Ballentine Problem 8.5 (angular momentum)

[EE18]

Homework Statement

Ballentine asks us "Show that the three-dimensional single particle state functions $\Psi_{m}(\mathbf{x}):= f(r)Y_l^{m}(\theta,\phi)$ and $\Psi_{-m}(\mathbf{x})$ have the same position and momentum distributions."

Relevant Equations

See below.

I am struggling with the latter, and think that I somehow need to assume $f$ is real-valued to proceed?

My work:
The position distributions are equal since
$$P_{-m}(\mathbf{x}) = |\Psi_{-m}(\mathbf{x})|^2 = |f(r)Y_l^{-m}(\theta,\phi)|^2 = |f(r)(-1)^m(Y_l^{m})^*|^2 = P_m(\mathbf{x})$$
if we recall that the magnitude of a complex conjugate is equal to the original complex number.

The momentum distributions are equal since (using (5.4) and inserting position space completeness twice)
$$P_{-m}(\mathbf{x}) = |\braket{\mathbf{p}}{\Psi_{-m}}|^2 \equiv \braket{\mathbf{p}}{\Psi_{-m}}\braket{\Psi_{-m}}{\mathbf{p}} =\frac{1}{(2\pi\hbar)^3}\int d\mathbf{x} e^{-i\mathbf{p}\cdot\mathbf{x}}f(r)Y_l^{-m}(\theta,\phi)\int d\mathbf{x}' e^{i\mathbf{p}\cdot\mathbf{x}'}f^*(r')(Y_l^{-m})^*(\theta',\phi')$$
$$ \stackrel{(1)}{=} \frac{1}{(2\pi\hbar)^3}\int d\mathbf{x} e^{-i\mathbf{p}\cdot\mathbf{x}}f(r)(-1)^m(Y_l^{m})^*(\theta,\phi)\int d\mathbf{x}' e^{i\mathbf{p}\cdot\mathbf{x}'}f^*(r')(-1)^mY_l^{m}(\theta',\phi')$$
where in (1) we have used (7.37) and its complex conjugate and that $(-1)^{2m} = 1$.

...But I can't go any further.

The referenced equations in my work are standard equations like the position representation of the momentum eigenstates. In particular, I think that I need to have that $f$ is real so that, after my last step, I can change variables in each integration to minus the given variable and so show the last equality. But I can't do that without being sure $f$ is real!

 

[anuttarasammyak]

I observe say f(r) is real
$$\Phi_{-m}(\mathbf{p}) :=\frac{1}{(2\pi\hbar)^3}\int d\mathbf{x} e^{-i\mathbf{p}\cdot\mathbf{x}}f(r)Y_l^{-m}(\theta,\phi)=\frac{(-1)^m}{(2\pi\hbar)^3}\int d\mathbf{x} e^{-i\mathbf{p}\cdot\mathbf{x}}f(r)Y_l^{m}(\theta,\phi)^*=(-1)^m\Phi^*_{m}(\mathbf{-p})$$
$$|\Phi_{-m}(\mathbf{p})|^2=|\Phi_{m}(\mathbf{-p})|^2$$
Is it helpful ?