I am a beginner in quantum field theory and I am learning from the lecture notes by David Tong. On Page 58, he gives an example of two nucleons scattering and says that we are only interested in $\langle f|S-1|i\rangle$ since this is the only relevant quantum amplitude for scattering events. I do not understand how subtracting $\langle f|i \rangle$ from $\langle f|S|i \rangle$ will exclude the non-scattering events and give us what we are looking for.
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2$\begingroup$ The operator that you call $M$ is usually denoted by $T$. This is just a way to separate the non-trivial part of the $S$-matrix, since $\langle \alpha \vert 1 \vert \beta \rangle=0 $ if $\vert \alpha \rangle \perp \vert \beta\rangle$. $\endgroup$– pppqqqCommented Dec 14, 2016 at 10:16
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$\begingroup$ @pppqqq NB: It's me that called it $\mathcal M$; I re-wrote the title since the question is not about nucleon scattering, and it is denoted by $T$ or $\mathcal M$, such as for example in Schwarz, or even $\mathcal A$. In hindsight, it would have been most appropriate to write it as $\mathcal A$ in the title, as this is what Tong uses later on, which the OP is reading. $\endgroup$– JamalSCommented Dec 14, 2016 at 10:28
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The decomposition $S=1+i\mathcal M$ is just a convenient way to separate out the "trivial" part of the $S$-matrix.
Recall that two quantum states are orthogonal if and only if there exists, at least in principle, a measurement which can distinguish beetween the two. Therefore the $1$ in $S=1+i\mathcal M$ is zero if "something happens":$$\langle \alpha \vert 1 \vert \beta \rangle =0 \iff \vert \alpha \rangle \perp \vert \beta \rangle.$$
Note that $\mathcal M$ satisfies the identities: $$\mathcal M^\dagger \mathcal M+i (\mathcal M-\mathcal M^\dagger)=0\\\mathcal M \mathcal M^\dagger +i (\mathcal M^\dagger-\mathcal M)=0$$which are equivalent to the unitarity of the $S$-matrix and ultimately lead to the optical theorem, see for example Weinberg S., "The quantum theory of fields, Vol.1".
Side note. As pointed out, there are several notations for the so called "transfer matrix" $S-1$. The one I encounter more frequently is $S=1+iT$.
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$\begingroup$ Thank you for your response but I do not understand why two quantum states are orthogonal if and only if a measurement can distinguish between the two. What does it mean to distinguish between the two states ? $\endgroup$ Commented Dec 14, 2016 at 14:05
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$\begingroup$ What I mean is that if the system is prepared into state $\vert \alpha \rangle$ XOR into state $\vert \beta \rangle$, you can in principle perform an experiment which determines the state ($\alpha$ or $\beta$) with certainty. In other words, there exists an observable $A$ which has value $1$ in the state $\vert \alpha \rangle$ and value $0$ in the state $\vert \beta \rangle$. (The observable might be $A=\vert \alpha \rangle \langle \alpha \vert$) $\endgroup$– pppqqqCommented Dec 14, 2016 at 14:28
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$\begingroup$ How does this relate to orthogonality ? Do you know of any resource where I could read more about this ? $\endgroup$ Commented Dec 14, 2016 at 16:53
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$\begingroup$ Do you agree with me that if it is possible, in a single experiment, to determine whether the system is in state $\alpha$ or $\beta$, then there must exist an observable for which $\alpha$ and $\beta$ are eigenvectors with distinct eigenvalues? This comes from the basic postulates of quantum mechanics. Now, if $\alpha$ and $\beta$ are eigenvectors with distinct eigenvalues, they are automatically orthogonal. All this is discussed, for example, in the first chapter of Sakurai, "Modern Quantum Mechanics". $\endgroup$– pppqqqCommented Dec 14, 2016 at 17:31
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$\begingroup$ I agree that if in a single experiment, we can determine that the state is $\alpha$ and also that it is not in state $\beta$ (by whatever means necessary), then the rest is as you said. So, how do we know that the initial state $ |i>$ is orthogonal to the final state $ |f>$ when there is scattering ? Are they distinct eigenstates of an operator ? If so, which operator and how do we know the eigenstates are distinct ? $\endgroup$ Commented Dec 14, 2016 at 19:04