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I was looking at problem $4.6$ of Gasiorowicz's Quantum Physics, where he asks to prove that the scattering matrix of a potential of the form

$$V(x)=\frac{\hbar^2}{2m}\frac{\lambda}{a}\delta(x-b)$$

is a certain unitary ($S^\dagger S=I$) but not symmetric matrix. I thought, as is referenced in the Wikipedia page on the $S$-matrix, that for any real potential $V(x)$, there is time reversal symmetry, meaning the $S$-matrix is symmetric. Where is my mistake?

Also, can we say that in the case where $V(x)$ is even then doing $x\to -x$ implies $$\psi_{in}=S\psi_{out}\implies S^2=Id$$

EDIT: For reference: enter image description here

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    $\begingroup$ Note that the potential is not symmetric: $V(-x) \propto \delta(x+b)$. $\endgroup$
    – Roger V.
    Nov 19, 2021 at 12:46
  • $\begingroup$ Indeed, the last question is more general and does not pertain to the potential aforementioned but is about S-matrices in general. $\endgroup$
    – Bcpicao
    Nov 19, 2021 at 15:04

1 Answer 1

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Tragically, the author of the book you reference uses a different convention for the $S$-matrix:

$$\tag{1} \left(\matrix{C \\B}\right)=S_G\left(\matrix{A \\D}\right) $$

Instead of

$$\tag{2} \left(\matrix{B \\C}\right)=S_{RW}\left(\matrix{A \\D}\right) $$

Where the subscript indicates "rest of world". The letters $A,B\dots$ and notation here follow the wiki page. On conjugating $\psi_L$ and $\psi_R$ we find

$$\tag{3} \left(\matrix{D^* \\A^*}\right)=S_G\left(\matrix{B^* \\C^*}\right) $$

As you can see, this is no longer in the same form as (1), the components of the in and out states have been flipped. We cannot continue to the next step of the proof on the wiki page because we cannot substitute the conjugate of (3) into (1). Instead we have

$$\tag{4} \left(\matrix{D \\A}\right)=S_G^*\left(\matrix{B \\C}\right) $$

On introducing the matrix $X=\left(\matrix{0 \ 1 \\1 \ 0}\right)$ we can write this as

$$\tag{5} X \left(\matrix{A \\D}\right)=S_G^* \ X \left(\matrix{C \\B}\right) $$

Into which may be substituted (1) to yield

$$\tag{6} (XS_G)^{-1}=(XS_G)^* $$

Where I have used $X^2=\mathbb{I}$. In this convention, the $S$-matrix does not generally satisfy $S^{-1}=S^*$ for real potentials; instead, (6) is the analogous statement about $S_G$.

The expression in the question is correct using the author's definition.

If the potential is symmetric, you may use the same method above (applying parity rather than complex conjugation, and relating the coefficients $A,B,...$) to show that

$$ \tag{7} S_{RW}=XS_{RW}X \implies S^{-1}_{RW}S_{RW}=\mathbb{I} $$

Furthermore, if the potential is even, the energy eigenstates may be chosen to be parity eigenstates. You can read more about that in eg. the notes by Tong or in chapter 16 of Quantum mechanics: a modern development by Ballentine

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  • $\begingroup$ Well spotted! I hadn't indeed realized the difference in definition. Any clues regarding my follow up question? $\endgroup$
    – Bcpicao
    Nov 20, 2021 at 13:21
  • $\begingroup$ Included in my answer now $\endgroup$
    – Sal
    Nov 20, 2021 at 15:59

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