3
$\begingroup$

The operator is

$$\hat{A} = -i \left(x \frac{d}{dx} + \frac{1}{2} \right).$$

Is it true that

$$\langle \hat{A} \psi_1(x)|\psi_2(x)\rangle = \langle \psi_1(x)|\hat{A}\psi_2(x)\rangle\ ?$$

Here, $\langle\ldots|\ldots\rangle$ is a scalar product defined as

$$\langle\psi_1(x)|\psi_2(x)\rangle = \int_{\Omega} \psi_1(x) \psi_2^*(x) \ dx.$$

$\endgroup$

2 Answers 2

4
$\begingroup$

Here is a different route. Consider the operators $\hat{x}$ and $\hat{p}$ where $$\hat{x}\psi(x)=x\psi(x)$$ and $$\hat{p}\psi(x)=-i\psi'(x).$$ Show that $\hat{x}$ and $\hat{p}$ are Hermitian operators. Also, show that $[\hat{x},\hat{p}]=\hat{x}\hat{p}-\hat{p}\hat{x}=i$.

It can be proven that $\hat{P}\hat{Q}+\hat{Q}\hat{P}$ is a Hermitian operator for any Hermitian operators $\hat{P}$ and $\hat{Q}$. The same situation applies to $\hat{x}$ and $\hat{p}$. We have $\hat{x}\hat{p}+\hat{p}\hat{x}$ is Hermitian, but $$\hat{x}\hat{p}+\hat{p}\hat{x}=2\hat{x}\hat{p}-[\hat{x},\hat{p}]=2\hat{x}\hat{p}-i=2\hat{A}.$$ So, $2\hat{A}$ is Hermitian, and so $\hat{A}$ is Hermitian, since $2$ is a real number.

$\endgroup$
0
3
$\begingroup$

The short answer is: Yes it is. You can see this simply by doing an integration by parts. Let us leave out the $-i$ and show that $x \frac{d}{dx} + \frac{1}{2}$ is antisymmetric instead.

\begin{equation} \int_{\Omega} \left( \left( x \frac{d}{dx} + \frac{1}{2} \right) \psi_1 \right) \overline{\psi_2} \: dx= -\int_{\Omega} \left( x \frac{d}{dx} \overline{\psi_2} \right) \psi_1 + \psi_1 \overline{\psi_2} \: dx + \frac{1}{2} \int_{\Omega} \psi_1 \overline{\psi_2} \: dx \end{equation} By integration by parts and since $\frac{d}{dx} \left( x \overline{\psi_2} \right) = x \frac{d}{dx} \overline{\psi_2} + \overline{\psi_2}$. Clearly this equals $- \int_{\Omega} \overline{\left( \left( x \frac{d}{dx} + \frac{1}{2} \right) \psi_2 \right)} \psi_1 \: dx$. Thus your operator is indeed symmetric (since multiplication by $i$ turns antisymmetric into symmetric operators and vice versa).

More importantly perhaps, you should note that this is an unbounded differential operator. Although you did not state it explicitly, you are probably thinking about it as acting on $L^2(\Omega)$, where it is densely defined. There the natural scalar product is the one you gave. If you choose the Sobolev space $H^1$ as its domain, the operator even becomes self-adjoint, a much stronger property of unbounded operators than just being symmetric.

You can learn more about the difference between symmetry and self-adjointness of unbounded operators for instance in the book of Teschl.

$\endgroup$
7
  • $\begingroup$ Then $\hat{A}$ is symmetric on $L^2$ and self-adjont(hermitian in physicicsts terms) on Sobolev space? $\endgroup$
    – user464980
    Nov 13, 2018 at 15:08
  • 1
    $\begingroup$ Not quite. The point is that you want to define the operator on $L^2$, but it does not make sense to differentiate a general $L^2$ function, and you cannot extend the operator continuously to all of $L^2$. Thus it is unbounded, i.e. only defined on a dense subspace. In a sense, the largest "sensible" subspace is $H^1$. If you choose this as domain (the domain is an inherent part of the definition of unbounded operator), it becomes self-adjoi t. If you only use say smooth L2 functions as domain, it us symmetric but not self-adjoint. $\endgroup$
    – Max
    Nov 13, 2018 at 15:24
  • $\begingroup$ Okay, I think I get it right. Then question out of a contest(kinda) -- when functions are in L2 i often use normalization contidtion in a form of $\int \psi_m \psi_n^* = \delta_{mn}$ where delta is a Kronecker's one. When functions doesnt belong to L2 i use normalization condition with Diracs delta $\delta(m-n)$. Am I able to normalize functions in H1 in this 'Diracs delta' way? $\endgroup$
    – user464980
    Nov 14, 2018 at 12:18
  • $\begingroup$ I‘m not quite sure what you mean by normalisation using Dirac delta... Could you explain tjat? $\endgroup$
    – Max
    Nov 14, 2018 at 20:19
  • $\begingroup$ Eigenfunctions of operator A are normalized with condition $\int_{\Omega} \psi_m(x) \psi_n(x) dx = \delta(m-n)$ $\endgroup$
    – user464980
    Nov 15, 2018 at 22:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .