# 3D Minimum uncertainty wavepackets

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 $$\sigma_\mathbf{r}\sigma_\mathbf{p} = \frac{\hbar}{2}.$$

• Well, I don't know geometric algebra but this seems definitely fishy. For one thing the object you should get as $[\mathbf r, \mathbf p]$ should be a bivector (if you've written this in indices, it should carry two of them) but you get just a scalar (and you dropped a minus sign too). Well, that bivector can be represented by an identity operator on a 3D vector space too (is this what you mean by the last row?) but you surely can't plug such an object into HUP. You need to get just numbers somehow. – Marek Mar 18 '11 at 22:58
• Dagnabit you're right! And you'd think after almost three years of being a math and physics major I wouldn't mess up on a simple dimensions thing. Oh well – Avi Steiner Mar 20 '11 at 4:14

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.

The minimum uncertainty for (< r^2 >< p^2 >)^0.5 is 3/2 hbar To see that we can use the differential calculus. Lets us write it explicitly (< x^2 >+< y^2 >+< z^2 >)*( px^2+py^2+pz^2) is more then what ? From one dimension we have immediately the diagonal terms < x^2 >< px^2 > > hbar^2/4 and so on for y an z. The tricky point is to minimize 3 cross terms like < x^2 >< pz^2 >+< z^2 >< px^2 > since the terms there are not independent. There are 3 of them. Since x pz are commuting the one part of those terms can be made as small as possible. Let us take < x^2 >< pz^2 > = epsilon. Then we have < z^2 > > hbar^2/4/ and < px^2 > > hbar^2/4/< x^2 > and there fore < x^2 >< pz^2 >+< z^2 >< px^2 > > epsilon + hbar^4/16/epsilon Now we only ask what epsilon minimizes the right side by calculating the derivative and putting to zero: 1 - hbar^4/16 /epsilon^2 = 0 so epsilon = hbar^2/4

So the total < r^2 >< p^2 > > 3 * hbar^2/ 2 + 3 * hbar^2/4 = 9/4 hbar^2

so < r > < p > > 3/2 hbar

• Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! – Gonenc Sep 30 '15 at 16:31

The minimum uncertainty for $(\langle r^2 \rangle\langle p^2 \rangle)^{0.5}$ is $\frac{3}{2} \frac{h}{2\pi}$

To see that we can use the differential calculus. Lets us write it explicitly $$(\langle x^2 \rangle+\langle y^2 \rangle+\langle z^2 \rangle)\times ( p_x^2+p_y^2+p_z^2)$$ is more than what?

From one dimension we have immediately the diagonal terms $$\langle x^2 \rangle\langle p_x^2 \rangle > \frac{1}{4}{(\frac{h}{2\pi})}^2$$ and so on for y an z.

The tricky point is to minimize 3 cross terms like $$\langle x^2 \rangle\langle p_z^2 \rangle+\langle z^2 \rangle\langle p_x^2 \rangle$$ since the terms there are not independent. There are 3 of them.

Since $x pz$ are commuting the one part of those terms can be made as small as possible.

Let us take $\langle x^2 \rangle\langle p_z^2 \rangle = \epsilon$. Then we have $$\langle z^2 \rangle > \frac{1}{4}(\frac{h}{2\pi})^2\langle p_z^2 \rangle$$ and $$\langle p_x^2 \rangle > \frac{1}{4}(\frac{h}{2\pi})^2\langle x^2 \rangle$$ and therefore

$$\langle x^2 \rangle\langle p_z^2 \rangle+\langle z^2 \rangle\langle p_x^2 \rangle >\epsilon + \frac{1}{16}(\frac{h}{2\pi})^{4}\epsilon$$ Now we only ask what epsilon minimizes the right side by calculating the derivative and putting to zero: $$1 - (\frac{h}{2\pi})^4 \frac{1}{16\epsilon^2} = 0 \ so\ \epsilon = (\frac{h}{2\pi})^2/4$$

So the total $$\langle r^2 \rangle\langle p^2 \rangle > 3 \times (\frac{h}{2\pi})^2/ 2 + 3 \times (\frac{h}{2\pi})^2/4 = 9/4 (\frac{h}{2\pi})^2$$

so $$\langle r \rangle \langle p \rangle > 3/2 (\frac{h}{2\pi})$$