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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*}

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

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 🙂

Heisenberg Uncertainty Relations (easy example n.1)

Today we will talk about the Heinsenberg Uncertainty Principle. In the next post you will find its mathematical derivation.

Ok let’s start.

The Heisenberg Uncertainty Principle/Relation (herein HUP) is one of the most important concepts in QM. Even is it seems difficult, at least from a purely mathematical point of view, it has a nice physical explanation.
Of course, the HUP is easy to misinterpret due to the counterintuitive laws of quantum physics but with some easy examples it becomes feasible. In the following I will put an example (and you will also find the reference at the end of the post).

Suppose that we are dealing with very small objects having only two properties: color and hardness. Moreover, for the sake of clarity, let’s restrict the colours to black and white only.
Of course, also the hardness could be expressed in such discrete way, i.e., an object could be hard or soft with no intermediate degrees of fluffiness ^_^

Given an object A, it can be black and soft, white and soft, and so on.

Now, we do also have two boxes (what an insane amount of two-something); the first measures the color and the second measures the hardness.

Color Box

Color Box

Hardness Box

Hardness Box

Of course we can suppose that a composition of these two boxes allows to measure both the properties. Let’s see if this works.
The first box, the one which measures the colour works like this: we provide it one of our misterious objects in input and we observe; if the object comes from the upper exit, then our object is surely white, otherwise it is black.
Same thing for the hardness. In this way we can exactly know if the object is hard or soft by knowing the output position.
An important property of these boxes is the repeatability: when one of our small objects of a certain value of a given property is given in input into a box measuring the value of that property, it comes out with the original value for that measured property. I.e., if the object comes from the color box and it is white, if I give it in input to another color box it will come out white again. The same thing works for the hardness as well.

Repeatability for Color

Repeatability for Color

Repeatability for Hardness

Repeatability for Hardness

Pretty intuitive, this works also in our “classical” world.

Now that we know that these properties are persistent we can ask if color and hardness are correlated.
We can test it!
If we measure one property (let’s say color) of our small objects and we take all the examples with a single value of that property (the white ones) and we measure the other property (hardness), the value of the other property is found to be probabilistically evenly split. What does this mean? If our objects that are known to all be white have their hardness measured, half will be soft and half will be hard; similarly, if the objects that are soft have their color measured, half will be black and half will be white.
So… the measure of the value of one property gives no predictive power on the other property.

Example of Uncorrelation

Example of Uncorrelation

Now that we know how to use these boxes, and how to compose them we can set an experiment slightly more sofisticated:

Experiment on Probabilities

Experiment on Probabilities

  • We measure the hardness of a beam of small objects;
  • we consider only the soft objects and we send them through the color box;
  • objects exiting the color box are half black and half white;
  • we send the half that are black through another hardness box, and the previous measurements found that the electrons entering this box were soft and black;
  • since measurement of hardness (and color) is repeatable, these objects should always come out as soft and never as hard.
  • So, which is the probability that our objects are hard? Reasonably, this probability should be 0.

    Half of the objects that were measured to only be soft are soft, and half now are black! The same thing works for any other pair of results from the first two boxes, if hardness and color were interchanged. Thus, the presence of the color box tampers with hardness, because without it, repeatability was ensured.
    As a consequence of this bizarre fact, we cannot build a box to simultaneously (and reliably) measure both color and hardness since the measurement of one disturbs the other one and vice versa!

    Simultaneous Measurement of Color and Hardness

    Simultaneous Measurement of Color and Hardness

    Simultaneously measuring hardness and color is disallowed! The general statement of this is the uncertainty principle, which is the idea that some measurable physical properties of real systems are incompatible with each other in the way that has been described thus far. This nice example was inspired by the course of Allan Adams, Matthew Evans, and Barton Zwiebach. 8.04 Quantum Physics I, Spring 2013. (Massachusetts Institute of Technology: MIT OpenCourseWare).. If you have time check it out. I have slightly modified it not talking of electrons or photons in order to be as general as possible at the beginning.

    Of course, this was an easy example to give you an idea of what are we talking about when we mention the Heisenberg Principle. Replace the color and hardness with two incompatible observables and we will do the trick.
    I know it just looks magic but it is how nature works.

    Next time I will introduce another nice example for the HUP, slightly more difficult, involving the famous double slit experiment.

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