Consider the eigenvalue equation:  $$\hat{Q}\Psi = q\Psi$$  where $q$ and $\Psi$ are eigenvalues and eigenfunctions of the hermitian operator $\hat{Q}$. If the spectrum of the hermitian operator is discrete, then the eigenfunctions lie in the Hilbert Space and constitute physically realizable states.  Why do discrete eigenvalues imply that their associated eigenfunctions are square-integrable (and hence live in the Hilbert Space)?
73092
These facts are taken for granted in a QM text I read. The purportedly guaranteed non-normalizability of eigenfunctions which correspond to a continuous eigenvalue spectrum is only partly justified by the author, who merely states that the non-normalizability is linked to the fact that such eigenfunctions do not tend to zero at infinity.  Not a very satisfying answer. What I'm really after is an explanation based in functional analysis. I believe there is a generalized result about inner products being finite for discrete spectra but infinite for continuous spectra.  Can anyone shed some light on this?
4403
Reading through David Tong lecture notes on QFT.  On pages 43-44, he recovers QM from QFT. See below link:    First the momentum and position operators are defined in terms of "integrals" and after considering states that are again defined in terms of integrals we see that  the ket states are indeed eigen states and the eigen values are therefore position and momentum 3-vectors.  What is not clear to me is the intermediate steps of calculations not shown in the lecture notes, in particular, the computation of integrals involving operators as their integrand, to obtain the desired results.
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I have for quite some time now tried to find a satisfactory answer to this, but I haven't yet. In perturbation theory, with small parameter $\lambda$, we expand the eigenstate as  $$| E \rangle=| E^{(0)} \rangle + \lambda | b \rangle + ...$$  Where $| E^{(0)} \rangle$ is an eigenstate of the unperturbed Hamiltonian. The problem is that when there is degeneracy, there is a choice of these eigenstates. I know the answer is that they are chosen to be eigenvectors of the perturbing Hamiltonian, but my question is why.
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Let us say I know that a given $N\times N$ matrix has all its eigenvalues as real, does it mean, it is hermitian. How do I prove (or disprove) that?
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A matrix is diagonalizable iff it has a basis of eigenvectors. Now, why is this satisfied in case of a real symmetric matrix ?
17594
In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being   $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$  (cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because   $$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$  (cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?   My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding   $$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$  (cf. second formula on p. 53).  References:   D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
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Let $A$ be a Hermitian square complex matrix of size $n$ then $\langle Ax,x \rangle=0$,   $\forall x \in \Bbb C^n$ if and only if $A=0$   I want to know if the condition that $A$ is Hermitian is required for the statement to still hold, please?  Thank you
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Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
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Let $V$ be an inner product space, finitely generated over $\mathbb{C}$, $T\in \operatorname{End}(V)$ that satisfies $TT^*=T^2$, show that $T$ is self-adjoint.    I know that $TT^*$ is positive so has positive square-root, thus the square root is positive definite. But I don't feel like this is a proof. Can anyone prove it without considering square root?
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I was wondering if anybody would explain (or please point me to a nice reference) as to why the ``discrete Laplacian" on a graph is actually called a Laplacian. Namely, how is it related to the standard Laplacian on $\mathbb{R}^n$ ? Is there a sense in which the discrete Laplacian on say, a square lattice, would converge to the standard Laplacian as the lattice spacing tends to $0$ ?
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Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') operators. The eigenvalues of the operator are the allowed values of the observable. Since Hermitian operators have a real spectrum, all is well.  However, there are non-Hermitian operators with real eigenvalues, too. Consider the real triangular matrix:  $$ \left( \begin{array}{ccc} 1 &amp; 0 &amp; 0 \\ 8 &amp; 4 &amp; 0 \\ 5 &amp; 9 &amp; 3 \end{array} \right)  $$  Obviously this matrix isn't Hermitian, but it does have real eigenvalues, as can be easily verified.  Why can't this matrix represent an observable in QM? What other properties do Hermitian matrices have, which (for example) triangular matrices lack, that makes them desirable for this purpose?
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Assume that we are working over a complex space $W$ of dimension $n$. When would an operator on this space have the same characteristic and minimal polynomial?    I think the easy case is when the operator has $n$ distinct eigenvalues, but what about if it is diagonalizable? Is that sufficient, or can there be cases (with repeated eigvals) when char poly doesn't equal min poly? What are the general conditions when the equality holds? Is it possible to define them without use of determinant? (I am working by Axler and he doesn't like it.)  Thanks.
109339
Consider a QFT consisting of a single, hermitian scalar field $\Phi$ on spacetime (say $\mathbb R^{3,1}$ for simplicity).  At each point $x$ in spacetime, $\Phi(x)$ is an observable in the sense that it is a hermitian operator (operator-valued distribution) on the Hilbert space of the theory, but is each such operator observable in a stronger, more physical sense?  Is there an experiment one could hypothetically perform to measure the value of such a field at a given spacetime point? This is one of those questions I glossed over while learning QFT, but now it's bugging me.  In particular, I think this point is central in preventing me from understanding certain basic assumptions in QFT such as microcausality which I also never really think about anymore.
62243
Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter.  The first part of his second postulate can be stated as:      Every dynamical variable may be represented by a Hermitian operator whose eigenvalues represent the result of carrying out a measurement of the value of the dynamical variable...   My question is: just how much can we deduce from saying that the some of the operators involved in Quantum Mechanics are Hermitian?  When I use the word Hermitian, I am referring to the property that $A=A^+$ where $A$ is an operator and $A^+$ is the adjoint of $A$.  Does an operator being Hermitian automatically imply that there will exist eigenfunctions and eigenvalues corresponding to this operator?  Does being Hermitian imply that, should there exist eigenfunctions corresponding to this operator, they will form a complete and orthonormal set?  That any wavefunction can be expanded in terms of these eigenfunction?  Just how much is hidden in the word "Hermitian"?
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For arbitrary non-diagonalizable square matrix $J$, can we always find a arbitrarily small perturbations matrix $\varepsilon A$ that $J+\varepsilon A$ is diagonalizable?  Using Jordan form as following, we can obtain that arbitrarily small pertubations matrix following a certain structure can make a matrix diagonalizable. But can we relax the form of the pertubations matrix?    Give any $J$, let $B$ be the Jordan form. That is, $J=U B U^{-1}$. For a pertubations matrix $\frac{1}{k} U\Delta U^{-1}$ where $\Delta$ is a diagonalizable matrix with different diagonal value, $J+\frac{1}{k} U\Delta U^{-1}$ is diagonalizable.   Can we relax the form of the pertubations matrix?
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For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar. $$  Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose $\hat{x}\psi=x\psi$ is the answer of the following question yes?  For the second one   $$(\hat{x}\hat{p}-\hat{p}\hat{x})\psi=x\hat{p}\psi-\hat{p}x\psi=i\hbar\psi, $$  but I don't see how can I say that $\hat{p}$ must be proportional to $\frac{\partial}{\partial x}$. I don't know if trying to see that $\hat{p}$ must satisfy the Leibniz rule and thus it should be proportional to the $x$ derivative could help. Or using the fact that $\hat{x}$ and $\hat{p}$ must be hermitian  Any hint will be appreciated.
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In Moore's introductory physics textbook Six Ideas that Shaped Physics, he describes a set of qualitative rules that first-year physics students can use to sketch energy eigenfunctions in a 1D quantum-mechanical potential.  Many of these are basically the WKB formalism in disguise;  for example, he introduces a notion of "local wavelength", and justifies the change in amplitude in terms of the classical particle spending more time there.  He also notes that the wavefunction must be "wave-like" in the classically allowed region, and "exponential-like" in the classically forbidden region.  However, there is one rule that he uses which seems to work for many (but not all) quantum potentials:     The $n$th excited state $\psi_n(x)$ of a particle in a 1D potential has $n$ extrema.   This is true for the particle in a box (either infinite or finite), the simple harmonic oscillator, the , and presumably a large number of other 1D quantum potentials.  It is not true, however, for a particle in a double well of finite depth;  the ground state, which has a symmetric wavefunction, has two maxima (one in each potential well) and one minimum (at the midpoint between the wells).  The following questions then arise:   Are there conditions can we place on $V(x)$ that guarantee the above quoted statement is true?  For example, is the statement true if $V(x)$ has only one minimum?  Is the statement true if the classically allowed region for any energy is a connected portion of $\mathbb{R}$?  (The second statement is slightly weaker than the first.) Can we generalize this statement so that it holds for any potential $V(x)$?  Perhaps there is a condition on the number of maxima and minima of $V(x)$ and $\psi_n(x)$ combined?   I suspect that if a statement along these lines can be made, it will come out of the orthogonality of the wavefunctions with respect to some inner product determined by the properties of the potential $V(x)$.  But I'm not well-enough versed in operator theory to come up with an easy argument about this.  I would also be interested in any interesting counterexamples to this claim that people can come up with.