Peskin & Schroeder: Free particle propagation

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

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?

EDIT:

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

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

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

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

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

You seem to think that

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

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

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

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

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

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

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

so your original integral is just

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

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

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

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

• 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.
– gh3
Aug 25, 2018 at 12:21
• @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. Aug 25, 2018 at 12:23

Note the integral of an arbitrary Gaussian function,

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

• Minor comment: qualified by $a>0$. Aug 24, 2018 at 17:04
• 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 $0$.
– gh3
Aug 24, 2018 at 17:20
• $\uparrow$ My answer is based on your short one. Aug 25, 2018 at 8:44
• @Frobenius : thanks for elaborating on my answer. I thought it would be sufficient! Aug 25, 2018 at 19:27

Hint :

Make the hypothesis that the integral of an arbitrary Gaussian function (see R.G.J's answer) $$\int_{-\infty}^{\infty} e^{\boldsymbol{-}ax^2\boldsymbol{+} bx} dx = \sqrt{\frac{\pi}{a}} ~e^{(b^2/4a)} \tag{01}\label{eq01}$$ 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 $$\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a \Vert\mathbf{p}\Vert^2\boldsymbol{+}\mathbf{b}\boldsymbol{\cdot}\mathbf{p}}\mathrm dp_1\mathrm dp_2\mathrm dp_3=\left(\dfrac{\pi}{a}\right)^{3/2}e^{(\Vert\mathbf{b}\Vert^2/4a)}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{04}\label{eq04}$$ or $$\int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{05}\label{eq05}$$ and $$\frac{1}{(2\pi)^3}\int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{m}{2\pi i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} \tag{06}\label{eq06}$$ So try to prove the hypothesis.