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What you call an operator theory is usually called the Heisenberg picture of quantum mechanics. What you call a wave function theory is usually called the Schroedinger picture of quantum mechanics. It is well-known that for every quantum mechanical model, the Heisenberg picture and the Schroedinger picture are fully equivalent through a dual description, ...


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The Schrödinger equation cannot be derived from classical physics. There are various consistency checks and motivations, such as its consistency with conservation of energy, but it is not derived from those considerations. However, that the Schrödinger equation conserves energy is built in when one knows that the Hamiltonian is the energy operator since $$ ...


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This is more of a comment than an answer, but I can't fit this into the amount of characters; Writing a quick bit of code, it looks to me like there's not much wrong with the method: The numerical and the analytical solution go on top of one another. N = 256 T = 256*128 L = 1. dt = 0.000001 x = linspace(0., N-1, N)*L/N psix = exp(1j*2*pi*x) psik = ...


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Quantum mechanics is not derived from classical mechanics or energy conservation, but there are "jumping off points" in classical mechanics that may serve to answer your question. If you study classical mechanics at a sufficiently advanced level you will discover the Hamiltonian formalism. The Hamiltonian for an isolated system with only conservative ...


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This subtlety is related to the fact that the momentum operator $\hat{P}$ (unlike the Hamiltonian $\hat{H}=\frac{\hat{P}^2}{2m}$) has no eigenfunctions compatible with the Dirichlet boundary conditions, and $\hat{P}$ is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, ...


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Quantum mechanics is a bit more than the Schrödinger equation. In particular, it says that all states evolve in time as given by the Schrödinger equation - there is the Hamiltonian operator $H$ and every time-dependent state $\lvert\psi(t)\rangle$ fulfills $\partial_t\lvert\psi(t)\rangle = -\mathrm{i}H\lvert\psi(t)\rangle$. In contrast, the potential ...


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Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...


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Let's make things a little more fun. How can we use the conservation of energy equation to derive Schrödinger equation in QM? Let's say we know the system's Hilbert space $\mathcal H$ and we know how to define a Hamiltonian $H:\mathcal H \rightarrow \mathcal H$ whose average value $\langle \psi | H | \psi\rangle$ provides the average energy in state ...


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I personally find most notation in the Heisenberg picture totally unsatisfactory for the exact same reason as you, its hard to look at the notation and not want to see a time dependent operator as a function from time to a matrix (or to an operator on Hilbert Space, or such). But it's similar to thermodynamics, a thermodynamic variable could be written as a ...


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In general, what we use for periodic boundary conditions is defined by $Y(0)=Y(L)$ and $\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition $Y(0) = Y(L)$ I believe is not sufficient to impose conditions on $k$. This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the ...


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Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2.



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