Particle on a ring (components in the postion basis)

Lambda96
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Hi,

I have problems with the task part b and g

To solve the task, we have received the following information

Bildschirmfoto 2023-06-14 um 19.14.36.png


Task b
Bildschirmfoto 2023-06-14 um 20.03.21.png


First, I wrote down what the state ##\psi## looks like

$$\psi=\frac{1}{\sqrt{N}} \sum\limits_{k}^{} \psi_k$$
$$\psi=\frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$

Then I to calculate ##\psi^j=\braket{\vec{e}_j|\psi}##.

$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$

$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{N} \sum\limits_{k}^{} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$
Now I unfortunately do not know how to proceed further. But I don't understand, if all momentums are equally probable, why the particle should be 100% at location N and not at other locations like 1 and 2 and so on. What makes the point N so special that the particle should be there in contrast to the other points?

To solve task g, we have received the following information
Bildschirmfoto 2023-06-14 um 19.25.16.png


Task g
Bildschirmfoto 2023-06-14 um 19.15.43.png

If I understood the task correctly, then the wave function is collapsed, to the eigenvector of the momentum operator, more precisely to ##\psi_0##. The wave function has with 100% the eigenvalue of ##\psi_0## after the uncertainty principle, the uncertainty would have to become extremely large concerning the position, which means that the particle can be everywhere on the ring and thus the probability for each position is equally large.
 
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Lambda96 said:
$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$
In the second sum on the right, you should change the summation index ##j## to some other symbol so that the summation index is not confused with the ##j## in ##\vec{e}_j^{\dagger}##. For example, you could write $$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{l}^{} e^{ikal} \vec{e}_l$$
 
Thanks TSny for your help 👍, I have now changed the index of the summation from ##j## to ##l## and have now calculated the following.

$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{l}^{} e^{ikal} \vec{e}_l$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \vec{e}_j^{\dagger} \cdot \vec{e}_l$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \delta_{jl}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} e^{ika} \delta_{j1}+e^{2ika} \delta_{j2}+ \ldots +e^{Nika} \delta_{jN}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \Bigl( e^{ia} \delta_{j1}+e^{2ia} \delta_{j2}+ \ldots +e^{Nia} \delta_{jN}+e^{2a} \delta_{j1}+e^{4ia} \delta_{j2}+ \ldots +e^{2Nia} \delta_{jN}+ \ldots + e^{Nia} \delta_{j1}+e^{2Nia} \delta_{j2}+ \ldots +e^{N^2ika} \delta_{jN} \Bigr)$$

$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \Bigl(\Bigl(e^{ia}+e^{2a} +\ldots+e^{Nia} \Bigr) \delta_{j1}+\Bigl( e^{2ia}+e^{4ia}+ \ldots +e^{2Nia} \Bigr)\delta_{j2}+ \ldots + \Bigl( e^{Nia}+e^{2Nia} + \ldots +e^{N^2ika} \Bigr)\delta_{jN}\Bigr)$$

Now I'm stuck 🙃

The individual entries in the brackets look almost like states ##\vec{\psi}_k##, i.e.

$$\braket{\vec{e}_j|\psi}=\frac{1}{\sqrt{N}} \Bigl(\frac{1}{\sqrt{N}} \Bigl(e^{ia}+e^{2a} +\ldots+e^{Nia} \Bigr) \delta_{j1}+\frac{1}{\sqrt{N}} \Bigl( e^{2ia}+e^{4ia}+ \ldots +e^{2Nia} \Bigr)\delta_{j2}+ \ldots + \frac{1}{\sqrt{N}} \Bigl( e^{Nia}+e^{2Nia} + \ldots +e^{N^2ika} \Bigr)\delta_{jN}\Bigr)$$

$$\braket{\vec{e}_j|\psi}=\frac{1}{\sqrt{N}} \Bigl(\vec{\psi}_1 \delta_{j1}+\vec{\psi}_2 \delta_{j2}+ \ldots + \vec{\psi}_N \delta_{jN}\Bigr)$$

Now does it mean that if, for example, ##\vec{e}_j=\vec{e}_2##, that only the state ##\braket{\vec{e}_2|\psi}=\frac{1}{\sqrt{N}} \Bigl(\vec{\psi}_2 \delta_{22}\Bigr)## remains and the particle is at location 2?
 
Lambda96 said:
$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \delta_{jl}$$
Good. Consider the sum $$ \sum\limits_{l}^{} e^{ikal} \delta_{jl}$$Since ##\delta_{jl}## equals zero for any ##l \neq j##, all of the terms in the summation are zero except for one term. So, the sum reduces to one term that can be written in terms of ##k##, ##a##, and ##j##.
 
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Likes Lambda96 and vanhees71
Thanks TSny for your help 👍👍

Then I can write the term as follows

$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} \sum\limits_{l}^{}e^{ikal} \delta_{jl}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\Bigl( e^{iaj}+e^{2iaj}+e^{3iaj}+ \ldots +e^{Niaj} \Bigr)$$
 
Lambda96 said:
$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} \sum\limits_{l}^{}e^{ikal} \delta_{jl}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}$$
$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\Bigl( e^{iaj}+e^{2iaj}+e^{3iaj}+ \ldots +e^{Niaj} \Bigr)$$
Good, except the sum over ##k## does not go from ##k = 1## to ##k = N##. Recall that the values of ##k## are ##k = 2\pi n /L## for ##n = 0, 1, 2, ... , N-1##.
 
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Likes Lambda96 and SammyS
Thanks again for your help TSny👍👍 👍, also thanks for the hint with the index ##k## 👍

Could I then write the second line as follows?$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}$$

$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{n=0}^{N-1} e^{\frac{2i \pi n a j}{L}}$$
 
Lambda96 said:
Thanks again for your help TSny👍👍 👍, also thanks for the hint with the index ##k## 👍

Could I then write the second line as follows?$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}$$

$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{n=0}^{N-1} e^{\frac{2i \pi n a j}{L}}$$
Yes, that looks right.
 
Thanks for your help TSny 👍👍👍
 
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You’re very welcome.
 
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