# Ladder operators in Klein -Gordon canonical quantisation

• spaghetti3451
In summary, the equations (2.25) and (2.26) from the Peskin and Schroeder text show the quantum Klein-Gordon field and its momentum density in Fourier space. Upon substitution of ##{\bf{p}} \rightarrow {\bf{-p}}##, the equations for the creation and annihilation operators appear to be inconsistent. However, upon further examination, it is clear that the creation and annihilation operators are independent of each other and the inconsistency is a result of a mistake in the equation for ##\pi({\bf{x}})##.
spaghetti3451
The quantum Klein-Gordon field ##\phi({\bf{x}})## and its momentum density ##\pi({\bf{x}})## are given in Fourier space by

##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)## and

##\pi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} (-i) \sqrt{\frac{ \omega_{{\bf{p}}}}{2}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} - a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##.

[These are equations (2.25) and (2.26) from the Peskin and Schroeder.]

Now, I used the substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained

##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = a^{\dagger}_{{\bf{p}}}##.

On the other hand, I used the same substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained

##a_{{\bf{p}}} = - a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = - a^{\dagger}_{{\bf{p}}}##.

Can someone explain what's going on?

Last edited:
There is a - missing in the second term of ##\Pi(\vec{p})##. That's seen in the Heisenberg picture and the relation ##\Pi=\dot{\phi}##.

Sorry, that's my own mistake. Peskin and Schroeder has the minus sign. I've made the edit now.

Actually, with the minus sign in the second term of ##\pi({\bf{x}})##, I get the inconsistency in the ladder operators.

I don't understand, how you get these equations in the first place. The ##\hat{a}_{\vec{p}}## are an independent set of annihilation operators. There shouldn't be a relation between ##\hat{a}_{\vec{p}}## and ##\hat{a}_{-\vec{p}}^{\dagger}##.

Well, under the substitution ##{\bf{p}} \rightarrow {\bf{-p}}##,

##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##

becomes

##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{-\bf{p}}}}} \big( a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{-{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} \big)##,

and then I compared the coefficients of the exponential ##e^{-i{\bf{p}} \cdot {\bf{x}}}## on both sides to obtain ##a^{\dagger}_{{\bf{p}}} = a_{-{\bf{p}}}## and the coefficients of the exponential ##e^{i{\bf{p}} \cdot {\bf{x}}}## on both sides to obtain ##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}##.

But now that you mention it, I'm beginning to think my argument is wrong. The creation and annihilation operators are, in fact, independent of each other, i.e. for instance, ##a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}}## is the same as ##a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}}## because the integration is over the entire range of momenta from ##-\infty## to ##\infty##.

Am I correct?

## 1. What are ladder operators in the Klein-Gordon canonical quantization?

Ladder operators are mathematical operators used in the Klein-Gordon canonical quantization to describe the energy levels of a quantum system. They are used to create and destroy particles and to determine the allowed energy states of a system.

## 2. How do ladder operators work in the Klein-Gordon canonical quantization?

Ladder operators work by acting on a quantum state to either increase or decrease its energy. They are represented by Hermitian operators and have corresponding eigenvalues that correspond to the energy levels of the system.

## 3. What is the significance of ladder operators in the Klein-Gordon canonical quantization?

The significance of ladder operators in the Klein-Gordon canonical quantization is that they provide a way to describe the energy levels of a quantum system. They also allow us to calculate the probabilities of different energy states and to understand the dynamics of a system.

## 4. Can ladder operators be used for other quantum systems?

Yes, ladder operators can be used for other quantum systems, such as the harmonic oscillator and the hydrogen atom. They are a general mathematical tool used to describe the energy levels of any quantum system.

## 5. How are ladder operators related to the concept of creation and annihilation operators?

Ladder operators are closely related to creation and annihilation operators. In fact, they are essentially the same thing, just expressed in different mathematical forms. Creation and annihilation operators describe the creation and destruction of particles in a system, while ladder operators describe the energy levels of the system.

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