# I Understanding the Uncertainty Principle

1. Sep 8, 2016

### majormuss

Hi everybody,
I was reading about the Uncertainty principle and I came across the statement:
"the position-space wavefunction of a minimum-uncertainty state is a Gaussian. Indeed, because it’s a minimum-uncertainty state, its momentum-space wavefunction is also a Gaussian."
Unfortunately it was stated without any mathematical proof or diagrams so I am a bit puzzled! Also,
what does it mean exactly that:
"The first energy level of a system is minimum uncertainty."
My guess it that it means for lower energies the uncertainty in the energy is minimum. However, due to the quantized nature of the energy levels, how is it possible to be uncertain about the energy of a particle?

2. Sep 8, 2016

### Lucas SV

A minimum-uncertainty state is one in which $\sigma_x \sigma_p = \frac{\hbar}{2}$, where $\sigma_x$ is the uncertainty in position and $\sigma_p$ is the uncertainty in momentum. $\hbar /2$ is the smallest possible value for the product of uncertainties in position and momentum, by the Heisenberg uncertainty principle.

I don't know if every minimum-uncertainty state is a gaussian, but it is certainly true that gaussians are minimum-uncertainty states. To show this, do problem 1.14 in Griffiths, Introduction to Quantum mechanics.

3. Sep 8, 2016

### Chandra Prayaga

The minimum uncertainty refers to the product: σxσp=ℏ/2, as stated above by Lucas SV. There is no uncertainty in energy in a stationary state, as the energy is known exactly.

4. Sep 9, 2016

### vanhees71

One proof of the uncertainy principle for two observables represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ goes as follows: Let $\lambda \in \mathbb{R}$. Now we use that the quadratic polynomial
$$P(\lambda)=\langle (\hat{A}+\mathrm{i} \lambda \hat{B}) \psi| (\hat{A}+\mathrm{i} \lambda \hat{B}) \psi \rangle \geq 0.$$
It's $=0$ if and only if (for some value of $\lambda$)
$$(\hat{A}+\mathrm{i} \lambda \hat{B}) |\psi \rangle=0. \qquad (*)$$
Without loss of generality we can assume that
$$\langle A \rangle=\langle{B} \rangle=0,$$
because otherwise we can consider the observables $\hat{A}-\langle A \rangle \hat{1}$ and $\hat{B}-\langle B \rangle \hat{1}$. Now due to the self-adjointness of the operators and since $\lambda \in \mathbb{R}$ by assumption, we have
$$P(\lambda)=\langle \psi|(\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A} + \mathrm{i} \lambda \hat{B})|\psi \rangle= \langle \psi |\hat{A}^2 + \lambda \hat{B}^2+\mathrm{i} \lambda [\hat{A},\hat{B} ]|\psi \rangle.$$
Since this polyonmial is $\geq 0$ always, it has either exactly one real root, and then for this root (*) is valid or it has two complex conjugate roots. This implies that always
$$\langle A^2 \rangle \langle B^2 \rangle \geq \frac{1}{4} \langle \mathrm{i} [\hat{A},\hat{B} ] \rangle.$$
This is the uncertainty relation.

Now (*) is the condition for the state to be of "minimal uncertainty", i.e., that the equality sign holds in the uncertainty relation. Now take $\hat{A}=\hat{p}$ and $\hat{B}=\hat{x}$. Now we use the position representation of the space ket and check the solution for (*), given any $\lambda \in \mathbb{R}$:
$$(\hat{p}+\mathrm{i} \lambda \hat{x})\psi(x)=0 \; \Rightarrow \; \mathrm{i} (-\partial_x + \lambda x) \psi(x)=0.$$
Here and in the following I use natural units, where $\hbar=1$. The above equation means
$$\psi'(x)=-\lambda x \psi(x)$$
with the solution
$$\psi(x)=A \exp \left (-\frac{\lambda}{2} x^2 \right ).$$
In order that $\psi \in L^2(\mathbb{R})$ we must have $\lambda>0$. So for any $\lambda>0$ the above Gaussian wave function represents a state of "minimal uncertainty", i.e., for which
$$\Delta x \Delta p=\frac{1}{2}.$$

Last edited: Sep 9, 2016
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