Quantum Physics and Stuff

Let's get random together

Tag: Wavefunction

Minimum Uncertainty States and Wavefunction Normalization

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.

Derivation of the Heisenberg Uncertainty Relation

Here I am. This time it took me a while but I think I have, sort of, a justification. I am planning a visiting at the Newcastle University, thus I had loads of paperwork to do and I had very little time to take care of my blog 🙁

Today I will derive the Heisenberg Uncertainty Principle in operatorial form. The target are those students that can’t grasp some steps that, if you are not an expert, may look a bit obscure. As usual, I tried to put and comment as many steps as possible, even the trivial ones.
At the end of the post you will find something interesting about the minimum uncertainty states.

So, this was an assignment for the Quantum Mechanics course. My QM professor reviewed it so I think it should be ok.

Let’s consider a generic state |\psi \rangle and an observable \hat{O} satisfying the following eigenvalues equation:

\hat{O} |o_n\rangle = \lambda_n |o_n \rangle

Since the observable is Hermitian, we can expand | \psi\rangle as follows:

| \psi\rangle = \sum_n {a_n |o_n\rangle}

where |o_n\rangle is the eigenstate of \hat{O} corresponding to the eigenvalue \lambda_n.
According to the measurement postulate, after we measure the observable \hat{O} (in the case that no degeneracy occurs, which means that there are no linearly independent eigenvectors correspondig to the same eigenvalue) given a quantum system in some state | \psi\rangle, as a result we obtain \lambda_n with probability |a_n| ^2. Moreover, right after the measurement, the state of the system collapses in the eigenstate |o_n\rangle.

The expectation value, i.e. the mean value in the long run for repeated measurements of \hat{O} on a set of identically prepared states |\psi \rangle is indicated as follows:

     \begin{align*}\langle \hat{O} \rangle _{\psi} := \langle \psi | \hat{O} | \psi \rangle  = \sum_n {\lambda_n |\langle o_n | \psi \rangle |^2}\end{align*}

where |\langle o_n| \psi \rangle |^2 represents the probability that the output \lambda_n will occur.

The measure of the spread of the results around the expectation value is called the standard deviation or root-mean-square, i.e. :

     \begin{align*}\Delta \hat{O} _{\psi} &:= \sqrt{\langle (\hat{O} - \langle \hat{O} \rangle)^2 \rangle} \nonumber \\ &= \sqrt{\langle \hat{O}^2 \rangle - \langle \hat{O} \rangle ^2} \end{align*}

Commuting Hermitian operators can be measured simultaneously, i.e., given two observables \hat{A} and \hat{B} if [\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}=0 then they share a common set of eigenstates in which both operators have a definite value at the same time; simplifying, we can state that \hat{A} and \hat{B} are simultaneously measurable if the result of measuring \hat{A} and then \hat{B} is the same as the one obtained measuring \hat{B} and then \hat{A}. In this case the observation of one operator does not affect the observation of the other one, hence the order of the measurements doesn’t matter. In the following example we consider two commuting Hermitian operators satisfying the following eigenvalue equations:

     \begin{align*}     \hat{A}|\psi_n\rangle=\alpha _n|\psi_n\rangle \nonumber \\     \hat{B}|\psi_n\rangle=\beta _n|\psi_n\rangle \nonumber \end{align*}

where \{ |\psi _n \rangle \} are a complete set of common eigenstates with respect to the eigenvalues \{\alpha _n\} , \{\beta _n \}.

Measurement of Commuting Observables

Measurement of Commuting Observables


In the above figure we can see an example of measurements of commuting operators while in the following there is represented the measurement of non-commuting operators. Recall that the two outcomes \alpha_n and \alpha_k are not necessarily the same.

Measurement of Non-Commuting Observables

Measurement of Non-Commuting Observables

The \Delta t are taken small enough to preserve the state from changes occuring during the time evolution.
This example can be generalized to deal with more than two observables considering also the case in which degeneracy occurs, i.e. in which we have two or more eigenstates corresponding to the same eigenvalue.

On the contrary, in the case of two operators that do not commute, i.e. their commutator has a non-zero value, it does not exist a complete set of common eigenstates, hence performing a measurement on them simultaneously is not possible. Moreover, the measurement of one operator affects the measurement of the other one, hence the order of the measurements is important.

We will see that the existence of non-commuting observables is at the basis of the Heisenberg’s uncertainty relation and is a quite important physical result.
To derive the uncertainty relation we will use the Schwarz Inequality:

     \begin{align*}         \langle \varphi| \varphi \rangle \langle \psi| \psi \rangle  \geq |\langle \varphi | \psi \rangle |^2         \label{eq: schwarz2} \end{align*}

with equality holding when |\psi \rangle  = \alpha |\varphi \rangle , for any two normalized states |\varphi \rangle, |\psi \rangle in a Hilbert space \mathcal{H}.

At first we will derive the uncertainty relation for a generalized pair of non-commuting Hermitian operators.

Let \hat{A} and \hat{B} be two non-commuting Hermitian operators, i.e. [\hat{A},\hat{B}] \neq 0 . We will prove that the following inequality holds:

(1)   \begin{equation*} \Delta\hat{A} _{\psi} \Delta\hat{B} _{\psi} \geq \frac{1}{2} | \langle [\hat{A},\hat{B}]\rangle _{\psi}| \end{equation*} To make the calculations easier we will express the latter inequality in term of variance, i.e.: \begin{equation*} (\Delta\hat{A})^2 (\Delta\hat{B})^2 \geq \frac{1}{4} | \langle [\hat{A},\hat{B}]\rangle | ^2 \label{eq: uncertainty-easier}\end{equation*}

Proof
Before starting the derivation of the uncertainty relation, we need some extra
definitions.

Let \hat{A}' and \hat{B}' be two non-commuting Hermitian operators defined as follows:

     \begin{align*}\hat{A}' &:= \hat{A} - \langle\hat{A}\rangle \\  \hat{B}' &:= \hat{B} - \langle\hat{B}\rangle \end{align*}

Those observables, when acting on the state |\psi\rangle give, respectively:

     \begin{align*} \hat{A}'|\psi\rangle &:= |\psi _{A'} \rangle \\ \hat{B}'|\psi\rangle &:= |\psi _{B'} \rangle \end{align*}

We can easily verify that the commutator of these operators is equal to the commutator between \hat{A} and \hat{B} :

     \begin{align*} [\hat{A}',\hat{B}'] &= [(\hat{A} - \langle\hat{A}\rangle)(\hat{B} - \langle\hat{B}\rangle)] - [(\hat{B} - \langle\hat{B}\rangle)(\hat{A} - \langle\hat{A}\rangle)] \nonumber\\ %&= \hat{A}\hat{B} - \hat{A}\langle \hat{B} \rangle - \langle \hat{A} \rangle \hat{B} + \langle \hat{A}\rangle\langle \hat{B} \rangle - \hat{B}\hat{A} + \hat{B}\langle \hat{A} \rangle + \langle \hat{B}\rangle \hat{A} - \langle \hat{B}\rangle\langle \hat{A} \rangle \nonumber \\  %&= \hat{A}\hat{B} - \hat{B}\hat{A} \nonumber \\  &= [\hat{A},\hat{B}] \end{align*}

We can write the product between \hat{A}' and \hat{B}' in the following way:

     \begin{align*} \hat{A}'\hat{B}' &= \frac{1}{2} \hat{A}'\hat{B}' + \frac{1}{2} \hat{A}'\hat{B}' + \frac{1}{2} \hat{B}'\hat{A}' - \frac{1}{2} \hat{B}'\hat{A}' \nonumber\\ &= \frac{1}{2} (\hat{A}'\hat{B}' + \hat{B}'\hat{A}') + \frac{1}{2} (\hat{A}'\hat{B}' - \hat{B}'\hat{A}') \nonumber \\  &= \frac{1}{2} \{\hat{A}',\hat{B}'\} + \frac{1}{2} [\hat{A}',\hat{B}'] \end{align*}

This way we have decomposed the above product in an anti-Hermitian part, i.e., [\hat{A}, \hat{B}]^{\dag} = - [\hat{A}, \hat{B}] and in an Hermitian one, i.e., \{\hat{A}, \hat{B}\}^{\dag} = \{\hat{A}, \hat{B}\}.

Now we can give the proof of the following inequality:
 \begin{center}\boxed{ (\Delta\hat{A})^2 (\Delta\hat{B})^2 \geq \frac{1}{4} | \langle [\hat{A},\hat{B}]\rangle | ^2 }\end{center}

     \begin{align*} (\Delta\hat{A})^2 (\Delta\hat{B})^2 &= \langle \psi| (\hat{A} - \langle \hat{A} \rangle) ^2 |\psi\rangle \langle \psi| (\hat{B} - \langle \hat{B} \rangle) ^2 |\psi\rangle \nonumber \\ &= \langle \psi | (\hat{A}') ^2 | \psi \rangle \langle \psi | (\hat{B}') ^2 | \psi \rangle \nonumber \\ &= \langle \psi _{A'}| \psi _{A'} \rangle \langle \psi _{B'}| \psi _{B'} \rangle \label{eq: derivation1} \end{align*}

The last equality follows from the hermiticity of \hat{A}' and \hat{B}'.
Using the Schwarz Inequality we obtain:

     \begin{align*} \langle \psi _{A'}| \psi _{A'} \rangle \langle \psi _{B'}| \psi _{B'} \rangle \geq |\langle \psi _{A'}| \psi _{B'} \rangle|^2 &= |\langle \psi| \hat{A}'\hat{B}'| \psi\rangle | ^2 \nonumber \\ &= |\langle \psi | \frac{1}{2} \{\hat{A}',\hat{B}'\} + \frac{1}{2} [\hat{A}',\hat{B}'] | \psi \rangle |^2 \nonumber \\ &\geq \frac{1}{4} |\langle \psi | [\hat{A}',\hat{B}'] | \psi \rangle |^2 \end{align*}

The decomposition in Hermitian and anti-Hermitian parts implies a decomposition also for the mean values of the product. Moreover, from the hermiticity of \hat{A}' and \hat{B}' it follows that \langle \psi |\{\hat{A}',\hat{B}'\} | \psi \rangle is real, while \langle \psi |[\hat{A}',\hat{B}'] | \psi \rangle is a pure imaginary number (the expectation value of an anti-Hermitian operator is \langle \hat{O}  \rangle =  -\langle \hat{O}\rangle^* ), hence the anti-commutator only strengthens our inequality because the modulus of a complex number is always greater than the modulus of its imaginary part only.
If we substitute this result we can prove that the inequality holds:

     \begin{align*} (\Delta\hat{A})^2 (\Delta\hat{B})^2 &= \langle \psi | (\hat{A}') ^2 | \psi \rangle \langle \psi | (\hat{B}') ^2 | \psi \rangle  \nonumber \\ &\geq \frac{1}{4} |\langle \psi |[\hat{A}',\hat{B}'] | \psi \rangle |^2 = \frac{1}{4} |\langle \psi | [\hat{A},\hat{B}] | \psi \rangle |^2  \end{align*}

Taking the square root of the last inequality we obtain:
 \Delta\hat{A} _{\psi} \Delta\hat{B} _{\psi} \geq \frac{1}{2} | \langle [\hat{A},\hat{B}]\rangle _{\psi}| \\ \qed

Recall that the equality holds when |\psi_{A'} \rangle = k |\psi_{B'}\rangle , and the coefficient [/latex]k[/latex] is a pure imaginary number such that [/latex]k = i \alpha[/latex] with \alpha \in \mathbb{R} . We need k to be purely imaginary due to the fact that the mean value in the right part of the last inequality is purely imaginary and we don’t want any of the considered expectation values to be zero.

We will use this in the following to find the states which minimize the uncertainty.

Uncertainty relations for Position and Momentum operators
Let’s now consider the uncertainty relation for the case in which we take:

    \begin{align*} \hat{A} &= \hat{q} \\ \hat{B} &= \hat{p} \end{align*}

where \hat{q} and \hat{p} are, respectively, the position and momentum operators. We want to prove that the following inequality holds:

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

This case is more interesting than the generalized one since the right part of the inequality does not depend on the state |\psi\rangle of the system.
The commutator between \hat{q} and \hat{p} is:

     \begin{align*} [\hat{q}, \hat{p}] = i \hbar \end{align*}

Substituting such commutator we obtain the following inequality:

     \begin{align*} (\Delta\hat{q})^2 (\Delta\hat{p})^2 \geq \frac{1}{4} |\langle [\hat{q},\hat{p}] \rangle |^2 &= \frac{1}{4} | \langle i \hbar \rangle | ^2  \nonumber \\ &= \frac{\hbar^2}{4} \label{eq: qp} \end{align*}

which implies that the inequality holds.

We can interpret this as a lower bound on the accuracy by which we can know both the position and the momentum of a particle given a state | \psi \rangle. Hence, once the lower bound is reached, the smaller is the variance of the position the higher is the uncertainty for the momentum and vice-versa.

It is possible now to guess which are the minimum uncertainty states for the position and momentum operators. You will find them here together with the wavefunction normalization.

If you want this post + minimum uncertainty states I made a PDF file that you can find here. As usual if you find any error or want to ask any question just write to me. I am happy to improve these little blog and its contents 🙂