1
$\begingroup$

In Quantum mechanics every wave funtion in position space is of the form $$\psi(x,t)=\frac{1}{\sqrt{2\pi \hbar}}\phi(p,t)e^{ip \cdot x/\hbar}dp,$$ and the corresponding wave function in momentum space is $$\phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}}\psi(x,t)e^{-ip \cdot x/\hbar}dx.$$

Both functions are probability amplitudes and due to Parsevals theorem:

$$\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\int_{-\infty}^{\infty}|\phi(p,t)|^2dp=1.$$

Average values in position space are:

$$\langle x_j \rangle = \int_{-\infty}^{\infty}x_j|\psi(x,t)|^2dx$$ and due to the transformation of derivative $$\langle p_k \rangle = \int_{-\infty}^{\infty}p_k|\phi(p,t)|^2dp=\int_{-\infty}^{\infty}-i \hbar \partial_k\psi(x,t)\overline{\psi(x,t)}dx.$$

This is all very clear, but how can we, for example, obtain the following average value:

$$\langle x_j \rangle \langle p_k \rangle = \int_{-\infty}^{\infty}-x_j \cdot i \hbar \partial_k\psi(x,t)\overline{\psi(x,t)}dx.$$

Above result is a consequence of the well-known Born's rule, but surely(?) we should be able to find it by direct calculation. It would be boring if we can calculate expectation values for $x$ and $p$, but not for their compositions without postulating the Born's rule.

$\endgroup$
1
$\begingroup$

The average of any observable $A$ is given by $\langle \Psi|A|\Psi\rangle$, which can be written in configuration space as

$$\langle A\rangle =\int_{\mathbb{R}^3} \bar \Psi(\vec r,t)\left(A\Psi(\vec r,t)\right)\,d^3\vec r$$

If $A=x_jp_k$, which in configuration space is $A=x_j\left(-i\hbar\frac{\partial}{\partial x_k}\right)$, then we have

$$\langle x_jp_k\rangle =\int_{\mathbb{R}^3} \bar \Psi(\vec r,t)x_j(-i\hbar )\frac{\partial \Psi(\vec r,t)}{\partial x_k}\,d^3\vec r$$

$\endgroup$
  • $\begingroup$ Yes, but how do we get that configuration space representation? $\endgroup$ – Hulkster Sep 19 '17 at 18:42
  • $\begingroup$ What do you mean by "How do we get ...?" The Probability density function is $\bar \Psi \Psi$ where $\Psi$ is the solution to the Schrodinger's Equation. $\endgroup$ – Mark Viola Sep 19 '17 at 18:43
  • $\begingroup$ Well, we do know how to get the fundamental operators position and momentum from wave functions, but for composition it's non-tirivial. $\endgroup$ – Hulkster Sep 19 '17 at 18:44
  • $\begingroup$ If the tensor $\vec r \vec p$ is an observable, then there is no issue. $\endgroup$ – Mark Viola Sep 19 '17 at 18:59
  • $\begingroup$ Apparently the Born's rule is just a postulate and we must work with those substitutions, so in that sense your answer is surely correct. Happy holidays and merry christmas :) $\endgroup$ – Hulkster Dec 18 '17 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.