College Math Teaching

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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