^{-iE_it} $$

So the exponential of the final state is complex conjugated. This "contains" morally the reason why the creation operator corresponds to exponention with negative sign and annihilation with positive sign. Of course, as the lecturer added that's not a formal proof but a motivation why this choice might be "reasonable".

Unfortunately, I was not clever enough to understand why this elementary observation on the integrand I sketched above provides the hint why the creation operator corresponds to exponention with negative sign and annihilation with positive sign and not in reversed way. I think that the essential ingredient to solve the problem is to understand if $\phi_i e^{-iE_it}$ is arbitrary initial state then what is

$$\hat{\phi}(\vec{x},t) \phi_i e^{-iE_it}~?$$

Assume that the initial state is $|0\rangle$. What is $\hat{\phi}(\vec{x},t) |0\rangle$? My hope is $\hat{\phi}(\vec{x},t) |0\rangle = |\vec{x}\rangle$ since the well known relation between eigen vectors of momentum and place operators gives $\langle p | |\vec{x} \rangle = e^{-i px}$. So if $\hat{\phi}(\vec{x},t) |0\rangle = |\vec{x}\rangle$ then indeed we can conlude that $\hat{a}(\vec{p})$ is the creation operator with $\hat{a}(\vec{p}) |0\rangle= |p \rangle$. But for this we need to verify that $\hat{\phi}(\vec{x},t) |0\rangle = |\vec{x}\rangle$ is true but that's not clear for me.

Does anybody have an idea what my lecturer possibly had in mind making this sketch and how this observation provide a hint/motivation why in the quantization of Klein-Gordon field the creation and annihilations operators were chosen in that way and not the reversed way? I have no idea how this sketch justifies the choice.

In physics.stackexchange I found a couple of questions dealing with similar problems like here, here or here. The motivation of my question is primarily to understand why the sketch by my lecturer which I tried to reproduce above gives a "reason" or at least a "hint" that answers my problem.

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2 Answers 2

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I haven't done the calculation, but you could try to interchange the definition of creation and annilation operators, and check if your equation still satisfies the commutation relation at equal time

$$ \left[\phi(\mathbf{x}),\pi\left(\mathbf{x}^{\prime}\right)\right]=i\delta^{(3)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) $$ which by definition the quantized field should satisfy. My belief is that your calculation will give an extra minus sign.

Notice that, when you have a $\phi(\mathbf{x},t)$, you can get the conjuagte momentum $\pi(\mathbf{x},t)=\dot{\phi}(\mathbf{x},t)$. To check whether the above commutation relation is satisfied, you should remember to take $t=0$ (or any other time), so that $\phi$ and $\pi$ are at the same time. In other words, you should check $[\phi(\mathbf{x},t=0),\pi\left(\mathbf{x}^{\prime},t=0\right)]$

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  • $\begingroup$ I should add that, of course, we should also require the creation and annilation operator to satisfy the correct commutation relation with $[a(p),a^{\dagger}(q)]=(2\pi)^3\delta(p-q)$ $\endgroup$
    – Liuke LYU
    Dec 27, 2021 at 22:09
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Since up to now nowbody posted an answer I would like present a heurstical hand wavy argument which came to me into mind which migth be exactly that one which my lecturer also intended to use. I would be thankful if somebody could look through it and tell me if what I'm write now make any sense.

Recall I asked why when we have our quantized KG-field

$$\hat{\phi}(\vec{x},t) = \int c \cdot d^3p\left[\hat{a}(\vec{p})\mathrm{e}^{+i(\vec{p}\cdot\vec{x}-E_pt)} + \hat{b}(\vec{p})\mathrm{e}^{-i(\vec{p}\cdot\vec{x}-E_pt))}\right] $$$$

the $\hat{a}(\vec{p})$ corresponds to creation operator and $\hat{b}(\vec{p})$ to annihilation. And the heuristical hint my lecturer gave in the course was to consider

$$(\phi_f e^{-iE_if})^* \hat{\phi}(\vec{x},t) \phi_i e^{-iE_it} = (\phi_f)^* e^{+iE_if}) \hat{\phi}(\vec{x},t) \phi_i e^{-iE_it} $$

How does it help me? Firstly, a general wave function $ |\varphi(t) \rangle$ is given by $e^{-iH_{KG}t}|\varphi(0) \rangle$ , note we have here minus in exponent. Expanding $ |\varphi(t) \rangle$ in momenta basis $\{|p \rangle \}$ we obtain $|\varphi(t) \rangle= \sum_p e^{-iH_{KG}t}c(p) |p \rangle =\sum_p e^{-iE_pt} c(p)|p \rangle$, $E_p >0$. $E_p >0$ means we consider unly positive energies and therefore no antiparticles are involved. Note that $c(p)$ not depend on time.

Now it is reasonable require that $\hat{\phi}(\vec{x},t) |0\rangle$ is a wave function in usual sense, that is the exponent in time evolution operator has negate sign. Therefore by above $\hat{\phi}(\vec{x},t) |0\rangle= \sum_p e^{-iE_pt} c(p)|p \rangle$.

Assume $\hat{b}(\vec{p})$ creates and $\hat{b}(\vec{p})$ annihilates. Then $\hat{a}(\vec{p}) |0 \rangle= |0 \rangle$ and $\hat{b}(\vec{p}) |0 \rangle= |p \rangle$ and thus $\hat{\phi}(\vec{x},t) |0\rangle= \sum_p \mathrm{e}^{-i(\vec{p}\cdot\vec{x}-E_pt))} |p \rangle= \sum_p e^{+iE_p t} c(p) |p \rangle$. Comparing the time dependent exponents we obtain a contradiction, therefore $\hat{a}(\vec{p})$ is creation. Doe my argument make sense?

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