Quantum Mechanics and Undergraduate Mathematics XII: position and momentum operators

August 11, 2011

Quantum Mechanics and Undergraduate Mathematics XII: position and momentum operators

Filed under: advanced mathematics, applied mathematics, physics, probability, quantum mechanics, science — collegemathteaching @ 1:52 am

Recall that the position operator is $X \psi = x\psi$ and the momentum operator $P \psi = -i\hbar \frac{d}{dx} \psi$ .

Recalling our abuse of notation that said that the expected value $E = \langle \psi, A \psi \rangle$ , we find that the expected value of position is $E(X) = \int_{-\infty}^{\infty} x |\psi|^2 dx$ . Note: since $\int_{-\infty}^{\infty} |\psi|^2 dx = 1,$ we can view $|\psi|^2$ as a probability density function; hence if $f$ is any “reasonable” function of $x$ , then $E(f(X)) = \int_{-\infty}^{\infty} f(x) |\psi|^2 dx$ . Of course we can calculate the variance and other probability moments in a similar way; e. g. $E(X^2) = \int_{-\infty}^{\infty} x |\psi|^2 dx$ .

Now we turn to momentum; $E(P) = \langle \psi, -i\hbar \frac{d}{dx} \psi \rangle = \int_{-\infty}^{\infty} \overline{\psi}\frac{d}{dx}\psi dx$ and $E(P^2) = \langle \psi, P^2\psi \rangle = \langle P\psi, P\psi \rangle = \int_{-\infty}^{\infty} |\frac{d}{dx}\psi|^2 dx$

So, back to position: we can now use the fact that $|\psi|^2$ is a valid density function associated with finding the expected value of position and call this the position probability density function. Hence $P(x_1 < x < x_2) = \int_{-\infty}^{\infty} |\psi|^2 dx$ . But we saw that this can change with time so: $P(x_1 < x < x_2; t) = \int_{-\infty}^{\infty} |\psi(x,t)|^2 dx$

This is a great chance to practice putting together: differentiation under the integral sign, Schrödinger’s equation and integration by parts. I recommend that the reader try to show:

$\frac{d}{dt} \int_{x_1}^{x_2} \overline{\psi}\psi dx = \frac{ih}{2m}(\overline{\psi}\frac{d \psi}{dx}-\psi \frac{d \overline{\psi}}{dx})_{x_1}^{x_2}$

The details for the above calculation (students: try this yourself first! 🙂 )

Differentiation under the integral sign:
$\frac{d}{dt} \int_{x_1}^{x_2} \overline{\psi} \psi dx = \int_{x_1}^{x_2}\overline{\psi} \frac{\partial \psi}{\partial t} + \psi \frac{\partial \overline{ \psi}}{\partial t} dt$

Schrödinger’s equation (time dependent version) with a little bit of algebra:
$\frac{\partial \psi}{\partial t} = \frac{i \hbar}{2m} \frac{\partial^2 \psi}{\partial x^2} - \frac{i}{\hbar}V \psi$
$\frac{\partial \overline{\psi}}{\partial t} = \frac{i \hbar}{2m} \frac{\partial^2 \overline{\psi}}{\partial x^2} + \frac{i}{\hbar}V \overline{\psi}$

Note: $V$ is real.

Algebra: eliminate the partial with respect to time terms; multiply the top equation by $\overline{\psi}$ and the second by $\psi$ . Then add the two to obtain:
$\overline{\psi} \frac{\partial \psi}{\partial t} + \psi \frac{\partial \overline{ \psi}}{\partial t} = \frac{i \hbar}{2m}(\overline{\psi} \frac{\partial^2 \psi}{\partial x^2} + \psi \frac{\partial^2 \overline{ \psi}}{\partial x^2})$

Now integrate by parts:
$\frac{i \hbar}{2m} \int_{x_2}^{x_1} (\overline{\psi} \frac{\partial^2 \psi}{\partial x^2} + \psi \frac{\partial^2 \overline{ \psi}}{\partial x^2}) dx =$

$\frac{ih}{2m} ((\overline{\psi} \frac{\partial \psi}{\partial x})_{x_1}^{x_2} - \int_{x_2}^{x_1} \frac{\partial \overline{\psi}}{\partial x} \frac{\partial \psi}{\partial x} - ( (\psi \frac{\partial \overline{\psi}}{\partial x})_{x_1}^{x_2} - \int_{x_2}^{x_1}\frac{\partial \psi}{\partial x}\frac{\partial \overline{\psi}}{\partial x}dx)$

Now the integrals cancel each other and we obtain our result.

It is common to denote $-\frac{ih}{2m}(\overline{\psi}\frac{d \psi}{dx}-\psi \frac{d \overline{\psi}}{dx}$ by $S(x,t)$ (note the minus sign) and to say $\frac{d}{dt}P(x_1 < x < x_2 ; t) = S(x_1,t) - S(x_2,t)$ (see the reason for the minus sign?)

$S(x,t)$ is called the position probability current at the point $x$ at time $t$ One can think of this as a "probability flow rate" over the point $x$ at time $t$ ; the quantity $S(x_1, t) - S(x_2, t)$ will tell you if the probability of finding the particle between position $x_1$ and $x_2$ is going up (positive sign) or down, and by what rate. But it is important that these are position PROBABILITY current and not PARTICLE current; same for $|\psi |^2$ ; this is the position probability density function, not the particle density function.

NOTE I haven’t talked about the position and momentum eigenvalues or eigenfuctions. We’ll do that in our next post; we’ll run into some mathematical trouble here. No, it won’t be with the position because we already know what a distribution is; the problem is that we’ll find the momentum eigenvector really isn’t square integrable….or even close.

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College Math Teaching

August 11, 2011