# Why aren't four-vectors used in the definition of a Klein-Gordon quantum field?

I am a beginner who is learning QFT. When I was going through the quantisation of Klein-Gordon real-field. I got confused about something:

The solution to Klein-Gordon equations are of the form $\psi(x^\mu) \sim e^{ik_\mu x^\mu}$. Now the solutions in Peskin and Schroeder have no time dependence. It gives a Fourier transform of this kind:

$$\psi(\vec x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{i\vec k. \vec x} + a^\dagger(\vec p) e^{-i\vec k. \vec x} \Bigr]$$

My problem here is, why is the solution is a superposition of $e^{i\vec k. \vec x}$ and not $e^{ik_\mu x^\mu}$ ?

[EDIT]

For example in these notes in equations 90 and 113 the solutions are superposition of $e^{ik_\mu x^\mu}$ and I don't where these two things disagree.

• We prefer to have one distinct question per post here. I've removed your second question from this post, but feel free to post it separately. – David Z Jan 17 '14 at 2:41
• Because they are working in the Schrodinger picture and so the operators are time-independent. Later in the chapter they will switch to the Heisenberg picture, and then four-vectors will be used. – Hunter Jan 17 '14 at 2:46
• I figured that out, still I am not able to convince myself, how the solutions are a superposition of $e ^{i \vec k . \vec x }$ and not a superposition of $e^ {i k_\mu x^\mu}$ – user35952 Jan 17 '14 at 2:49
• You can use $k^\mu x_\mu$ in the definition, but then you integrate over with $\delta(p^2 - m^2)$ which gives you the usual definition. – Slereah Dec 4 '17 at 8:18

As I mentioned in the comments, P&S are working in the Schrodinger picture which means that the operator fields are time-independent. Of course, in the Heisenberg picture, the solution of the Klein-Gordon equation is dependent on time (and then it will have four-vectors). In order to see this, let us write down the Klein-Gordon equation: $$\left(\partial^2 + m^2 \right) \phi(x)=0$$ where $g=\mathrm{diag}(+1,-1,-1,-1)$. Then the solutions in the Heisenberg picture can be written as: $$\phi(x) = e^{\pm i p_\mu x^\mu}$$ which can be easily verified: \begin{aligned} \partial^2 \phi & = \partial_\mu \partial^\mu \left(e^{\pm i p_\nu x^\nu}\right) \\& = \partial_\mu \left(\pm i p^\mu\right) \left(e^{\pm i p_\nu x^\nu}\right) \\& = \left(\pm i p^\mu\right) \left(\pm i p_\mu\right) e^{\pm i p_\nu x^\nu} \\& = - p_\mu p^\mu e^{\pm i p_\nu x^\nu} \\& = -(E^2 - \mathbf{p}^2) e^{\pm i p_\nu x^\nu} \\& = -m^2 e^{\pm i p_\nu x^\nu} \\& = -m^2 \phi \end{aligned} and so: $$\left(\partial^2 +m^2\right)\phi = \left(-m^2 +m^2\right)\phi = 0$$ It is normal to write the solution in terms of positive frequency solutions and negative frequency solutions: $$\phi(x)=\phi_+(x) + \phi_-(x) = a e^{- i p_\nu x^\nu} + b e^{+ i p_\nu x^\nu} \tag{1}$$ Of course, we also need to sum over all energy-momentum values $p_\mu$ (because equation $(1)$ is a solution for any value of $p_\mu$). Hence, the general solution is: $$\phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + b(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right]$$ where $N$ is a normalization constant.

In order to see how to switch between the Dirac and Schrodinger picture, I refer you to section $2.4$ of P&S.

Edit I couldn't help my self and will quickly add this:

P&S are discussing the real Klein-Gordon field, which means: $$\phi = \phi^*$$ and so: $$\int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i p^\mu x_\mu} + b(\mathbf{p}) e^{ip^\mu x_\mu}\right] = \int \frac{\mathrm{d}^3 \mathbf{p}}{N^*} \; \left[ a^*(\mathbf{p}) e^{ ip^\mu x_\mu} + b^*(\mathbf{p}) e^{-i p^\mu x_\mu}\right]$$ which implies: $$\begin{array}{cc} a(\mathbf{p}) = b^*(\mathbf{p}) \; ,& b(\mathbf{p}) = a^*(\mathbf{p}) \end{array}$$ and $N$ must be a real. So the real field can be written as: $$\phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + a^*(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right]$$

• Ok, so if I am getting this right. The factor $\frac{1}{\sqrt{2\omega_p}}$ shows up when we are switching between these two modes. Thanks for the detail. – user35952 Jan 17 '14 at 3:06
• @user35952 Could you clarify what you mean by the factor $\frac{1}{\sqrt{2 \omega_{\mathbf{p}}}} and which equation you are referring to? – Hunter Jan 17 '14 at 3:14 • The equation for$ \psi (x) $in my question. – user35952 Jan 17 '14 at 3:16 • @user35952 I'm going to see if I understand what you mean tomorrow as I am going to bed now. – Hunter Jan 17 '14 at 3:22 • Ok, thanks for your inputs anyway. Will try my best as well !! – user35952 Jan 17 '14 at 3:31 After the comments and answer from Hunter, I think that the difference between these two things lie in the fact that, in Schrodinger picture the$ \psi (\vec x) $satisfies the equation $$\bigg (\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2 \bigg)\psi (x) = 0$$ where$ \omega_p = \sqrt{|\vec p|^2 + m^2} $. $$\hat\psi(\vec x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{i\vec k. \vec x} + a^\dagger(\vec p) e^{-i\vec k. \vec x} \Bigr]$$ In the case of Heisenberg picture however, the Schrodinger ladder operators transform like (refer P&S section 2.4) $$a_h(\vec p) = e^{iHt}a(\vec p)e^{-iHt} = a(\vec p)e^{-iE_pt}$$ $$a^\dagger_h(\vec p) = e^{iHt}a^\dagger(\vec p)e^{-iHt} = a(\vec p)e^{iE_pt}$$ Now putting this into the expression for obtaining Heisenberg representation of the field, $$\hat\psi_h(\vec x ,t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{ip_\mu x^\mu} + a^\dagger(\vec p) e^{-ip_\mu x^\mu} \Bigr]$$ where$ p^0 = E(\vec p) \$

Now this operator satisfies the equation

$$i\frac{\partial}{\partial t} \hat\psi_h = [\hat\psi_h,\hat H]$$