Tag Info

New answers tagged

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

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 ...


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 ...


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 ...


0

Wavenumber k is the number of waves per metre. Frequency w is number of waves per second. The number w/k is the speed of the wave.


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.


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

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 ...


0

The question you ask is essentially: How to solve the measurement problem? As you can see from that article (although I wouldn't say it's a very good one), there are several approaches to either get a theory where no collapse occurs (thereby rendering the question futile) or to explain the collapse. So far, nothing has been so satisfactory as to be a ...


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

In theory, every example of wavefunction collapse should be explainable through a mechanism of normal time evolution of the wave function. The apparent collapse is only an illusion. People who agree with this idea like to talk about "decoherence", which is a fancy sounding word, but it doesn't really tell you anything. In fact, there is very little interest ...


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 $H$ ...


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 ...


0

@Suresh Doesn't the normalization condition follow directly from the fourier transform relation for delta function? I'm not clear about the need for box normalization and PBC. Sorry I'm writing this as an answer, I don't have enough reputation to comment. Thanks.


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 ...


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 ...


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 ...


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 ...


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 $\psi$: ...


0

Let's calculate Δx as the fellow with Curious Mind said. Shall we? But, for simplicity of formulas let me introduce the usual notation 2α^2 = σ^2, s.t. the Gaussian takes the form (1/πσ^2) ∫exp[−(1/2) (x/σ)^2] dx . Also for simplicity let's take k_0 = 0. To pass to k_0 ≠ 0 is not difficult. The standard deviation is σ, and introducing in the Gaussian x = ...


-1

In short, “a wave traveling in the $+x$ direction” has nothing to do with actual motion of a wave packet. In spite of some mathematical similarities, wave function isn’t anything physically like a gravity wave on water surface. In quantum “waves”, there is no water (or gas, other 3D continuum, string, or anything else that can convey a physical meaning to ...



Top 50 recent answers are included