There are two definitions of S operator (or S matrix) in quantum field theory. Are they equivalent? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-16T06:26:16Z https://physics.stackexchange.com/feeds/question/105152 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/105152 9 There are two definitions of S operator (or S matrix) in quantum field theory. Are they equivalent? 346699 https://physics.stackexchange.com/users/34669 2014-03-26T08:15:05Z 2017-04-01T04:55:14Z <p>I read several textbooks of QFT and found that there are two kinds of definition of $S$ operator (or <a href="http://en.wikipedia.org/wiki/S-matrix" rel="nofollow noreferrer">S matrix</a>).</p> <ul> <li><p>First kind:</p> <p>Define $\hat{S}$ is map from out space to in space $$\hat{S}\left|\beta,\text{out}\right\rangle：=\left|\beta,\text{in}\right\rangle,$$ so that $$S_{\beta\alpha}:= \left \langle \beta,\text{out} | \alpha,\text{in}\right\rangle= \left \langle \beta,\text{out} \middle |\hat{S}\middle | \alpha,\text{out}\right\rangle= \left \langle \beta,\text{in}\middle |\hat{S}\middle | \alpha,\text{in}\right\rangle.$$ I understand that all these vectors are defined in the Heisenberg picture.</p></li> <li><p>Second definition: $$S_{\beta\alpha}:={}_I \left \langle \beta \middle |\hat{S}\middle | \alpha \right\rangle_I$$ where subscript $_I$ means vector are in interacting picture. In this definition, then, $$\hat{S}=U_I(+\infty,-\infty),$$ where $U_I(+\infty,-\infty)$ is the evolution operator in interacting picture.</p></li> </ul> <p>Are these two definitions equivalent? I am confused about it.</p> <p>Remark: I konw that the matrix element $S_{\beta\alpha}$ is the same in these two pictures, what I want to ask is whether the operator $\hat{S}$ is same in these two definitions. Thanks!</p> https://physics.stackexchange.com/questions/105152/-/105233#105233 4 Answer by Siva for There are two definitions of S operator (or S matrix) in quantum field theory. Are they equivalent? Siva https://physics.stackexchange.com/users/3998 2014-03-26T18:45:41Z 2014-03-26T18:45:41Z <p>I don't completely understand the two sets of statements you've written, but I think I understand the essence of your question. Maybe this helps:</p> <hr> <p>The <em>S-matrix</em> (operator) is a transfer function from <em>in</em> states to <em>out</em> states. </p> <ol> <li><p>If your states are not evolving (Heisenberg/Interaction picture) then you need to insert an evolution operator between the states.</p></li> <li><p>If your states are in the Schrodinger picture and they're evolving with time, then $| out, t = \infty \rangle = (\textrm{evolution operator}) \; | out , t = 0 \rangle$ </p></li> </ol> <p>So the definition (convention) for the S-matrix depends on your convention for defining the Hilbert space at late times (whether it is the same Hilbert space as initial times, or if it is the time-evolved Hilbert space). This is equivalent to whether you're in the Schrodinger or the Heisenberg picture. Physically, I hope it's now clear why both descriptions/conventions are the same object.</p>