Shyan said:
Consider two hermitian operators A and B and a system in state |a\rangle which is an eigenstate of A with eigenvalue \lambda
So we have:
<br />
\langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0<br />
Shyan said:
There is still a point here.
We can't tell that the assumption that A has an eigenstate,causes the paradox because even if P doesn't have an eigenstate,one still can find an operator that does and do the calculation for that operator and its eigenstate and again arrive at the paradox.So I think sth else should be
wrong.
I've been trying to understand it via reading the thread that Goerge suggested but I can't follow the discussion because I don't know enough math.
Can somebody present the final result of that thread in a simple language?
thanks
There is a problem with the step
\left\langle a \vert AB \vert a\right\rangle =\left\langle Aa \vert Ba\right\rangle
To see the problem in terms of domains, let's work through in detail a fairly elementary example for which the eigenstates and eigenvalues above actually exist. In this example, subtleties with domains definitely come into play.
First, consider something even more elementary, real-valued functions of a single real variable. The domain of such a function f is the (sub)set of all real numbers x on which f is allowed to act. Suppose f is defined by f\left(x\right) = 1/x. The domain of f cannot be the set of all real numbers \mathbb{R}, but it can be any subset of \mathbb{R} that doesn't contain zero. Take the domain of f to be the set of all non-zero real numbers. Define g by g\left(x\right) = 1/x with domain the set of all positive real numbers. As functions, f \ne g, because f and g have different domains, i.e., it takes both a domain and an action to specify a function. As functions, f = g only when f and g have the same actions and the same domains.
Let the action of the momentum operator be given by P=-id/dx (for convenience, set \hbar =1). On what wave functions can P act, i.e, what is the domain, D_{P}, of P? Since P operates the Hilbert space H of square-integrable functions, D_{P} must be a subset of the set of square-integrable functions. The action of P has to give as output something that lives in the Hilbert space H, i.e., the output has to be square-integrable, and thus D_{P} must be subset of the set of square-integrable functions whose derivatives are also square-integrable. Already, we see that the domain of P cannot be all of the Hilbert space H.
As an observable, we want P to be self-adjoint, i.e, we want P=P^{\dagger }. As in the case of functions above, this means that the actions of P and P^{\dagger } must be the same, and that the domains (the states on which P and P^{\dagger } act) D_{P} and D_{P^{\dagger }} must be the same. For concreteness, take wave functions on the interval with endpoints x=0 and x=1. The adjoint of the momentum operator is defined by
<br />
\begin{align}<br />
\left\langle P^{\dagger }g \vert f\right\rangle &=\left\langle g \vert Pf\right\rangle \\<br />
&=-i\int_{0}^{1}g* \frac{df}{dx}dx \\<br />
&=-i\left( \left[ g* f\right] _{0}^{1}-\int_{0}^{1}\frac{dg}{dx}* fdx\right) \\<br />
& =-i\left[ g* f\right] _{0}^{1}+\left\langle Pg \vert f\right\rangle ,<br />
\end{align}<br />
where integration by parts has been used.
Consequently, the actions of P and P^{\dagger } are the same as long as the first term in the last line vanishes, i.e., as long as
<br />
\begin{align}<br />
0 &= g* \left( 1\right) f\left( 1\right) -g* \left( 0\right) f\left( 0\right) \\<br />
\frac{f\left( 1\right) }{f\left( 0\right) } &= \frac{g* \left( 0\right) }{g* \left( 1\right) }<br />
\end{align}<br />
for non zero f\left( 0\right) and g\left( 1\right). Now, f\left( 1\right) /f\left( 0\right) is some complex number, say \lambda, so f\left( 1\right) = \lambda f\left( 0\right). Hence,
<br />
\begin{align}<br />
\lambda & =\frac{g* \left( 0\right) }{g* \left( 1\right) }\\<br />
\lambda * & =\frac{g\left( 0\right) }{g\left( 1\right)} \\<br />
g\left( 1\right) & =\frac{1}{\lambda * }g\left( 0\right) .<br />
\end{align}<br />
From the relation \left\langle P^{\dagger }g \vert f\right\rangle =\left\langle g \vert Pf\right\rangle, we see that g\in D_{P^{\dagger }} and f\in D_{P}. These domains can be made to be the same if the same \lambda restrictions are placed on f and g, i.e, if \lambda =1/\lambda*, or \lambda* \lambda =1. This means that \lambda can be written as \lambda =e^{i\theta } with \theta real. Different choices of \theta correspond to different boundary conditions, with \theta =0 corresponding to the periodic boundary condition used by Dickfore above (with L=1). Let's use this choice, so that f is in D_{P^{\dagger }}=D_{P} if f is square-integrable, f' is square-integrable, and f\left( 1\right) =f\left( 0\right). Note that this works even if 0=f\left( 0\right). With this choice, P=P^{\dagger }, and we can write \left\langle Pg \vert f\right\rangle =\left\langle g \vert Pf\right\rangle. We cannot use this when P (on either side) acts on a wave function h that is not in D_{P^{\dagger }}=D_{P}. This is the problem with the "proof" above.
In \left\langle a \vert AB \vert a\right\rangle, take A=P and B=X. Take f\left( x\right) =e^{2\pi ix}. Then, \left( Pf\right) \left( x\right) =2\pi f\left( x\right) and \left( PXf\right) \left( x\right) =P\left( xf\left( x\right) \right). However, h\left( x\right) =xf\left( x\right) =xe^{2\pi ix} does not satisfy the boundary condition h\left( 1\right) =h\left( 0\right), so h is not in the domain D_{P}, and we cannot just slide [/itex]A=P[/itex] to the left.