# Fourier transform and expectation values in QM?

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.

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

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

• Yes, but how do we get that configuration space representation? – Hulkster Sep 19 '17 at 18:42
• 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. – Mark Viola Sep 19 '17 at 18:43
• Well, we do know how to get the fundamental operators position and momentum from wave functions, but for composition it's non-tirivial. – Hulkster Sep 19 '17 at 18:44
• If the tensor $\vec r \vec p$ is an observable, then there is no issue. – Mark Viola Sep 19 '17 at 18:59
• 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 :) – Hulkster Dec 18 '17 at 23:08