3
$\begingroup$

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 $\phi(x)$:

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

Improve this question
$\endgroup$
3
  • $\begingroup$ 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 $t=0$. $\endgroup$
    – Valter Moretti
    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

Reset to default
4
$\begingroup$

Hint:
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
$\phi(p)=\frac{2\pi}{\sqrt{2E_{\mathbf p}}}[\delta(p^0-E_{\mathbf p})a_{\mathbf p}+\delta(p^0+E_{\mathbf p})b_{\mathbf{-p}}^{\dagger}]$ .
Thus
$\phi(x)=\int{\frac{dp^4}{(2\pi)^4}\phi(p)e^{-ip\cdot x}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-ip\cdot x}+b_{\mathbf{p}}^{\dagger}e^{ip\cdot x}]}$.
Obviously this is just the last equation in your question.
Then the inverse Fourier transforms are
$\phi(p)=\int{d^4x\phi(x)e^{ip\cdot x}}$,
and
$\phi(\mathbf p,t)\equiv \frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]=\int{d^3\mathbf x\phi(x)e^{-i\mathbf{p\cdot x}}}$.

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).

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.