# 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. 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 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 $(\langle r^2 \rangle\langle p^2 \rangle)^{0.5}$ 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 $$\langle r \rangle \langle p \rangle > 3/2 (\frac{h}{2\pi})$$

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! Sep 30 '15 at 16:31