A step towards defining the path of electron

Hi learners,

As the famous Heisenberg's uncertainty principle declares the inability of accurately measuring the properties of a quantum system, hence we can not determine the path of electron in the atom. The main aim of my discussion is to present you guys, a hypothesized procedure and an attempt to approximately find out the path of an electron in the atom. In a nutshell this is an ambitious project.

Note :- I am no scientist, professor or teacher, I am just 15 year old. Neither I have done a course or degree in quantum mechanics . So please do correct me if I am wrong in any line or point. Your valuable comments and suggestions will be entertained. So let's begin....

Consider a hypothetical isolated atom of hydrogen (though we can never isolate the atom) having an electron in its 1s-subshell . As the orbital is included here , we'll talk of only probability. We may take different quantum states of this particular atom as : \[\left| {{\Psi _1}} \right\rangle ,\left| {{\Psi _2}} \right\rangle ,\left| {{\Psi _3}} \right\rangle , \ldots \ldots \ldots \ldots ,\left| {{\Psi _n}} \right\rangle \] These quantum states of electron contain their different positions and momenta, As the electron can have many positions and momenta at the same time, hence we are at a risk of double counting. In order to avoid this, we will collapse the wavefunction of their individual quantum states to obtain one position and one momentum . Finally just by taking the summation of the collapsed wavefunction of the different quantum states obtaining 'n' positions and 'n' momentum , discarding 'n' momenta and considering 'n' positions we'll get the approx. path of the electron in the hydrogen atom.

                                                                                  OR

Another SUPPOSED way of doing this is to let that the individual quantum state of the electron is the probability of finding it in a particular region of the 1s - electron cloud (which is in a superposition of many positions and momenta). This means that a particular region of the 1s - electron cloud (the region in which the electron is likely to be found is called the electron cloud) the electron of the quantum state \[\left| {{\Psi _1}} \right\rangle \] is found. Similarly, the electron may be found in the \[\left| {{\Psi _2}} \right\rangle ,\left| {{\Psi _3}} \right\rangle , \ldots \ldots \ldots \ldots ,\left| {{\Psi _n}} \right\rangle \] quantum states of the corresponding regions in the electron cloud of the 1s - subshell. Next, as the wavefunction of the corresponding quantum states change with time and its distance from the nucleus (because the motion of is not fixed hence its distance should change). then by partially differentiating the wavefunction w.r.t. time and distance from the nucleus (let it be 'x' ) (though I haven't that much knowledge of partial differentials and their integration). It MAYBE like this :-

\[\frac{{\partial {\Psi _1}}}{{\partial ({x_{1,}}{t_1})}} = {1^{st}}{\text{ position and }}{{\text{1}}^{st}}{\text{ momentum}}\]

\[\frac{{\partial {\Psi _2}}}{{\partial ({x_{2,}}{t_2})}} = {2^{nd}}{\text{ position and }}{{\text{2}}^{nd}}{\text{ momentum}}\]

\[\frac{{\partial {\Psi _3}}}{{\partial ({x_{3,}}{t_3})}} = {3^{rd}}{\text{ position and }}{{\text{3}}^{rd}}{\text{ momentum}}\]

\[\begin{gathered} \vdots \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ \end{gathered} \]

\[\frac{{\partial {\Psi _n}}}{{\partial ({x_{n,}}{t_n})}} = {n^{th}}{\text{ position and }}{{\text{n}}^{th}}{\text{ momentum}} \]

And another important thing to be noted here is that these quantum states must be indeterminate ( hence can't be counted), hence we'll take the limit of the summation of derivative of wavefunction of each quantum quantum state and this, eventually, be the path of the electron .

\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{\partial {\Psi _i}}}{{\partial ({x_{i,}}{t_i})}}} \Rightarrow \int\limits_0^\infty {\partial ({x_{i,}}{t_i}) = } {\Psi _i}\]

Most importantly, in order to avoid confusion regarding my assumption of the quantum states in the particular region of the so-called electron cloud, I will be showing a handwritten diagram as follows :

In the end, I would like to thank you all for reading my article . Have great day !

Note by Amit Panghal
1 week, 1 day ago

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Hi @Amit Panghal, this is certainly full of interesting questions.

One immediate wonder I have is what you mean by this:

These quantum states of electron contain their different positions and momenta, As the electron can have many positions and momenta at the same time, hence we are at a risk of double counting. In order to avoid this, we will collapse the wavefunction of their individual quantum states to obtain one position and one momentum . Finally just by taking the summation of the collapsed wavefunction of the different quantum states obtaining 'n' positions and 'n' momentum , discarding 'n' momenta and considering 'n' positions we'll get the approx. path of the electron in the hydrogen atom.

It sounds like you feel we can simply collapse the wave function and get something that's an average path. I think most people would want to hear details of how you see that process happening. Usually, "collapsing" the wave function is simply settling into one of the many possible states. Using your notation, a quantum state might be a superposition along the lines of \(\ket{\psi} = \alpha_1\ket{\Psi_1} + \alpha_2\ket{\Psi_2} + \cdots + \alpha_n\ket{\Psi_n}.\) And collapsing in the normal sense means it would adopt the state \(\ket{\Psi_i}\) with probability \(\lvert\alpha_i\rvert^2.\)

In your second model, it sounds like you are proposing to split up the physical space encompassed by, for example, the 1s orbital, and assigning a different wave function to each subvolume? I am not sure I fully understand the idea here.

On the whole, I am curious how you got the idea for these theories?

Josh Silverman Staff - 5 days, 19 hours ago

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That's a nice and interesting article. Good work Amit Panghal

Ram Mohith - 1 week, 1 day ago

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Thank you very much for appreciation @Ram Mohith

Amit Panghal - 7 hours ago

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