I always like to show the following to those who have just learned quantum theory. The commutation relation between poisition and momentum is [tex]$[x,p]=i hbar$[\/tex] or [tex]$xp-px=i hbar$[\/tex]. Now act on an eigenstate of the [tex]$x$[\/tex] operator [tex]$|x_0rangle$[\/tex] and you get [tex]$(xp-px)|x_0rangle=xp|x_0rangle- p x_0 |x_0rangle$[\/tex]. Take the inner product of this state with the ket [tex]$langle x_0|$[\/tex]: [tex]$langle x_0 | (xp|x_0rangle- p x_0 |x_0rangle)= langle x_0| (x_0 p – x_0 p)|x_0rangle=0$[\/tex]. But if we carry out the same procedure on the right hand side of the commutation relation we get [tex]$langle x_0| i hbar |x_0rangle=ihbar$[\/tex], which, last time I checked, was not zero. Snicker. It’s so mean to give this to those who’ve just learned quantum theory, but shucks, it’s also pretty fun to watch them squirm and figure out what went wrong.<\/p>\n","protected":false},"excerpt":{"rendered":"
I always like to show the following to those who have just learned quantum theory. The commutation relation between poisition and momentum is [tex]$[x,p]=i hbar$[\/tex] or [tex]$xp-px=i hbar$[\/tex]. Now act on an eigenstate of the [tex]$x$[\/tex] operator [tex]$|x_0rangle$[\/tex] and you get [tex]$(xp-px)|x_0rangle=xp|x_0rangle- p x_0 |x_0rangle$[\/tex]. Take the inner product of this state with the ket … <\/p>\n