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In quantum scattering theory, we imagine the incident particle to be incoming plane wave $\psi(z)=Ae^{ikz}$ and assume that the scattered wave is of the form of a spherical wave for large r. So we begin by looking for solutions to the Schrodinger equation of the form $$\psi(r,\theta)\approx A\left[e^{ikz}+f(\theta)\frac{e^{ikr}}{r} \right] \tag{1}$$

I have a few issues with the above equation. Firstly and most obviously, both terms in the above expression are not normalizable. This in-and-of itself does not bother me too much. The incoming plane wave can easily be replaced with a normalizable wave packet (which is not done simply due to mathematical difficulty and inconvenience). The spherical wave on the other hand only blows up at the origin but the above equation only applies for large r and so presumably near the origin the wave function no longer takes the form of a spherical wave but rather some other form which is normalizable. My issue though is that we are told to believe that the differential cross section is $D(\theta)=\frac{d\sigma}{d\Omega}=|f(\theta)|^2$ and that this corresponds to "the probability of scattering in a given direction $\theta$" (to quote Griffiths). How are we able to intepret $|f(\theta)|^2$ as a probability, if in general the function it is multiplied by is not normalizable? In the 1 dimensional case, we determine the reflection and transmission coeffecients by taking ratios of the the amplitudes of the reflected wave and the transmitted wave with the incoming waves amplitude. In this case, none of the plane waves are normalizable but its still reasonable to assume that these ratios will meaningfully provide relative probabilities because we can imagine that if there was a normalization constant, it would simply cancel out when we take the ratio. In the 3d case, that clearly can't occur because we never take any ratios, we simply use $|f(\theta)^2|$ as a measure of the probability.

So what exactly is going on here? Why are we able to interpret anything in equation 1 as a probability if nothing in the equation is normalizable and we never take any ratios?

Any help on this issue would be most appreciated!

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  • $\begingroup$ One needs quantum field theory and Feynman diagrams to real tackle how crossection is calculated in elementary particle physics. Look at Veltman lectures here cds.cern.ch/record/276704 $\endgroup$ – anna v May 17 at 16:02
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    $\begingroup$ Why do you say the function you multiply $|f(\theta)|^2$ is not normalisable? When you integrate over a solid angle there is an area element $r^2 \sin\theta$ which cancels with the $1/r^2$ $\endgroup$ – RogerJBarlow May 17 at 16:39
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The first thing to note is that just because the incoming and outgoing wave functions live in the continuous spectrum, it does not mean they can't be normalized, one just need to use the continuous spectrum normalization. It can be verified, e.g. in [1], that it is okay to normalize a free particle plane wave along the $z$ axis to be of the form $\psi_{in} = e^{ikz}$.

Consider a particle moving along the $z$ direction, with wave function $$\psi_{in} = A e^{ikz}, $$ which then scatters off a central scattering center. We will come back to $A$, but one can assume $A=1$ for now. Note $p = \hbar k = m v$ tells us the particle moves with velocity $v = \hbar k/m$. The amplitude of the scattered wave is assumed to depend only on $\theta$ and decrease radially so it can be modelled as a spherical wave of the form $$\psi_{out} = f(\theta) \frac{e^{ikr}}{r}.$$

The probability for measuring the scattered particle in some volume element $dV$, is $$dP_{out} = |\psi_{out}|^2 dV.$$ In spherical coordinates we can write $dV$ as $$dV = dS dr = dS v dt$$ for $v$ the incoming particle velocity magnitude along the radial direction, is $$dP_{out} = |\psi_{out}|^2 dV = |\psi_{out}|^2 dS \cdot dr = |\Psi_{out}|^2 dS \cdot v dt.$$ From this the probability per unit time is $$\frac{dP_{out}}{dt} = |\Psi_{out}|^2 dS \cdot v = \frac{v}{r^2}|f(\theta)|^2 dS = v |f(\theta)|^2 d \Omega.$$ The value of this, however, changes depending on the behavior of the incoming particles, different incoming particle configurations result in different values of $v$, the rest is intrinsic to the scattering problem. This is nothing but the usual motivation for setting up a cross section, we need to divide out the quantity which depends on the incoming configuration to get an intrinsically measurable quantity which turns out to have units of area, denoted $d \sigma$ $$d \sigma = \frac{dP_{out}}{v dt} = |f(\theta)|^2 d \Omega.$$ We would like to interpret $v = v/V = \rho v = j$, for a unit incoming volume $V = 1$ (i.e. $\rho = 1/V = 1$), as the current density of incoming particles in a unit volume, but the only way this is possible is if $<\hat{j}_{in}> = v$ i.e. we must get $j = v$ when we insert $\psi_{in} = A e^{ikz}$ into $\hat{\mathbf{j}} = \frac{i \hbar}{2m} ( \psi_{in} \nabla \psi_{in}^* - \psi_{in}^* \nabla \psi_{in})$. Checking this you should get that $A = 1$ meaning we can interpret $\psi_{in} = e^{ikz}$ as the wave function of an incoming particle whose current density is equal to the particle velocity. Now we have that the cross section is the ratio of the 'transition probability per unit time' to the incident flux $$d \sigma = \frac{dP_{out}/dt}{j_{inc}} = |f(\theta)|^2 d \Omega.$$ 'Quantum field theory' scattering involves the same thinking since one is examining continuous spectrum initial and final free particle systems at $\pm \infty$, the main difference is you have to get to $d \sigma$ starting from the time-dependent Schrodinger equation, usually in the second quantization formalism.

It is interesting to note that if one considers time-dependent perturbation theory in the continuous spectrum, e.g. Fermi's Golden Rule to first order, where the initial wave function is also in the continuous spectrum, it is the fact that the incoming wave function must now have a dimension that implies we can no longer interpret 'Fermi's Golden rule' as having the correct dimensions of a transition probability that it usually has, instead leading one to being able to interpret the result as a cross section, the Born approximation formula for the cross section. For this, the above, and more on the continuous spectrum, see [1].

References:

  1. Landau, Lifshitz - Quantum Mechanics, 3rd Edition.
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  • $\begingroup$ Thanks for the great response! Just a few things before I accept your answer. First, when you use $dS$ in the above, you mean that $dS=r^2\sin(\theta) d\theta d\phi$ correct? Then finally, the only thing I don't really understand in your answer is when you initially state that free particle plane waves are normalizable. Do you mean that a continuous superposition of free particle waves is normalizable (i.e a wave packet) or do you mean that an actual plane wave can be normalized because as far as i'm aware $\int_\infty ^\infty e^{ikx}=\infty$ ? $\endgroup$ – SalahTheGoat May 19 at 6:59
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    $\begingroup$ Regarding $dS$, yes. Regarding plane waves, for continuous spectrum wave functions we want $\psi(q) = \int a_f \psi_f(q) df$ to act as close to the way $\psi(q) = \sum_n c_n \psi_n(q)$ acts as possible, e.g. $|c_n|^2$ gets replaced by $|a_f|^2 df$, $\sum_n |c_n|^2 = 1$ by $\int |a_n|^2 df = 1$, $c_n = \int \psi(q) \psi_n^*(q) dq$ by $a_{f'} = \int dq \psi(q) \psi_{f'}^*(q) dq$, and $\int dq \psi_n^*(q) \psi_m(q) = \delta_{mn}$ by $\int \psi_{f'}^*(q) \psi_f(q) dq = \delta(f - f')$, etc... This last condition is how one normalizes in the continuous spectrum. $\endgroup$ – bolbteppa May 19 at 16:35
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    $\begingroup$ For example, with $\psi_{\mathbf{p}}(\mathbf{r}) = C e^{i \mathbf{p} \cdot \mathbf{r}/\hbar}$, you can check that $\int \psi_{\mathbf{p}'}(\mathbf{r})\psi_{\mathbf{p}}(\mathbf{r}) d^3 \mathbf{r} = \delta(\mathbf{p}'-\mathbf{p})$ fixes $C = 1/\sqrt{(2 \pi \hbar)^3}$. This ensures that $\psi(\mathbf{r}) = \int a(\mathbf{p}) \psi_{\mathbf{p}}(\mathbf{r}) d^3 \mathbf{p}$ acts as a wave function should. For more on this see sections 5, 15 and 19 of [1]. $\endgroup$ – bolbteppa May 19 at 16:36

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