MHB What Are the Eigenvalue Properties of Hermitian Matrices in Quantum Mechanics?

[bugatti79]

Hi Folks,

I am looking at Shankars Principles of Quantum Mechanics.

For Hermitian Matrices M^1, M^2, M^3, M^4 that obey

M^iM^j+M^jM^i=2 \delta^{ij}I, i,j=1...4

Show that eigenvalues of M^i are \pm1
Hint: Go to eigenbasis of M^i and use equation i=j. Not sure how to start this?
Should I consider a 2*2 Hermitian Matrix such as \begin{bmatrix}1 & -i\\ -i& 1\end{bmatrix} and evaluate the LHS?

 

[I like Serena]

Hi Folks,

I am looking at Shankars Principles of Quantum Mechanics.

For Hermitian Matrices M^1, M^2, M^3, M^4 that obey

M^iM^j+M^jM^i=2 \delta^{ij}I, i,j=1...4

Show that eigenvalues of M^i are \pm1
Hint: Go to eigenbasis of M^i and use equation i=j. Not sure how to start this?
Should I consider a 2*2 Hermitian Matrix such as \begin{bmatrix}1 & -i\\ -i& 1\end{bmatrix} and evaluate the LHS?

Hi bugatti!

Let's set $i=j$, then we get:
$$(M^i)^2 = I$$

If we have $M^i v = \lambda v$ then it follows that $(M^i)^2 v = \lambda^2 v$ and also that $(M^i)^2 v= I v = v$.
For which values of $\lambda$ would these equations be satisfied?

 

[bugatti79]

Hi bugatti!

Let's set $i=j$, then we get:
$$(M^i)^2 = I$$

If we have $M^i v = \lambda v$ then it follows that $(M^i)^2 v = \lambda^2 v$ and also that $(M^i)^2 v= I v = v$.
For which values of $\lambda$ would these equations be satisfied?

We would then have $$I v= \lambda^2 v$$
How do we work out the eigenvalues if we don't know the value of M..?
To me, the eigenbasis are determined by first finding the eigenvalues and eigenvectors. Not sure what he means by " going to the eigenbasis of M"!

 

[I like Serena]

We would then have $$I v= \lambda^2 v$$
How do we work out the eigenvalues if we don't know the value of M..?
To me, the eigenbasis are determined by first finding the eigenvalues and eigenvectors. Not sure what he means by " going to the eigenbasis of M"!

I don't understand either what is intended with: "going to the eigenbasis of M".
However, picking $i=j$ yields the eigenvalues directly using $I v= \lambda^2 v$.

 

[bugatti79]

I don't understand either what is intended with: "going to the eigenbasis of M".
However, picking $i=j$ yields the eigenvalues directly using $I v= \lambda^2 v$.

But i calculate \lambda to be the square root of the identity matrix which is the identity matrix to which its eigenvalues \lambda_1=\lambda_2=1 from the characteristic polynomial x^2-2x+1.
i don't see where \pm 1 comes from

 

[I like Serena]

But i calculate \lambda to be the square root of the identity matrix which is the identity matrix to which its eigenvalues \lambda_1=\lambda_2=1 from the characteristic polynomial x^2-2x+1.
i don't see where \pm 1 comes from

Those are the eigenvalues of the identity matrix, but that's not what that equation is for.
What we have, is:
$$v = \lambda^2 v$$
where $v$ is a non-zero vector (an eigenvector of $M^i$).
This can only be true if $\lambda^2 = 1$.
What do you get if you solve this?Alternatively, we can interpret your results, where we still need to apply a square root somehow, which you've left out.
So yes, the identity matrix has only eigenvalue 1.
But since we had $\lambda^2$ in the equation instead of $\lambda$, that means that $\lambda^2 = 1$.

 

[bugatti79]

Ok, I see it.
Thank you!