# One dimensional scattering cross section

My quantum mechanics book (http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf, pages 94-95 of the pdf, or 86-87 of the book) considers the one dimensional step potential $$V(x)=V_0,|x|<a,$$$$V(x)=0, |x|>a$$ It says to consider the case where the total energy of an incoming particle $E>V_0$ and $V_0<0$. Then it says that in this case, we will have reflected and transmitted particles. It says the transmitted particles can be divided into two cases, those that go straight through without being disturbed, or those that are trapped and later transmitted, i.e scattered. All the reflected particles have been scattered. This makes sense.

Now it defines the scattering cross section as the total probability for scattering. This is the sum of the probability for reflection, $|R|^2=\frac{|A_{ref}|^2}{|A_{in}|^2}$, where $A_{in}$ represents the amplitude of the incoming wavefunction which will be $A_{in}e^{ikx}$ and $A_{ref}$ represents the amplitude of the reflected wavefunction which will be $A_{ref}e^{-ikx}$, and the probability for transmission by scattering, which is where my problem arises.

So again we can say we have some transmitted wavefunction $A_{trans}e^{ikx}$, but now my book states that we should write $A_{trans}=(1+T)A_{in}$, where the one represents the possibility of a particle passing through and being transmitted without being disturbed, and the $T$ represents the possibility of a particle being trapped and subsequently scattered. Then the probability for transmission by scattering is $|T|^2$ and the scattering cross section is $$\sigma=|R|^2+|T|^2$$.

So basically I don't really understand how defining $T$ in terms of those two amplitudes ensures $|T^2|$ is the probability to be transmitted via a scattering process, and nor why the one represents being undisturbed and the $T$ being scattered. Thanks for any help in advance :)

Edit: As an extra question (but not really important relative to the above), surely you could define such a cross section even if $V_0>0$ so long as $E>V_0$ and the particle was free, because then you would always get some trapping of particles from repeated reflections in the step region. Is this right?

• What book, and what page or section? Mar 31 '15 at 12:27
– Watw
Mar 31 '15 at 12:33

There has to be some amplitude for transmission with scattering and some amplitude for transmission without scattering. The 1 in the $1+T$ term is the amplitude of the wave that does not scatter, this is just a matter of definition. The $T$ term is the amplitude for transmission by definition. The part of the wave that scatters has square amplitude $\sigma$ and the wave has to be either scattered or reflected, so $\sigma = |R|^2 + |T|^2$, where $R$ is the reflection amplitude. Given those definitions you can work out $R$ and $T$. You could use some other symbols if you like.

You can define the cross section for $V_0 > 0$, regardless of whether $E>V_0$. There is a non-zero probability for transmission even if $E<V_0$.

• You're saying that it's like this because of definition, but there must be some logic behind the definitions so that $|T|^2$ works out as the correct probability. My problem is I can't see that logic.
– Watw
Mar 31 '15 at 13:03
• Do you understand that the probability of a system being in a particular state is a square amplitude of that state? Mar 31 '15 at 13:28
• Yeah, I can't see how to use that here...
– Watw
Mar 31 '15 at 13:40
• $T$ is the amplitude that the system ended up in $x>a$ by transmission through the barrier. So the probability that it ended up in $x>a$ by transmission through the barrier is $|T|^2$. Mar 31 '15 at 14:31
• Surely $1+T$ is the amplitude it was transmitted?
– Watw
Mar 31 '15 at 15:30

As I understand - you start with a barrier ($E$ being higher or lower the $V_0$) and you get a left part (incoming wave and reflected wave) and a right part (transmited wave). Suddenly - there is a new condition: $V_0 \lt 0$, which means a hole and you have one more situation =direct passage of the particle. ok.

You have a reflected part with factor $R$, $A_{refl}=R \times A_{in}$ similar to what you had in previous case - and - the right-hand part. But let us call this r-hand factor S, not $T+1$. $S = \frac{1}{2}(e^{2i\phi^\prime} - e^{2i\phi})$ here. Quite symmetric situation to $R$. The condition $P_{trans}+P_{refl}=1$ is still valid, as you cannot loose anything from the original amplitude, $P_{trans}=|S|^2$, $P_{refl}=|R|^2$.

The logic is - as there is a hole ($V_0 \lt 0$), the right-hand part now also can contain whatever just passes through without an interaction. Let us just define $S=1+T$ to have a tool to remove that non-interacting part. Once you speak about a cross section, you need it.

The question is - why $1$? I think, this is paving a road to 3D case with a real potential. Focus on $S$, it is a complex number. If there is no interaction (e.g.no potential) $S=1$, cross section is 0, $T=0$ (phase shift$=0$), all your incoming amplitude=outgoing amplitude. In a case of interaction your $T$ and $S$ start to be non-zero, non-one and your cross section goes as $|T|^2$.

What I miss is $R$ should be actually somehow incorporated in $T$ as it is (for me) a case of interaction. If $R$ is not inside $S=1+T$, you could have an impression that in no interaction case (no potential) you still have $\sigma=|R|^2$ and your out-amplitude looks bigger than $1 \lt |R|^2 + |1|^2$. There is no such $R$ term in real potential life.

• Fig. 5.17 on page 99 seems to have $\sigma$ greater than one, but I thought it was a probability. That seems weird.
– Watw
Mar 31 '15 at 21:00
• Yes, interesting. I have an impression that this is a step further. At certain $k_0$ it can happen ,that $\sigma_0 = \frac{4\pi}{k^2_0}$, $E_0 \sim k^2_0$ is called resonance energy. The picture resembles me this (and the typical dependence on $1/k^2$). However I agree, that the previously defined $\sigma$ looks more like something to be less than 1. Apr 2 '15 at 6:57
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