Skip to main content
Physics

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

7
  • $\begingroup$ If somebody will explain meaning standing behind |r1 r2> notation, I will be very grateful. $\endgroup$ Commented Apr 25, 2013 at 12:11
  • $\begingroup$ $\newcommand\ket[1]{\left|#1\right>} \ket{x_1x_2}$ is a common shorthand for $\ket{x_1}\otimes\ket{x_2}$. In general, what is written inside a ket is just a label: its meaning is heavily context dependent. Mixing the two representations is not uncommon, as long as there is no confusion. $\endgroup$ Commented Apr 26, 2013 at 6:04
  • $\begingroup$ ok, I totally agree with that. But let us concern a particular case from the question above, i.e. the case when $x_1$ and $x_2$ are quantum numbers labeling the eigenstates of the positional operator $\hat{x}$. What is the meaning of decomposing $\ket{x_1}\otimes\ket{x_2}$ ? Also, I am curious if there is any references containing such notation for coordinate space. It would be interesting to read more about. $\endgroup$ Commented Apr 26, 2013 at 7:16
  • $\begingroup$ I untastood $x_1$ and $x_2$ as labelling the position of the two fermions. But then, of course the ket $\ket{x_1x_2}$ is not antisymmetrized and does not correspond to an allowed state. Only $\ket{x_1x_2} -\ket{x_2x_1}$ is a valid state. $\endgroup$ Commented Apr 26, 2013 at 7:29
  • 1
    $\begingroup$ With Fermions, the situation is complicated by the fact that you impose a global symmetry. $\newcommand\ket[1]{\left|#1\right>}$ $\ket{x1x2}\ket{\uparrow\uparrow}$ is not a valid state, but $\ket{x1x2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}$ is $\endgroup$ Commented Apr 26, 2013 at 10:13