Based on the 1D case mentioned in Griffiths, I decided to try looking at the features of 3D Gaussian wavefunctions, i.e. (position basis) wavefunctions of the form $\psi(\mathbf{r}) = Ae^{-\mathbf{r}^\dagger\mathsf\Sigma\mathbf{r}/4}$, where A is a normalization constant, r is position, Σ is a positive-definite symmetric matrix (which by a suitable change of coordinate basis can be made diagonal), and † denotes the conjugate transpose. Applying standard results for Gaussian integrals, I was able to get
- $\langle \mathbf{r} \rangle = 0$
- $\langle r^2\rangle = \operatorname{Tr}\mathsf\Sigma$
- $\langle \mathbf{p} \rangle = 0$
- $\langle p^2\rangle = \frac{\hbar^2}{4}\operatorname{Tr}\mathsf\Sigma^{-1}$
So, substituting into Heisenberg's uncertainty principle and rearranging terms, it follows that, in order to get minimum uncertainty with respect to $\mathrm{r}$ and $\mathrm{p}$, we need to have
$(\operatorname{Tr}\mathsf\Sigma)(\operatorname{Tr}\mathsf\Sigma^{-1})=1$.
Here's where I'm running into a difficulty. As I mentioned before, the matrix Σ can always be assumed to be diagonal. Then the only possible solution for Σ is
$\mathsf\Sigma = \begin{pmatrix} 1 & 0 & 0\\ 0 &-1 &0\\ 0 &0 &1\end{pmatrix}\times\mathrm{constant}$
But this contradicts the fact that Σ is positive-definite (the -1 would imply that one of the coordinates has negative uncertainty, an absurdity).
Assuming I did all the calculations correctly, this seems to imply that a Gaussian wavefunction is not the minimum uncertainty wavefunction with respect to r and p. On the other hand, it's comparatively trivial to show that it is the minimum uncertainty wavefunction with respect to x and px, y and py, and z and pz individually.
Is there a wavefunction which is the minimum unceratinty wavefunction with both respect to the individual coordinates (e.g. x and px) and with respect to r and p?
Edit It was asked by marek what I meant by "minimum uncertainty with respect to $\mathbf{r}$ and $\mathbf{p}$". To answer this, recall that the generalized uncertainty principle takes the form of
$$ \sigma_A\sigma_B \geq \frac{1}{2}\left|\langle[A,B]\rangle\right|.$$
Although I'm not entirely sure it's valid to do so, I assumed that to calculate the commutator $[\mathbf{r},\mathbf{p}]$ I could use the formalism of geometric algebra (see Geometric algebra). Then
$$\begin{align*}
[\mathbf{r},\mathbf{p}]f &= \frac{\hbar}{i}\mathbf{r}\nabla f – \frac{\hbar}{i}\nabla(f\mathbf{r})\\
&= \frac{\hbar}{i}\sum_{jk} \left[x^j\hat{\mathbf{e}}_j\frac{\partial f}{\partial x^k}\hat{\mathbf{e}}^k – \frac{\partial}{\partial x^k}\left(fx^j\hat{\mathbf{e}}_j\right)\hat{\mathbf{e}}^k\right]\\
&= \frac{\hbar}{i}\sum_{jk} \left[ x^j\frac{\partial f}{\partial x^k} \hat{\mathbf{e}}_j\hat{\mathbf{e}}^k – \frac{\partial f}{\partial x^k}x^j\hat{\mathbf{e}}_j\hat{\mathbf{e}}^k – f{\delta^j}_k\hat{\mathbf{e}}_j\hat{\mathbf{e}}^k\right]\\
&= \frac{\hbar}{i} f,
\end{align*}$$
where $f$ is an arbitrary function, $x^1,x^2,x^3$ are the position coordinates, and $\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2,\hat{\mathbf{e}}_3$ are the standard Cartesian basis vectors. Thus, the uncertainty principle for $\mathbf{r}$ and $\mathbf{p}$ takes the form
$$\sigma_\mathbf{r}\sigma_\mathbf{p} \geq \frac{\hbar}{2},$$
which means that the minimum uncertainty wavepacket with respect to $\mathbf{r}$ and $\mathbf{p}$ must satisfy
$$\sigma_\mathbf{r}\sigma_\mathbf{p} = \frac{\hbar}{2}.$$
Best Answer
It seems that problem here is with mishandling vector quantities. We want to compute things such as $\left<p^2\right>$ but these are in fact $\sum_i \left<p_i^2\right>$ and so the problem decomposes into components where the standard HUP and minimality conditions can be applied. But what you've done is that you applied one-dimensional HUP to $\left<x^2\right>$ and $\left<p^2\right>$ which just can't be right. The correct form of HUP in this case would be $$\sum_i \left<x_i^2\right>\left<p_i^2\right> \geq 3 {\hbar^2 \over 4}$$
So, to reiterate, there is really nothing new to solve in more dimensions as the problem decomposes completely and you can write your solution as $\Psi(x,y,z)$ = $\psi_x(x)\psi_y(y)\psi_z(z)$ with each $\psi_{\alpha}$ a Gaussian from the one-dimensional variant of this problem.