Hello everyone!
In the last one I derived the Heisenberg Uncertainty Principle -ok ok I am not the first to do it ^_^ but at least I tried to put it as simple as possible!

The last topic were the Uncertainty Relations for the position and momentum operators. Now we will see which are those states that minimize this uncertainty. What does this mean? These states (that, when we will talk about the quantum harmonic oscillator, we will call coherent) are the states that behave -almost- classically.

It is possible to build states having the minimum possible spread of \hat{p} and \hat{q}, i.e., states that make this inequality  \Delta\hat{q}\Delta\hat{p} \geq \frac{\hbar}{2} become an equality:

     \begin{align*} \Delta\hat{q}\Delta\hat{p} = \frac{\hbar}{2} \end{align*}

This equality holds if and only if the two vectors involved are proportional to each other. Moreover, as we have already seen in the previous post, we require that the coefficient of proportionality is a complex number with the imaginary part only.

     \begin{align*} [ \hat{q} - \langle \hat{q} \rangle ]| \psi \rangle &= k  [ \hat{p} - \langle \hat{p} \rangle ] | \psi \rangle  \ \ \ \ \ \ \emph{where} \ k = i\alpha \nonumber \end{align*}

The latter equation, since we have used the coordinate representation, becomes:

     \begin{align*} [q - \langle \hat{q} \rangle ] \psi (q) &= i\alpha  [ - i \hbar \frac {d}{dq}- \langle \hat{p} \rangle ] \psi (q) \end{align*}

We can re-write it as:

 \alpha \hbar \frac{d}{dq} \psi (q) - q \psi (q) + (\langle \hat{q} \rangle - i\alpha \langle \hat{p} \rangle) \psi (q) = 0

Which can be simplified as follows:

     \begin{align*} \alpha \hbar \frac{d}{dq} \psi &= \psi (q - \langle \hat{q} \rangle + i \alpha \langle \hat{p} \rangle) \nonumber \\ \Rightarrow \frac{d \psi}{\psi} &= \frac{q + i \alpha \langle \hat{p} \rangle - \langle \hat{q} \rangle}{\alpha \hbar} \ dq \nonumber \\ \Rightarrow \int{\frac{d \psi}{\psi}} &= \int{\frac{q + i \alpha \langle \hat{p} \rangle - \langle \hat{q} \rangle}{\alpha \hbar} \ dq} \nonumber \\ \Rightarrow \ln{\psi} &= \frac{q^2} {2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle q}{\alpha \hbar}  \nonumber + C. \\ &= \frac{q^2 - 2 \langle \hat{q} \rangle q + \langle \hat{q} \rangle ^2 - \langle \hat{q} \rangle ^2}{2 \alpha \hbar} + i \frac{\langle \hat{p} \rangle q}{\hbar}+ C. \nonumber \\ \Rightarrow \psi (q) &=  A \exp{\Big[\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} \Big]}\\ &= A \exp{\Big[ - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} \Big] } \exp{\Big[\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle \hat{q}}{\hbar} \Big]} \end{align*}

Where A is a constant to be determined by normalizing the state.
This equation represents the form of minimum uncertainty states that, as we can see, take the standard Gaussian form.
It is important to note that any system in which there is a stationary state (with expectation value = 0) that has a Gaussian wavefunction will minimize the uncertainty relation.

Wavefunction Normalization
In the following I made a step-by-step normalization of the wavefunction \psi (q) for those who are not familiar with it. Since a Gaussian function looks difficult if you are not trained, I decided to display all the calculations here. Recall that A is in the form e^C and\alpha \leq 0.

     \begin{align*}\int_{- \infty}^{+\infty} | \psi (q)| ^2 dq = 1\end{align*}

Now let’s expand it:

     \begin{align*}\int_{- \infty}^{+\infty} | A e^{(\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} )}| ^2 dq = 1\end{align*}

We can now take |A|^2 out of the integral (remember that A = e^C) and rewrite the modulus square that remains inside the integral in this way:

     \begin{align*}|A|^2 \int_{- \infty}^{+\infty} e^{(\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} )} e^{(\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} - \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} )} dq = 1 \end{align*}

We can simplify the terms \frac{i \langle \hat{p} \rangle q}{\hbar} and -\frac{i \langle \hat{p} \rangle q}{\hbar} and make the remaining calculations, obtaining:

     \begin{align*}|A|^2 \int_{- \infty}^{+\infty} e^{(\frac{(q - \langle \hat{q} \rangle)^2}{\alpha \hbar} - \frac{\langle \hat{q} \rangle ^2 }{\alpha \hbar} )} dq = 1 \end{align*}

The term \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} is a constant, you can see that it does not depend from q, so now we have an integral in the form \int_{- \infty}^{+\infty} e^{-ax^2 + bx + c} that we are able to solve. If you are not able to solve it check here \textbf{link}} and the result is:

     \begin{align*}|A|^2 \sqrt{\frac{\pi}{\frac{1}{|\alpha | \hbar}}} \ e^{-\frac{\langle \hat{q} \rangle ^2}{\alpha \hbar}} = |A|^2 \sqrt{\pi|\alpha | \hbar} \ e^{-\frac{\langle \hat{q} \rangle ^2}{\alpha \hbar}}= 1\end{align*}

Now it’s trivial to determine the value of A:

    \begin{align*} |A|^2 &= e^{\frac{\langle \hat{q} \rangle ^2}{\alpha \hbar}} \ \frac{1}{\sqrt{\pi|\alpha | \hbar}}\\ A &= e^{\frac{\langle \hat{q} \rangle ^2}{2 \alpha \hbar}} \ \frac{1}{\sqrt[4]{\pi|\alpha | \hbar}} \end{align*}

And the value of the wavefunction \psi (q):

    \begin{align*}\psi (q) &= \frac{1}{\sqrt[4]{\pi|\alpha | \hbar}} \ e^{\frac{\langle \hat{q} \rangle ^2}{2 \alpha \hbar}} \ e^{(\frac{(q - \langle \hat{q} \rangle)^2}{2 \alpha \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar} - \frac{\langle \hat{q} \rangle ^2 }{2 \alpha \hbar} )}\\ &= \frac{1}{\sqrt[4]{\pi|\alpha | \hbar}} \ e^{-\frac{(q - \langle \hat{q} \rangle)^2}{2 |\alpha| \hbar} + \frac{i \langle \hat{p} \rangle q}{\hbar}} \end{align*}

Again if you need the PDF version of this, together with the first part, you can download it here. If you want to use this for your work just cite me, it is kind a nice thing to do.

For a bit more of theory about how to solve these kind of integrals wait for the next post! You will understand the reasons behind some steps.