Quantum Mechanics: Eigenvaules, and orthogonality

[Baggio]

Hey, I've been trying to solve this problem it sounds simple but i don't know where to start:

If [tex]\phi_{1}[/tex] and [tex]\phi_{2}[/tex] are normalised, have the same eigenvalue and obey [tex]\int \phi_{1}*\phi_{2}d\tau = c[/tex] find the linear combination that is normalised and orthogonal to [tex]\phi_{1}[/tex]

Thanks

 

[Baggio]

NB: * denotes a complex conjugate,

Thanks

 

[M98Ranger]

for reaching out! It sounds like you are working with some concepts from quantum mechanics, specifically eigenvectors and orthogonality. Eigenvectors are special vectors that, when operated on by a linear operator, only change by a scalar factor. In quantum mechanics, operators correspond to physical observables, such as position or momentum. Eigenvalues are the corresponding scalar factors that represent the possible values of the observable.

In your problem, you are given two normalized eigenvectors, \phi_{1} and \phi_{2}, with the same eigenvalue. This means that they are both possible states of the same observable. The condition \int \phi_{1}*\phi_{2}d\tau = c is known as the orthogonality condition, which means that the two eigenvectors are perpendicular to each other in the vector space.

To find the linear combination that is normalized and orthogonal to \phi_{1}, we can use the Gram-Schmidt process. This process takes a set of linearly independent vectors and constructs an orthogonal basis from them.

First, we start by defining a new vector \psi = \phi_{2} - \frac{\int \phi_{1}*\phi_{2}d\tau}{\int \phi_{1}*\phi_{1}d\tau}\phi_{1}. This vector is orthogonal to \phi_{1} by construction, and we can easily verify that it is also normalized.

Next, we can use the Gram-Schmidt process to construct a new vector \chi that is orthogonal to both \phi_{1} and \psi. This is done by subtracting the projection of \psi onto \phi_{1} from \psi. The resulting vector is then normalized and orthogonal to both \phi_{1} and \psi.

The process can be repeated to construct additional vectors that are orthogonal to all previous vectors, until we have a complete orthogonal basis. In your case, since we only need one vector, we can stop after finding \chi.

To summarize, the linear combination that is normalized and orthogonal to \phi_{1} is given by \chi = \phi_{2} - \frac{\int \phi_{1}*\phi_{2}d\tau}{\int \phi_{1}*\phi_{1}d\tau}\phi_{1}. I hope this helps you get started on solving your problem!