According to my lecture notes, the inverse Fourier transform of an operator $\phi(p)$ is given by

$$\phi(x)=\int \frac {d^4p}{(2\pi)^4}\phi(p)e^{-ip\cdot x}.$$

As @WenChern pointed out below, Peskin has a somewhat different formula on p.20:

$$\phi(\mathbf{x},t) = \int \frac {d^3 \mathbf{p}}{(2\pi)^3} \phi(\mathbf{p},t) e^{i\mathbf{p} \cdot \mathbf{x}}.$$

I am trying to see how these two formulas are equivalent and what would be the corresponding expansions of $\phi(p)$ oand $\phi(\mathbf p, t)$ as a Fourier transform of $\phi(x)$. Also, I'd like to know why one does not take these integrals over the mass shell in the same way one does in the following other definition of $\phi(x)$:

$$\phi(x)=\int \frac {d^3\mathbf p} {(2\pi)^32E_{\mathbf{p}}}[a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}]\biggr\vert_{p_0 = E_{\mathbf p}}.$$

  • $\begingroup$ What you write cannot hold in any sense, there must be some mistake in your source. The point is that $\phi(x)$ satisfies a second order differential equation and thus it is determined by two initial conditions: $\phi(0, \vec{x})$ and $\partial_t\phi(t, \vec{x})|_{t=0}$. Instead, your formulae say that $\phi(p)$, which completely determines $\phi(x)$, only depends on the value of $\phi(x)$ at $t=0$. $\endgroup$ Commented Nov 26, 2014 at 13:33
  • $\begingroup$ @ValterMoretti good point! I edited my question. $\endgroup$
    – Rodrigo
    Commented Nov 26, 2014 at 15:30
  • 1
    $\begingroup$ physics.stackexchange.com/q/83260 $\endgroup$
    – Simon
    Commented Nov 27, 2014 at 7:37

1 Answer 1


1. $\phi(x,t)$ at different times are not independent.
2. $\int{d^4p\delta(p^2-m^2)}=\int{d^4p\frac{\delta(p^0-E_p)}{2p^0}}$. The left side of this equation is Lorentz invariant.

This time your question is much clearer.
If $\phi(x)$ is an arbitrary function of $x$, there's nothing confusing. If $\phi(x)$ is constrained by the Klein-Gordon equation, we have
$0=(\square+m^2)\phi(x)=\int{\frac{dp^4}{(2\pi)^4}(m^2-p^2)\phi(p)e^{-ip\cdot x}}$.
Since $e^{-ip\cdot x}$s are linearly independent, $\phi(p)$ must vanish everywhere except on the mass shell $p^2=m^2$. Then the most general form of $\phi(p)$ should be
$\phi(p)=\frac{2\pi}{\sqrt{2E_{\mathbf p}}}[\delta(p^0-E_{\mathbf p})a_{\mathbf p}+\delta(p^0+E_{\mathbf p})b_{\mathbf{-p}}^{\dagger}]$ .
$\phi(x)=\int{\frac{dp^4}{(2\pi)^4}\phi(p)e^{-ip\cdot x}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-ip\cdot x}+b_{\mathbf{p}}^{\dagger}e^{ip\cdot x}]}$.
Obviously this is just the last equation in your question.
Then the inverse Fourier transforms are
$\phi(p)=\int{d^4x\phi(x)e^{ip\cdot x}}$,
$\phi(\mathbf p,t)\equiv \frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]=\int{d^3\mathbf x\phi(x)e^{-i\mathbf{p\cdot x}}}$.

Due to the limitation fo the length of characters, I add the comments below.
The first identity in the last line is the definition of $\phi(\mathbf p,t)$. The second identity in it is the inverse 3-dimensional Fourier transform of $\phi(x)=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}$. Direct comparison of $\phi(\mathbf p,t)$ and the general form of $\phi(p)$ shows that $\phi(p)$ contains aditional delta functions, while $\phi(\mathbf p,t)$ is free of delta functions. Beides, since $\phi(p)$ is the 4-dimensional Fourier transform of $\phi(x)$, it is not a function of $t$. I don't think that $\phi(p)$ can be understood as "a particle whith 4-momentum $p$". It onlly make sense mathematically. The square root is just a matter of convention which can be absorbed by $a_{\mathbf p}$ and $b_{\mathbf p}$ (see, Peskin p21).

  • $\begingroup$ I understand 2., but can you elaborate on 1.? I think that one should start with an integral over the whole spacetime constrained by a delta function, like $\int d^4 x\delta(x_0)$ but I don't really see how the delta function $\delta(x_0)$ is analogous to $\delta(p^2-m^2)$ $\endgroup$
    – Rodrigo
    Commented Nov 26, 2014 at 13:12
  • $\begingroup$ Sorry, I misunderstood your question at the first time. $\endgroup$
    – Wen Chern
    Commented Nov 26, 2014 at 13:59
  • $\begingroup$ I think it may be helpful to write the second equation in your original question in the form $\phi(\mathbf{p},t)=\int{d^3\mathbf{x}\phi(\mathbf{x},t)e^{-i\mathbf{p.x}}}$. Thus we see that $\phi(\mathbf{p},t)$ is not a function of $p$ with $p$ constrained by the mass shell condition, but a function of $\mathbf{p}$ and $t$. $\endgroup$
    – Wen Chern
    Commented Nov 26, 2014 at 14:08
  • $\begingroup$ Besides, the factor $\frac{1}{2E_{\mathbf{p}}}$ should be excluded from the first equation in your question. See, for instance, Peskin, An Introduction to Quantum Field Theory, p20. $\endgroup$
    – Wen Chern
    Commented Nov 26, 2014 at 14:18
  • $\begingroup$ Thank you for your answer! But can you explain how you got to the last line? I'm still a little confused about the difference between $\phi(p)$ and $\phi(\mathbf{p},t)$. The former (what's the word?) a particle with momentum $p$, the latter (...) a particle with momentum $\mathbf{p}$ at time $t$. But since a momentum $\mathbf{p}$ determines $p$ by $p=(E_{\mathbf p},\mathbf p)$, it seems that the latter is more "specific". Also, I think that you shouldn't have the square root in the factor $\frac 1 {2E_{\mathbf p}}$. $\endgroup$
    – Rodrigo
    Commented Nov 27, 2014 at 4:30

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