0
$\begingroup$

In page 32 of R. Shankar's Principles of Quantum Mechanics is given the eigenvalue problem:

We begin by rewriting Eq. (1.8.2)as $$(\Omega - \omega I)|V\rangle =|0\rangle \tag{1.8.3}$$ Opening both sides with $(\Omega - \omega I)^{-1}$, assuming it exists, we get $$|V\rangle =(\Omega - \omega I)^{-1}|0\rangle \tag{1.8.4}$$ Now any finite operator(an operator with finite matrix elements) acting on the null vector can only give us a null vector. It therefore seems that in asking for a nonzero eigenvector $|V\rangle$, we are trying to get something for nothing out of Eq. (1.8.4). This is impossible. It follows that our assumption that the operator $(\Omega - \omega I)^{-1}$ exists(as a finite operator) is false. So we ask when this situation will obtain. Basic matrix theory tells us (see Appendix $\mathrm A. 1$) that the inverse of any matrix $M$ is given by $$M^{-1}= \frac{\mathrm{cofactor} M}{\mathrm{det}(M)} \tag{1.8.5}$$ Now the cofactor of $M$ is finite if $M$ is. Thus what we need is the vanishing of the determinant. The condition for nonzero eigenvectors is therefore $$\mathrm{det}(\Omega - \omega I) =0 \tag{1.8.6}$$

Why is a vanishing determinant necessary to find nonzero eigenvectors?

$\endgroup$
4
  • 3
    $\begingroup$ Isn't this just stating that if the determinant of a matrix is zero, it doesn't have an inverse? $\endgroup$
    – Tyberius
    May 17, 2018 at 16:20
  • $\begingroup$ You can define the determinant of a linear operator in an infinite dimensional Hilbert space only in some suitable situation (e.g. trace class operators that are bigger than one). You should use proper operator theory to study eigenvectors and eigenvalues. $\endgroup$
    – yuggib
    May 17, 2018 at 16:28
  • 1
    $\begingroup$ @Tyberius, you are right. I reason that $\Omega - \omega I$, which is finite, cannot have an inverse, for we do not want a zero eigenvector. For this to happen, the det should be zero. We then find the eigenvalues of the operator whose eigenvector is $|V\rangle$ $\endgroup$
    – R004
    May 17, 2018 at 17:56
  • $\begingroup$ You can also think of the determinant as being the product of the eigenvalues of an operator. Equation 1.8.3 tells you that the operator $\Omega- \omega I$ has an eigenvalue of zero, thus the determinant must be zero. $\endgroup$
    – user111476
    May 17, 2018 at 18:18

2 Answers 2

3
$\begingroup$

The operator $(\Omega - \omega I)$ can be seen as a matrix. But to be inversible, a matrix needs its determinant to be non-zero. Why does it need to be non-zero ?

Let's take a matrix A, which we'll assume is inversible and let's demonstrate that its determinant can't be zero : $$ det(A) \cdot det (A^{-1}) = det(A \cdot A^{-1}) = det (I) = 1 $$ (with the property of multiplication of determinants) thus $det(A)$ can't be zero.

$\endgroup$
1
$\begingroup$

For, for notational clarity, you want to understand $\vert 0\rangle$ as the $0$ vector.

Start with $$ (\Omega-\omega I)\vert V\rangle=\vert 0\rangle \, . \tag{1} $$ If $(\Omega-\omega I)$ as a matrix has an inverse - call it $(\Omega-\omega I)^{-1}$ as you do, then multiply both sides of (1) by this inverse to get $$ (\Omega-\omega I)^{-1}(\Omega-\omega I)\vert V\rangle = \vert V\rangle =\vert 0\rangle \tag{2} $$ so that your eigenvector $\vert V\rangle $ is the $\vert 0\rangle$ vector. If you do not want $\vert V\rangle$ to be $\vert 0\rangle$, the assumption that $(\Omega-\omega I)$ has an inverse must be false. For this to be false, it must be that $$ \hbox{Det}\left(\Omega-\omega I\right)=0\, . $$ Alternatively, if the determinant is NOT $0$, then the inverse will exists and you find from (2) that $\vert V\rangle$ is the $0$ vector.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.