The discussion focuses on the commutation relation between the Hamiltonian operator ##\hat{H}## and the momentum operator ##\hat{P}## for a particle in a one-dimensional box. It is established that the commutator ##[\hat{H}, \hat{P}] = [\frac{\hat{P}^2}{2m}, \hat{P}] + [V(x), \hat{P}]##, where the first term is zero, but the second term is non-zero due to the potential's dependence on position. The potential ##V(x)## is identified as troublesome because it is not differentiable at the walls of the box, leading to the conclusion that ##\hat{H}## and ##\hat{P}## cannot be simultaneously diagonalized.
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Does it? We have ##[H,P]=[P^2/2m,P]+[V(x),P]##; the first term is trivially zero but there’s no reason to expect the second one to be.hokhani said:At the first glance it seems that they commute
I don't see why the potential is "troublesome"; it's a function of ##x## whose derivative won't in general be zero, so it won't commute with ##p##.pines-demon said:Well usually ##[f(x),p]=i\hbar \partial f/\partial x ##, for a function ##f(x)##, but the potential here is kind of troublesome.
In which case it's certainly not zero. So either way the OP's expectation of being able to find a common diagonalization for ##H## and ##P## is false.dextercioby said:You may discover that the commutator is not even defined...
The potential is not a function that is differentiable in the usual way (at least not for the infinite well, for the finite well you might use Dirac deltas).PeterDonis said:I don't see why the potential is "troublesome"; it's a function of ##x## whose derivative won't in general be zero, so it won't commute with ##p##.
Very nice separation of the Hamiltonian. If we take ##\psi(x)=\sum_m a_m \psi_m## where ##\psi_m## are the eigenfunction of the Hamiltonian, then it seems that we have ##[V(x),P]\psi(x)=0## since inside the box (##|x|<L##) the potential is zero ##V(x)=0## and for ##|x| \geq L## we have ##\psi_m=0##!Nugatory said:Does it? We have ##[H,P]=[P^2/2m,P]+[V(x),P]##; the first term is trivially zero but there’s no reason to expect the second one to be. what is ##[V(x),P]\psi(x)## when you expand the commutator?
For infinite walls I did that in the post #8.dextercioby said:Finite walls or infinite walls? Start down by writing the Hilbert space, then the two operators. You may discover that the commutator is not even defined...
Take ##\psi(x)## to be an arbitrary function and expand the commutator. You will end up with a term proportional to ##\frac{dV}{dx}##, which works just fine when ##V## is continuous and is in general non-zero (but is zero where ##V## is constant, which creates the "at first glance" illusion in your original post).hokhani said:If we take ##\psi(x)=\sum_m a_m \psi_m## where ##\psi_m## are the eigenfunction of the Hamiltonian, then it seems that we have ##[V(x),P]\psi(x)=0## since inside the box (##|x|<L##) the potential is zero ##V(x)=0## and for ##|x| \geq L## we have ##\psi_m=0##!
The usual textbook demonstration for ##~[V(X),P]=iV'(X)~## assumes that ##~V(x),~V'(x)~## can be expanded as power series. Can you generalize it to other differentiable (let alone continuous) functions?Nugatory said:which works just fine when ##V## is continuous
So you assume ##~\infty\cdot0=0~## ?hokhani said:Very nice separation of the Hamiltonian. If we take ##\psi(x)=\sum_m a_m \psi_m## where ##\psi_m## are the eigenfunction of the Hamiltonian, then it seems that we have ##[V(x),P]\psi(x)=0## since inside the box (##|x|<L##) the potential is zero ##V(x)=0## and for ##|x| \geq L## we have ##\psi_m=0##!
What does "derivable" mean? If you mean sometimes you can't derive its form from first principles, so what? That's true of lots of physics problems; you just have to guess the form of a function.pines-demon said:The potential is not a function that is derivable in the usual way (at least not for the infinite well, for the finite well you might use Dirac deltas).
My bad, I meant differentiable.PeterDonis said:What does "derivable" mean? If you mean sometimes you can't derive its form from first principles, so what? That's true of lots of physics problems; you just have to guess the form of a function.
If you mean its derivative isn't always well-defined, yes, that's true--I addressed that case in post #6.
Right, it is undefined. However, still another problem remains:JimWhoKnew said:So you assume ##~\infty\cdot0=0~## ?
The "pathology" of the infinite potential reduces the Hilbert space from ##~\text{L}^2(-\infty,\infty)~## to ##\text{L}^2(-L,L)~,~\psi(\pm L)=0~## . Therefore we should no longer expect to span the functions outside the box.hokhani said:Right, it is undefined. However, still another problem remains:
The eigenfunctions of the particle in box, $\psi_m(x)$, should form a complete basis and any function, $\psi(x)$, seems to be expanded as $\psi(x)=\sum_m a_m \psi_m(x)$ while $\psi_m(x)$ is zero outside the box!
I look for the error in the term ##\langle x |P| x \rangle=-i\hbar \frac{\partial}{\partial x} \delta(0)## which seems undefined.JimWhoKnew said:While you are at it, try to find the error in the following argument:$$i\hbar \delta(0)=\langle x |i\hbar| x \rangle=\langle x |[X,P]| x \rangle=\langle x |XP| x \rangle-\langle x |PX| x \rangle=x\langle x |P| x \rangle-\langle x |P| x \rangle x=0$$
BTW: use of the "preview" option before posting is always recommended.