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37

I) Well, one can identify a complex-valued observable with a normal operator $$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$ A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator. Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. ...

30

1. Definitions Let's consider the nondimensionalized Hamiltonian $$\hat H=\frac{\hat{p}^2}2-\frac1r.\tag1$$ Its standard eigenfunctions diagonalize operators $\hat H$, $\hat L_z$ and $\hat L^2$. Laplace-Runge-Lenz operator can be defined as $$\hat{\vec A}=\frac{\vec r}r-\frac12\left(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p}\right).\... 28 One problem with the given 3\times 3 matrix example is that the eigenspaces are not orthogonal. Thus it doesn't make sense to say that one has with 100% certainty measured the system to be in some eigenspace but not in the others, because there may be a non-zero overlap to a different eigenspace. One may prove^{1} that an operator is Hermitian if and ... 17 The expression you wrote for \sigma_z^1 + \sigma_z^2 is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces. The spin state of a single spin-\frac{1}{2} particle is ... 17 As a mathematical structure, the field of complex numbers does not admit an order relation which is an extension of the order we have in \mathbb{R}. This means that there is absolutely no way of saying if 5+3i is bigger or smaller than 5+6i for example. We just know it is not equal and we have to stop here. Therefore it is physically really hard (... 14 As far as most textbooks on (non-relativistic) quantum mechanics go, we do not consider the solution for n=0 because it gives us a trivial solution (and we interpret it to mean that there is no particle inside the box/well). However, if there were a ground state with zero energy for a square well potential, it would imply that (since the particle has ... 13 If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space? Yes, that's basically correct. If the value of the field at a point is observable, the eigenvalues of the operator representing it are the ... 13 A maximally classical state should have minimum and equally distributed uncertainty in X and P. In other words the uncertain in X should equal the uncertainty in P and this uncertainty should be as small as possible. This leads to$$ (X-\langle X\rangle)|\alpha\rangle=-i(P-\langle P\rangle)|\alpha\rangle $$or if we rearrange the equation$$ \alpha|\...

12

Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body, and wrt. to an arbitrary choice of Cartesian coordinates $x$, $y$, and $z$), then the diagonal elements $I_{xx}$, $I_{yy}$, and $I_{zz}$ of the inertia tensor satisfy the triangle inequality, $$I_{xx} +I_{yy} ~\geq~ I_{zz}, \qquad I_{yy} +I_{zz} ... 12 Imaginary numbers can be represented as pairs of real numbers. You can also make a device, which mixes the measurement outcomes of two reals on hardware level to produce complex "amplitude" and "phase" as outcomes, which you further might call as measuring a complex number. More generally, any measurement is eventually reading off the values on the ... 12 It is sufficient to prove that the vectors |n\rangle form a Hilbert basis of L^2(\mathbb R). This fact cannot be completely established by using the ladder operators. To prove that the span of the afore-mentioned vectors is dense in the Hilbert space, one should write down the explicit expression of the wavefunctions of the said vectors recognizing that ... 12 Yes. There is a simple explanation. The classical harmonic oscillator has a well defined frequency \omega, independent of initial conditions. This can only happen if the quantum system has precisely equally spaced energies with gap \hbar \omega. The reasons for this is that motion in a quantum system can only happen if more than one energy level is ... 11 Let us look at a one-dimensional example: Recall that f(x) = \dot{x}, so f encodes the time evolution of x. If f < 0, then x will move to the left. If f > 0, then x will move to the right. If f = 0, x will not move at all, this is why f(x_0) = 0 is the equilibrium condition. Now, look what happens if you perturb the equilibria ... 11 Write an arbitrary state as$$|\Psi\rangle = \sum_{n=0}^{\infty} c_n |n\rangle \,.$$Now apply the raising operator$$ \begin{align} a^\dagger |\Psi\rangle &= a^\dagger \sum_{n=0}^{\infty} c_n |n\rangle \\ &= \sum_{n=0}^{\infty} c_n \sqrt{n+1} |n+1\rangle \\ &= \sum_{n=1}^{\infty} c_{n-1} \sqrt{n} |n\rangle \end{align} $$If |\Psi\rangle is ... 11 In answer to the title question: no, you can't always decompose an L_2 function in terms of only the bound spectrum of hydrogen. This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions |\chi_{E,l,m}⟩ of the ... 11 The precise theorem is that suppose you take the zeroes of the Riemann zeta function as 1/2+i \gamma_n. Then you look at the normalized differences between neighboring zeroes. If you pretend these were drawn from a random variable and calculate the correlations, you get the same result as if you just took a random Hermitian matrix like the kind that is ... 10 Several reasons: Orthogonal functions arise naturally in the study of Sturm-Liouville theory which includes many classical and quantum system mathematical models; More generally, it is the class of normal operators (and an important special case self adjoint operators) which the spectral theorem most readily works and is most complete for. The eigenvectors ... 10 What is your Hilbert space? In L^2(\mathbb R) your eigenfunction would have infinite norm. If you dealt instead with a bounded set L^2([a,b]), your operator would not be Hermitian unless you impose suitable boundary conditions to discard boundary terms. These boundary conditions, however, would rule out your candidate eigenvector! 10 As \phi(f) and \pi(f), which are self adjoint, satisfy the same commutation relations as X and P, the closure of the space generated by polynomials of the former pair of operators applied to \lvert 0\rangle is isomorphic to L^2(\mathbb R). Therefore the spectrum of \phi(f) and \pi(f), is purely continuous and coincides to \mathbb R and ... 10 A coherent state is, amongst other interesting things, an eigenstate of the annihilation operator. It is not an eigenstate of the creation operator; hence, I'm not sure this "eigenvalue problem" makes much sense. This is easy to realize. You can quickly see that \langle0|a^\dagger|\alpha\rangle=0, whereas \langle0|\alpha\rangle\neq0. If you really want ... 10 To add to Innisfree's correct answer, I'd like to emphasize something that the OP does not seem to know and that is that the creation operator has no eigenvectors (nor, therefore, eigenvalues). It is easy to see this: write a general state as a row vector (\psi_0,\,\psi_1,\,\cdots) of superposition weights for the number states |0\rangle,\,|1\rangle,\,\... 10 By saying X|x\rangle = \lambda |x\rangle and then integrating over x without allowing for the fact that \lambda depends on x, you're essentially saying that the action of X on all |x\rangle states is the same, so that X is a constant:$$ X = X\int|x\rangle\langle x|\mathrm dx = \int X|x\rangle\langle x|\mathrm dx = \int \lambda|x\rangle\langle ...

10

The language of the set-piece you quote leaves a lot to be desired. For someone with a reasonable command of QM it's clear what the intention was, but it is still phrased in an ambiguous way and the writer is at fault for any confusion that ensues. To be clear, the quote is asking you to show that the given wavefunction is an eigenfunction of the ...

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I’m not sure what you mean by “$a$ and $b$ are unique” but clearly if $A=UBU^\dagger$ and $U$ is unitary, $A$ and $B$ have the same eigenvalues but it doesn’t mean $U$ doesn’t do anything. For instance, the Pauli matrices $\sigma_{x,y,z}$ all have the same eigenvalues, are related by a unitary transformation $U$, but are certainly different. The ...

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You are right. $|l=0,m=0\rangle$ (i.e., $Y^0_0$ ) is a simultaneous eigenvector of $L^2$ and $L_x,L_y,L_z$ with eigenvalue $0$. What is impossible is the existence of a common basis of eigenvectors of $L_x$ and $L_y$: the two operators would be diagonal with respect to that common basis and therefore they would commute in contrast to the commutation ...

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This is a nice exercise! It can be completely solved with relatively elementary mathematical techniques. Let us start by assuming $\hbar:=1$, defining the formally selfadjoint differential operator over smooth functions $$D := \frac{1}{2}(XP +PX) = -i \left(x \frac{d}{dx}+\frac{1}{2}I\right)\:,$$ and proving that it admits a unique selfadjoint extension ...

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No. It is a good thought but no. In short, you are correct that we can prepare states whose average energy takes on any value between the lowest and highest energy state. But this is different than creating a state which is "between" the energy eigenstates. Let $|n\rangle$ be the $n$th energy eigenstate. This means that we have that: \hat{H}|n\rangle = ...

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The spectrum of a bounded linear operator $A$ is by definition the set of numbers $\lambda$ where $A-\lambda$ is not invertible. In the finite dimensional case, this means $A-\lambda$ is neither injective nor surjective, and the former statement is just a fancy way of saying that there exists an eigenvector; as this eigenvector is by definition a part of the ...

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