An eigenstates, eigenvectors and eigenvalues question

pigletbear
Messages
2
Reaction score
0
Good evening :-)

I have an exam on Wednesday and am working through some past papers. My uni doesn't give the model answers out, and I have come a bit stuck with one question. I have done part one, but not sure where to go from here, would be great if someone could point me in the right direction:

S2) Show that the state vectors |Sx+> = \frac{1}{\sqrt{}2} times a 2x1 matrix (1,1) and |Sy+> = \frac{1}{\sqrt{}2} times a 2x1 matrix (1,-1) are eigenvectors of Sx = h/2 times a 2x2 matrix (0 1, 1 0) with respective eigenvalues plus and minus h/2...

This I can do by using |A - λI| = 0, finding the eigenvalues, then using A.v=λv and setting up simutaneous quations to find the eigenvalues

Part two... Of what operator is the state \frac{1}{sqrt{}2}[/itex]/(|S<sub>x</sub>+&gt; + |S<sub>y</sub>+&gt;) and eigenstate, and with what eigenvalue...<br /> <br /> Any help would be great and much appreciated
 
Physics news on Phys.org
pigletbear said:
Good evening :-)

I have an exam on Wednesday and am working through some past papers. My uni doesn't give the model answers out, and I have come a bit stuck with one question. I have done part one, but not sure where to go from here, would be great if someone could point me in the right direction:

S2) Show that the state vectors |Sx+> = \frac{1}{\sqrt{}2} times a 2x1 matrix (1,1) and |Sy+> = \frac{1}{\sqrt{}2} times a 2x1 matrix (1,-1) are eigenvectors of Sx = h/2 times a 2x2 matrix (0 1, 1 0) with respective eigenvalues plus and minus h/2...

This I can do by using |A - λI| = 0, finding the eigenvalues, then using A.v=λv and setting up simutaneous quations to find the eigenvalues

Part two... Of what operator is the state \frac{1}{sqrt{}2}[/itex]/(|S<sub>x</sub>+&gt; + |S<sub>y</sub>+&gt;) and eigenstate, and with what eigenvalue...<br /> <br /> Any help would be great and much appreciated
<br /> <br /> Hello!<br /> My suggestion is to try to explicitly add up the two states in matrix notation... <br /> The answer should then be obvious to you :)<br /> <br /> Edit:<br /> The answer you gave to the first question is right. Nevertheless it is more time consuming then it was necessary and time is precious during exams :) <br /> You have to <i>show</i> that those are egenvectors with given eigenvalue, so you could simply show that <br /> <br /> S_x |S_x ± \rangle = ± \frac{h}{2}|S_x ± \rangle, <br /> <br /> without solving the egenvalue equation <br /> <br /> |S_x-\lambda I|=0.<br /> <br /> Ilm
 
Last edited:
Ilmrak said:
Hello!
My suggestion is to try to explicitly add up the two states in matrix notation...
The answer should then be obvious to you :)

Edit:
The answer you gave to the first question is right. Nevertheless it is more time consuming then it was necessary and time is precious during exams :)
You have to show that those are egenvectors with given eigenvalue, so you could simply show that

S_x |S_x ± \rangle = ± \frac{h}{2}|S_x ± \rangle,

without solving the egenvalue equation
Ilm

|S_x-\lambda I|=0. is a lot easier and faster to solve ;)

about the second question, it works in the same way:

A(|Sx+> + |Sy+>) = λ (|Sx+> + |Sy+>)

=> (A-λI)(|Sx+> + |Sy+>) = 0

A-λI = 0; Fill in λ = 1/sqrt(2) and the matrix will be (1/sqrt(2) 0; 0 1/sqrt(2) )

It's been a while since I did this, I could be wrong ofcourse.. (and it looks a bit to easy, but hey?)edit: I think I'm wrong...
 
Dreak said:
|S_x-\lambda I|=0. is a lot easier and faster to solve ;)[...]

Yes it is easy but no, it isn't faster.
And if it were a bigger matrix (or worst a differential operator) it wouldn't be so easy to solve that equation while it would still be easy to let a matrix (or a differential operator) act on a vector (or a function).
To solve an equation is (almost) always more difficult than checking one solution ^^

Ilm
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top