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This operator has been studied in the paper by Miloslav Znojil, PT-symmetric square well, Phys. Lett. A, 285 (2001) 7-10, and then by the same and many other authors. However I did not find out whethe …

Your formula for the first Dirichlet eigenvalue cannot be correct: for $a=1/\sqrt{2}$, the first eigenvalue is $5$. Indeed, the eigenfunction $$y(x)=e^{-x^2/2}(2x^2-1)$$ is positive on $(-1/\sqrt{2},1 …

This is easier to explain on the finite dimensional example (though the idea is the same). The spectral theorem says that for every self-adjoint (same as Hermitian, in finite dimensional case) operato …

Yes. Let the size of your matrix be $n$. Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$. Assume without loss of generality that thi …

It depends on what you mean by "analytical result". Even quartic oscillator $q(x)=x^4+ax^2$ was studied VERY much. See, for example, Bender, Carl M.; Wu, Tai Tsun Anharmonic oscillator. Phys. Rev. (2 …

A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to th …

Yes, this follows from the Maximin characterization of eigenvalues of symmetric matrices. If $A$ is an $n\times n$ symmetric matrix, form the Rayleigh ratio $R(x)=(x,Ax)/(x,x),$ where $(.,.)$ is the s …

Under your assumption on the potential, there is indeed such a unique solution, (I assume you mean $x\to+\infty$ in your boundary condition. This is proved by reducing your differential equation to an …

Consider two operators $L_1w=-w''+U(x)w$ with eigenvalues $\lambda_k$ and $L_2w=-w''+V(x)w$ with eigenvalues $\mu_k$. If $U\geq V$ then $\lambda_k\geq \mu_k$. To prove this consider the Rayleigh ratio …

If $V(x)\to+\infty$ as $x\to\pm\infty$ then the spectrum is discrete, $\lambda_n\to+\infty$, and the ground state (the eigenfunction corresponding to the smallest eigenvalue) does not change sign. Mor …

The answer is yes. Suppose your $V$ equals to a polynomial $P$ when $|x|$ is large. Then there is a constant $c$ such that we have $P-c<V<P+c$, which implies that $\lambda_k^\prime-c<\lambda_k<\lambda …

Solutions of the equation you wrote are not polynomials. Legendre polynomials are solutions of the equation $$(1-x^2)y''-2xy'+l(l+1)y=0.$$ This equation is written in the same article you refer to. It …

The number of bound states is indeed finite. This can be proved as follows. First of all, for a bound state your eigenvalue $-k^2$ must be real. This is because your operator with real $u$ and zero bo …

It will be better if you write the definition of your matrix in a more readable way. From what you wrote, it seems that your matrix satisfies $M(k,k+1)M(k+1,k)\geq 0$. With this condition, all eigenva …

A good reference is Whittaker Watson, vol. 2.