In Peskin & Schroeder Ch. 2, p. 14, in proving that the NRQM propagation amplitude for a free particle is nonzero everywhere, they move from \begin{equation} U(t)~=~ \frac{1}{(2\pi)^3} \int d^3p \hspace{2pt} e^{-i(\mathbf{p}^2/2m)t} \cdot e^{i\mathbf{p}\cdot(\mathbf{x} - \mathbf{x_0})} \end{equation} to the end result: \begin{equation} U(t)~=~ \left(\frac{m}{2\pi i t}\right)^{3/2} \hspace{2pt} e^{im(\mathbf{x} - \mathbf{x_0})^2/2t}. \end{equation}

I don't quite get all of the steps in between. In evaluating the first integral, I first put it in polar coordinates, with z along $x - x_0$, but then I eventually end up with a gaussian integration that looks like it should be zero. How do I get from the first equation to the second?


The next step I do after the above is to rewrite the integral:

\begin{equation} \frac{1}{(2\pi)^2} \int_0^{\infty} \int_{1}^{-1} dp \hspace{2pt} d\cos \theta \hspace{2pt} p^2 e^{-i(\mathbf{p}^2/2m)t} \cdot e^{ip\Delta x \cos \theta} \end{equation}

where $\Delta x \equiv |x - x_0|$ and I did the integration over $\phi$. From here, I get

\begin{equation} \frac{1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} \left( e^{-ip\Delta x} - e^{ip\Delta x} \right). \end{equation}

It seems like this integration should give $0$, unless I'm making a mistake somewhere. Where's my mistake?


3 Answers 3


You seem to think that

\begin{equation} \frac{1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} \left( e^{-ip\Delta x} - e^{ip\Delta x} \right) = 0 \end{equation}

probably because the exponential functions kinda look like they cancel, but this is not so. Note that

$$ e^{-ip\Delta x} - e^{ip\Delta x} = -2i\sin(p\Delta x)$$

which means that you want to compute

\begin{equation} \frac{-2}{(2\pi)^2\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t}\sin(p\Delta x) \end{equation}

and it is quite obvious that this will not vanish. In fact we can do a little work on the second term in the original integral

\begin{equation} \frac{-1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} e^{ip\Delta x} \end{equation}

substitute $p' = -p$, then this equals

\begin{equation} \frac{1}{(2\pi)^2i\Delta x} \int_{-\infty}^{0}dp' \hspace{2pt}p'\hspace{2pt}e^{-i(\mathbf{p'}^2/2m)t} e^{-ip'\Delta x} \end{equation}

so your original integral is just

\begin{equation} \frac{1}{(2\pi)^2i\Delta x} \int_{-\infty}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} e^{-ip\Delta x} = \frac{1}{(2\pi)^2i\Delta x} i \frac{d}{d\Delta x}\int_{-\infty}^{\infty}dp \hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} e^{-ip\Delta x} \end{equation}

this is just the derivative of a normal gaussian integral. Using the general formula for gaussian integrals

\begin{equation} \int_{-\infty}^{\infty} e^{-ax^2 + bx} dx = \sqrt{\frac{\pi}{a}} ~e^{\frac{b^2}{4a}} \end{equation}

which R.G.J. has already provided in his answer we immediately obtain

\begin{equation} (\frac{m}{2\pi it})^{3/2} \exp(i \frac{\Delta x^2 m}{2t}) \end{equation}

which is your desired result.

  • $\begingroup$ Great, thanks very much. My problem was getting the coefficient of p out of the integral, and I didn’t think to do it this way. $\endgroup$
    – gh3
    Commented Aug 25, 2018 at 12:21
  • $\begingroup$ @gh3 this is actually a very useful trick called differentiation under the integral sign and will be very useful for all kinds of gaussian integrals. I'd recommend you check it out somewhere in more detail. $\endgroup$ Commented Aug 25, 2018 at 12:23

Note the integral of an arbitrary Gaussian function,

\begin{equation} \int_{-\infty}^{\infty} e^{-ax^2 + bx} dx = \sqrt{\frac{\pi}{a}} ~e^{\frac{b^2}{4a}} \end{equation}

  • 1
    $\begingroup$ Minor comment: qualified by $a>0$. $\endgroup$
    – K7PEH
    Commented Aug 24, 2018 at 17:04
  • $\begingroup$ Yep, got that. My problem is that in doing the derivation I end up with a factor of $p$ in front of the gaussian, still integrating from negative infinity to positive infinity, which means the integral evaluates to $0$. $\endgroup$
    – gh3
    Commented Aug 24, 2018 at 17:20
  • $\begingroup$ $\uparrow$ My answer is based on your short one. $\endgroup$
    – Frobenius
    Commented Aug 25, 2018 at 8:44
  • $\begingroup$ @Frobenius : thanks for elaborating on my answer. I thought it would be sufficient! $\endgroup$
    – R.G.J
    Commented Aug 25, 2018 at 19:27

Hint :

Make the hypothesis that the integral of an arbitrary Gaussian function (see R.G.J's answer) \begin{equation} \int_{-\infty}^{\infty} e^{\boldsymbol{-}ax^2\boldsymbol{+} bx} dx = \sqrt{\frac{\pi}{a}} ~e^{(b^2/4a)} \tag{01}\label{eq01} \end{equation} is valid for $\;a,b\;$ pure imaginary numbers and for our case \begin{align} a &=i\left(\dfrac{t}{2m}\right) \tag{02.1}\label{eq02.1}\\ b_k &=i\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_k, \quad k=1,2,3 \tag{02.2}\label{eq02.2} \end{align} Then \begin{align} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_1^2\boldsymbol{+}b_1 p_1}\mathrm dp_1 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_1^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_1\vert^2/2t} \tag{03.1}\label{eq03.1}\\ \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_2^2\boldsymbol{+}b_2 p_2}\mathrm dp_2 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_2^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_2\vert^2/2t} \tag{03.2}\label{eq03.2}\\ \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_3^2\boldsymbol{+}b_3 p_3}\mathrm dp_3 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_3^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_3\vert^2/2t} \tag{03.3}\label{eq03.3} \end{align} Multiplying above 3 equations side by side we have \begin{equation} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a \Vert\mathbf{p}\Vert^2\boldsymbol{+}\mathbf{b}\boldsymbol{\cdot}\mathbf{p}}\mathrm dp_1\mathrm dp_2\mathrm dp_3=\left(\dfrac{\pi}{a}\right)^{3/2}e^{(\Vert\mathbf{b}\Vert^2/4a)}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{04}\label{eq04} \end{equation} or \begin{equation} \int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{05}\label{eq05} \end{equation} and \begin{equation} \frac{1}{(2\pi)^3}\int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{m}{2\pi i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{06}\label{eq06} \end{equation} So try to prove the hypothesis.


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