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6

The very minimum that a wavefunction needs to satisfy to be physically acceptable is that it be square-integrable; that is, that its $L_2$ norm, $$ \int |\psi(x)|^2\mathrm d x, $$ be finite. This rules out functions like $\sin(x)$, which have nonzero amplitude all the way into infinity, and functions like $1/x$ and $\tan(x)$, which have non-integrable ...


4

Your idea that a quantum fluctuation created the universe is a misinterpretation of the suggestions that I have heard. Explaining why requires introducing a few ideas, so bear with me while I do this. The object we think of as the universe is made up of two bits: a manifold equipped with a metric = spacetime some matter/energy The manifold and metric ...


4

No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian. The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous ...


3

Any Hilbert space $\mathcal{H}$ with the notion of unitary time evolution also possesses the notion of Hamiltonian. If $\mathcal{U}(t) : \mathcal{H} \to \mathcal{H}$ is the time evolution operator for every $t \in \mathbb{R}$, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element $H$ ...


3

If you are talking about the time independent Schrödinger equation, it's not a trivial question as it may seem, as the comments suggest. I will restrict the answer to the one-dimensional case, since multiply connected domains in higher dimensions give some additional problems. Not all functions $\psi$ that are solutions of the equation ...


3

This question is about Guassian wave-packet propagation and the corresponding Green's function in ordinary quantum mechanics. Assuming $\hbar=m=1$ for simplicity, consider the solution (with its initial condition) to the following Schrodinger equation: $$i\partial_t G=-\partial^2_x G \\ G(t=0,x)=\delta(x) $$ Now assume the Fourier ansatz for $\psi$: ...


3

Technically, $\omega^2/1^2-k^2/c^2=0$ is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form $\omega^2/a^2 - k^2/b^2 = 1$ for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than $\omega = ck$, like e.g. plasmas.


2

The freespace dispersion equation is $\omega^2 = k^2\,c^2$ and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation $$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1}$$ which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...


2

I agree in full with Marty Green except the explanations of chemistry in which I was unable to follow so well (that doesn't say that I disagree with them). But, let me put the things in short. The collapse is a phenomenon that is supposed to occur when a quantum object comes in contact with a quantum system. For instance, a quantum particle falls on a ...


1

I am going to answer this in a hurry because the question is on the edge of being closed. Quantum mechanics isn't just about "wavefunctions", it is also about "observables". An observable is something like: energy, position, momentum... i.e. it includes all the properties of classical physics. The wavefunction (or state vector or quantum state) is the ...


1

#What is a wavefunction? A wave function is a mathematical solution of one of the basic quantum mechanical equations: Schrodinger, Klein Gordon, Dirac. By the postulates of quantum mechanics the square of this wavefunction gives the probability of finding the system under study when looking at (x,y,z,t) or (p_x, p_y, p_z,E) or similar four vector spaces. ...


1

Strictly speaking a photon cannot be localized and the single particle "wavefunction" (as well as it's corresponding position operator $\hat{r}$) only exists in an approximate sense. The reason for this is quantum electrodynamics (QED), which is the theory that contains photons, is a quantum field theory (QFT) rather than the (non-relativistic) quantum ...


1

This seems like a reasonable homework-like question, so I'll provide a hint. Realize that, for $x>0$, both the simple and semi oscillators have the same potential $V(x)$. For the simple oscillator, draw out the first few of the energy eigenfunctions $\psi_n(x)$. Now for the semi-harmonic oscillator, think about what the boundary condition on any ...


1

The formal analogy between a mode of the radiation field and a particle in a harmonic potential stems from the fact that both systems have the Hamiltonian (in appropriate units) $$ H = \frac{1}{2}P^2 + \frac{1}{2}\omega^2 X^2,$$ where the variables $X$ and $P$ obey canonical commutation relations $[X,P] = \mathrm{i}\hbar$. For the radiation field, these ...



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