I have recently studied scattering theory on a formal level and I think I understand the subject quite well by now. However what I often struggle with is to translate the abstract identities into explicit representations and solve problems with it. I have condensed this issue down to the following example problem, which requires a little bit of algebra but is rather instructive in my opinion. I have provided most of the formulae already, there is probably a conceptual mistake somewhere though.

Consider the one-dimensional Schrödinger equation $$\left(-\frac{1}{2}\frac{d^2}{dx^2} + V(x)\right)\psi(x) = E\psi(x)$$ with the finite square well potential that is terminated by an infinite barrier on one side

$$ V(x) = \begin{cases} \infty, & \text{for } x \leq -L \\ V_0, & \text{for } -L \leq x \leq 0 \\ 0, & \text{for } 0 \leq x. \end{cases}$$ For simplicity assume $V_0<0$.

One set of scattering states for this problem is easily found as $$\psi^{(+)}(E,x) = \frac{1}{\sqrt{2\pi}}\begin{cases} \frac{I(E)\beta}{\alpha}\sin\left(\alpha(1+\frac{x}{L})\right), & \text{for } -L \leq x \leq 0 \\ e^{-i\sqrt{2E}x} + S(k) e^{i\sqrt{2E}x}, & \text{for } 0 \leq x. \end{cases}$$ Here, $S(k)$ is the scattering matrix (just a number since there is only the reflection channel here) $$S(E)=-\frac{\alpha\cot(\alpha) + i\beta}{\alpha \cot(\alpha) - i\beta}$$ and the remaining coefficients are $$I(E) = - \frac{2i\alpha}{\alpha \cos(\alpha) - i\beta \sin(\alpha)},$$ $$\alpha = \sqrt{\beta^2-2V_0L^2},$$ $$\beta = L \sqrt{2E}.$$

So far so good. Now from formal scattering theory we know that there is also a T-matrix defined by (see e.g. Eq. (7.40) in Newton's book (available on Springer Link)) $$T(E) = \langle\psi_0(E)|V|\psi^{(+)}(E)\rangle.$$

Importantly, the T-matrix is related to the scattering matrix, which in the single channel case takes the simple form (see Eq. (7.58) in Newton's book) $$S(E) = 1 - 2\pi i T(E).$$

Here, $\psi_0$ is an eigenstate of the free Hamiltonian (i.e. with $V=0$), in our example with the boundary condition at $x=-L$ we get $$\psi_0(E,x)=\sqrt{\frac{2}{\pi}}\sin\left(\beta(1+\frac{x}{L})\right).$$

Now for our example the overlap integral for the T-matrix can be evaluated in the position representation $$T(E) = V_0 \int_{-L}^{0} dx \psi_0(E,x) \psi^{(+)}(E,x)$$ and we can plug in our formulae for that. However when substituting the result into the relation to the scattering matrix, it does not hold. I have checked this using Mathematica and manual calculation.

I am clearly doing something wrong. But what? My suspicion is that I have plugged in the wrong states, but I don't know what the right ones are.

EDIT: Following the discussion with TwoBs, here is some more insight on which states should be used. As far as I understand $\psi_0(E,x)$ can just to be an eigenstate of the free Hamiltonian; $\psi^{(+)}(E,x)$ is an eigenstate of the full Hamiltonian but also defined uniquely by the Lippmann-Schwinger equation: $$|\psi^{(+)}(E)\rangle = |\psi_0(E)\rangle + G^{(+)}(E) V |\psi^{(+)}(E)\rangle,$$ with $G^{(+)}(E) = \frac{1}{E-H_0 + i0^+}$.

The explicit formula I gave for $|\psi^{(+)}(E)\rangle$ above was just some eigenstate of the full Hamiltonian, so the mistake is probably that it does not fulfill the Lippmann-Schwinger equation with the $|\psi_0(E)\rangle$ I used. But which state does?

  • 1
    $\begingroup$ As a tentative resolution, I would think that you should be using actual free solutions (plane waves) where the whole potential, not just $V_0$, is removed. In other words, the S-matrix isn't defined with respect to the free propagation on the whole real line, rather than just a semi-line? Incidentally, the free wave should have the same energy of the exact solution. $\endgroup$
    – TwoBs
    Commented Aug 24, 2018 at 8:52
  • $\begingroup$ @TwoBs That is a great idea! To clarify, would you use something like $\psi_0(E,x) \propto e^{\pm i\sqrt{2E}x}$? I actually tried these before and have 3 problems with it: a) I can't get it to give a T-matrix that fulfills the relation to the S-matrix. b) From an abstract point of few I think this would give a non-unitary S-matrix for reflection, since can not avoid opening the transmission channel if you use these solutions. c) I think that instead of an infinite potential at $x<-L$ you could also consider a boundary condition at $x=-L$. And the free states should adhere to the BCs... $\endgroup$ Commented Aug 24, 2018 at 9:05
  • $\begingroup$ @TwoBs nevertheless I think you are on the right track!! Especially because if you look at the form of the $\psi^{(+)}$ state, asymptotically you get the $e^{\pm i\sqrt{2E}x}$ states with the relative magnitude being the scattering matrix. Due to the above arguments my suspicion is, however, that the free state might be correct and the $\psi^{(+)}$ state wrong. I do not want to exclude either option though. $\endgroup$ Commented Aug 24, 2018 at 9:09
  • $\begingroup$ Not sure, another way (orthogonal to my previous comment :-) ) is perhaps to change your definition of the S-matrix and call $S$ the relative coefficient of free solutions on the half-line $(-L,+\infty)$ (rather than on the whole line) that are needed to reproduce the case with $V_0\neq 0$. So that both with $V_0=0$ and $V_0\neq 0$ the wave functions satisfy proper BC's at $x=-L$. $\endgroup$
    – TwoBs
    Commented Aug 24, 2018 at 20:06
  • $\begingroup$ @TwoBs might be an option, do you think you can actually show how the identity is fulfilled with that approach? Just one word of warning: I suspect this won’t work either. Reason: I didn’t define the S-matrix via relative coefficients, but via the proper overlap integral. I know its form from an entirely different calculation and the relative coefficient in what i called psi^+ happens to be the same. $\endgroup$ Commented Aug 25, 2018 at 8:43

1 Answer 1


This is mostly correct. To make it work, a few inconsistencies in the definitions and notation need to be taken care of.

First, you are defining the Hamiltonian with $V_0=0$ to be the no scattering situation. With no scattering, $T=0$, so the $S=1+2i\pi T$ equation was derived with the convention that $S = 1$ when there is no scattering. Instead with $V_0=0$, your S contains the phase shift of the wave traveling from $x=0$ to $x=-L$ and back along with a sign change from the reflection. To be consistent, you should define $S$ without this extra phase shift: \begin{equation} \psi^{(+)}(E,x) = \frac{1}{\sqrt{2\pi}} \left \{ \begin{array}{cc} \frac{I(E)\beta}{\alpha}\sin\left (\alpha(1+\frac{x}{L})\right), & -L \leq x \leq 0\\ e^{-i\sqrt{2E}x}-S(E)e^{i\sqrt{2E}2L}e^{i\sqrt{2E}x} \end{array} \right . \end{equation} with \begin{equation} S(E) = \frac{\alpha\cot\alpha+i\beta}{\alpha\cos\alpha-i\beta} e^{-i\sqrt{2E}2L} \end{equation}

The Green's function for the $\psi^+(E,x)$ state that satisfies the Lippmann-Schwinger equation contributes only outgoing waves at infinity. So this tells you that for large $x$ it should be in the form $\psi_0(E,x)+Ce^{i\sqrt{2E}x}$, where the $\psi_0(E,x)$ contributes all of the incoming waves. Your scattering solution has the incoming wave $e^{-i\sqrt{2E}x}$ multiplied by $\sqrt\frac{1}{2\pi}$, but in your $\psi_0(E,x)$ this term is multiplied by $-\frac{i}{\sqrt{2\pi}}e^{-i\sqrt{2E}L}$. So the simplest workaround is to redefine $\psi_0(E,x)$ to be \begin{equation} \psi_0(E,x) = i\sqrt{\frac{2}{\pi}}e^{-i\sqrt{2E}L}\sin\left (\beta(1+\frac{x}{L})\right) \,, \end{equation} so that it is consistent with $\psi^{(+)}(E,x)$ and the Lippmann-Schwinger equation as TwoBs suggested. Finally, the on-shell $T$ matrix $T(E)$ should be defined from the half on-shell $T$ matrix $T(E',E)=\langle \psi_0(E')|V|\psi^{(+)}(E)\rangle$ by $\int dk' \delta(\frac{k'^2}{2}-E)T(\frac{k'2}{2},E)$, where the integral is given by how you form the completeness relation with your normalization. Therefore $T(E)= \frac{1}{\sqrt{2E}} \langle \psi_0(E')|V|\psi^{(+)}(E)\rangle$ See, for example S. Weinberg, Quantum Theory of Fields, Cambridge University Pres, 1995, Eq. 3.2.7.

With these changes, integrating

\begin{equation}T(E)=\frac{V_0}{\sqrt{2E}}\int_{-L}^0 \psi_0(E,x)\psi^{(+)}(E,x)\end{equation}

will give $S(E)=1-2\pi i T(E)$.


Here is a more detailed explanation of half on-shell comment above: If you start with \begin{equation} (H_0+V)|\psi^\pm_\alpha\rangle=E_\alpha|\psi^\pm_\alpha\rangle\,, \end{equation} where $\alpha$ labels the particular eigenstate $|\phi_\alpha$ and solve with the boundary conditions that all incoming (+) or outgoing (-) waves come from the initial state $|\phi_\alpha\rangle$, you have \begin{equation} |\psi^\pm_\alpha\rangle = |\phi_\alpha\rangle + \frac{1}{E_\alpha-H_0+i0^\pm} V|\psi^\pm_\alpha\rangle\,. \end{equation} Here $|\phi_\alpha\rangle$ is an eigenstate of $H_0$ with energy $E_\alpha$. If you insert a complete set of such eigenstates of $H_0$, you define the half on-shell T-matrix \begin{equation} T_{\beta\alpha}= \langle \phi_\beta|\frac{1}{E_\alpha-E_\beta+i0^+} V|\psi^+_\alpha\rangle\,. \end{equation} It is half on-shell because $E_\alpha$ is the eigenstate energy, but $E_\beta$ is not. You can also solve this where $E_\alpha$ is replaced with an arbitrary complex number, so that neither energy is on-shell. The S-matrix is given by \begin{equation} S_{\beta\alpha} = \delta(\alpha-\beta)-i2\pi \delta(E_\alpha-E_\beta) T_{\beta\alpha}\,. \end{equation} With the delta function, only terms with both energies equal contribute, so these give the on-shell T-matrix. Typically, you solve for the all the components of the half on-shell T-matrix, but its on-shell components give you the scattering.

The above delta function is the one I referred to above. It is the delta function you would have in Fermi's golden rule if you were to calculate the transition rate (in lowest order Fermi's golden rule uses the potential matrix elements, but in higher order these are replaced by the T-matrix elements). The cross section is the transition rate divided by the incoming flux and you would get the same sort of relation.

  • $\begingroup$ wow, finally an answer, thank you so much! The solution looks great, I will need some time to think through the consistency and do the integrals. If that works out I will come back with an upvote and a green tick ;) $\endgroup$ Commented Oct 26, 2018 at 6:24
  • $\begingroup$ right, I finally got to think about your answer in a bit more detail. Here are my thoughts: 1. Criticism first, I took your formulae and simply plugged everything in. Unfortunately the formula still does not seem to hold. 2. Now some praise: The missing phase factor is a good point, I think that is the main bit in finding the solution. 3. I don't understand the comment about energy normalization and the half on-shell T matrix, could you elaborate on that? Summary: Almost there, but the formula still doesn't hold. Any ideas why? Maybe just a typo? $\endgroup$ Commented Nov 9, 2018 at 15:43
  • $\begingroup$ Yes. I had a chance to look again, and I had written the wrong sign for the phase factor of psi_0. I edited the answer. $\endgroup$
    – user200143
    Commented Dec 2, 2018 at 2:33
  • $\begingroup$ I also added an appendix describing what is meant by half on shell, etc. $\endgroup$
    – user200143
    Commented Dec 2, 2018 at 21:23
  • $\begingroup$ excellent, the sign fixes the problem. Also thanks for the appendix! I prefer to not take the diversion to half-onshell scattering matrices, but the normalisation you point out can also be obtained in a direct derivation of $S=1-2\pi T$ (see e.g. Newton). I didn't know this before your answer, so thanks for that too! $\endgroup$ Commented Dec 4, 2018 at 10:22

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