The value of $\langle p^2 \rangle$ for $\psi(x)=e^{-|x|}$, where $\hat{p}:=i\frac{d}{dx}$ is the momentum operator.

Question

Consider the Hilbert space $H$ of square-integrable complex functions on the real line equipped with the inner product $$\langle \phi,\psi\rangle:=\int \phi^*\psi d\mu,$$ where $\mu$ is the Lebesgue measure and $*$ denotes complex conjugation. For a given $\psi\in H$, with $\|\psi\| := \sqrt{\langle \psi, \psi\rangle} = 1$, define the expectation value $\langle Q \rangle$ of an operator $\hat{Q}$ on $H$ as $\langle Q \rangle:=\int \psi^* \hat Q\psi d\mu$.

Clearly, the function $\psi(x) = e^{-|x|}$ is in $H$ and the momentum operator $\hat{p}:=-i\frac{d}{dx}$ is a symmetric operator on (a subset of) $H$. The computation for $\langle p^2 \rangle$ is as follows:

\begin{align*} \langle p^2 \rangle &= \int_{-\infty}^\infty e^{- |x|} \left(-\frac{d^2}{dx^2}\right)e^{- |x|} dx\\ &= -\int_{-\infty}^\infty e^{- |x|} \frac{d^2}{dx^2}e^{- |x|} dx = -\left(\int_{-\infty}^0 e^{ x} \frac{d^2}{dx^2}e^{ x} dx + \int^\infty_0 e^{- x} \frac{d^2}{dx^2}e^{- x} dx\right)\\ &= -\left(\int_{-\infty}^0 e^{2 x} dx + \int^\infty_0 e^{-2 x} dx\right) = -2 \int^\infty_0 e^{-2\alpha x} dx = -1. \end{align*}

Which I find strange because by symmetry we should have $\langle p^2\rangle = \langle \psi,\hat p^2\psi\rangle = \langle \hat p\psi,\hat p\psi\rangle\geq0$. I have tried to figure out what might be the problem and have come to the following conclusions:

1. $\psi$ is not differentiable at $x=0$, but that should not be an issue under the integral sign since $\{0\}$ is a set of measure zero.

2. My computation is wrong and have checked it numerous times.

Please help.

Answer 1

The domain of the differentiation operator is the space of all complex-valued functions $f$ absolutely continuous on $\Bbb{R}$ such that $f\in L^2(\Bbb{R})$ and $f'\in L^2(\Bbb{R})$ where $f'$ is the derivative a.e. of $f$.

The function $\psi(x)=\exp(-|x|)$ meets these conditions, so $\psi$ is in the domain of the differentiation operator.

But what is $\psi'$? $\psi'(x)=-\exp(-x)$ for $x>0$ and $\psi'(x)=\exp(x)$ for $x<0$. So $\psi'$ is discontinuous at $0$. The right limit value at zero is $-1$ while the left limit value at zero is $1$. Hence $\psi'$ is not in the domain of the differentiation operator. $\hat{p}^2\psi$ is undefined.

You can still compute $\langle\hat{p}\psi,\hat{p}\psi\rangle$ because $\hat{p}\psi\in L^2(\Bbb{R})$. But this isn't equal to $\langle\psi,\hat{p}^2\psi\rangle$. The latter value is undefined.

Edit: An informal way to view this is that the second derivative $\psi''$ is a sum of a multiple of a delta-function with a function smooth on $\Bbb{R}\setminus\{0\}$. Since a delta-function is not a member of $L^2(\Bbb{R})$, neither is $\psi''$.

Answer 2

As already explained in the answer given by Chad K, the momentum operator $P$ is defined on the dense domain $D(P) = \{\psi \in L^2(\mathbb{R})| \psi \; \text{absolutely continuous and } \psi^\prime \in L^2(\mathbb{R}) \} $, whereas the domain of $P^2$ is given by $D(P^2)= \{\psi \in L^2(\mathbb{R}) |\psi, \, \psi^\prime \; \text{abs. continuous and} \; \psi^\prime \in D(P) \}$, so $\exp(-|x | ) \in D(P)$, but $\exp(-|x|) \notin D(P^2)$.

A possible "dirty" way out, often chosen in theoretical physics, is to leave the framework of the Hilbert space $L^2(\mathbb{R})$, regarding $\exp(-|x|)$ simply as a distribution. In such an approach, differentiating $\exp(-|x|)$ twice (in fact, arbitrarily many times) poses no problem, $$\frac{d}{dx} e^{-|x|} = e^x \Theta(-x)-e^{-x}\Theta(x), \quad \frac{d^2}{dx^2} e^{-|x|}=e^{-|x|} -2 \delta(x),$$ where $\Theta(x)$ denotes the Heaviside step function. Evaluating now the integral $$\int\limits_{-\infty}^{+\infty} \! \! dx \, e^{-|x|} \left(-\frac{d^2}{dx^2} \right) e^{-|x|}=\int\limits_{-\infty}^{+\infty} \! \! dx\, e^{-|x|} \left(-e^{-|x|}+2\delta(x)\right)=1 $$ gives a positive result, coinciding at the same time with the outcome for $\langle P \phi | P \phi \rangle $ with $\phi(x)= \exp(-|x|) \in D(P)$ in the Hilbert space approach.

Answer 3

An alternative way of calculating $\langle p^2 \rangle$ is doing it in momentum space. Using the Dirac notation, we have $$ \langle p^2 \rangle=\langle \psi|\hat{p}^2 |\psi\rangle =\int_{-\infty}^{\infty}\langle \psi|\hat{p}^2|p \rangle \langle p|\psi\rangle\,dp =\int_{-\infty}^{\infty}p^2|\langle p|\psi \rangle|^2\,dp. \tag{1} $$ Now $$ \langle p|\psi \rangle=\int_{-\infty}^{\infty}\langle p|x \rangle \langle x|\psi\rangle\,dx=\int_{-\infty}^{\infty}\frac{e^{ipx}}{\sqrt{2\pi}}e^{-|x|}\,dx =\sqrt{\frac{2}{\pi}}\frac{1}{p^2+1}, \tag{2} $$ hence $$ \langle p^2 \rangle=\frac{2}{\pi}\int_{-\infty}^{\infty}\frac{p^2}{(p^2+1)^2}\,dp =1, \tag{3} $$ which agrees with the result obtained by @Hyperon.

Answer 4

The exponential tail is the outer solution of any bound state in a 1d-square well potential of width a and depth -V.

$$-\frac{1}{2}\ \psi_n''(x) -V *\ \left(\theta(x+\frac{a}{2})-\theta(x-\frac{a}{2})\right)\ \psi_n(x) = -E_n\ \psi_n(x)$$

The possible eigenvalues are determined by stitching together trigonometric eigenfunctions for positive kinetic energy above the potential bottom $$-f''=2(E-V)\ f , \quad V<E<0$$ inside the square well and real decreasing exponentials with negative kinetic energy parameter $$- f'' = 2 E\ f, \quad E<0 $$.

Algeebrically, this is the famous transfer - and reflexion coefficient problem, that even for a negative wall yields reflexions, strongly dependent on interference with bound states.

Numerically the solutions are determined solving the ODE with the boundary values $f,f'$, of the increasing exponential $e^{k x}$ at the left boundary of the well and integrating over the right boundary, where a linear combinatiíon of increasing and decreasing exponentials continue the solution to $x=\infty$. Eigenvalues of bound states occur if the coefficient of the growing exponential vanishes. The local energy density of the tails is negative, lowering the eigenvalues with respect to a reflecting wall with knots of the trigonometrics.

The possibility of negative kinetic energies contributing to total energy in areas of classically forbidden volumes is the critical point in history, where all classical analogues of quantum theory break down. Intimately connected the two other facts: tunneling through walls and the reflexion at holes.

Bottom line wrt the question: the states $e^{- k |x|}$ are the outer tails of the single bound state in a square well, that remains in the limit of a $-\delta$ potential. Trivially, the energy is the negative value $$-\frac{k^2}{2} -f(0)^2$$.

Its a standard exercise in any quantum theory course, see eg. Ballantine, exercise 4.9. The difficult point to show is, that in the limit width -> 0 and depth $-\infty$ the $\delta$-potential retaines one bound state.