# Tag Info

## New answers tagged wavefunction

1

The problem is considering an incoming right-mover for $x<0$ and asks how it scatters of a step potential into a reflected outgoing left-mover for $x<0$ and a transmitted outgoing right-mover for $x>0$. The last possibility -- an incoming left-mover for $x>0$ -- is not present in this scattering experiment. That's the answer to OP's question.

0

This is a scattering problem and is therefore somewhat different from the bounded solution problems (e.g. potential well, harmonic oscillator) you may have encountered. In this case we want to model a wave function that travels from the left to the potential barrier. Part of it is scattered back towards the left and part of it is transmitted through the ...

0

Note as x approaches +∞ the wavefunction vanishes. Apply $x=+∞$ boundary condition. $Ae^{-ik(+∞)}$+$Be^{ik(+∞)}$= 0 This will apply only and only if B=0!!! Please check on how to apply boundary conditions and understand the implied meaning

0

I have found the answer. A wave functional of a field $\Psi[\phi]$ can be sharply peaked at a particular function of $\phi$. And hence it will correspond to a classical field f. For example: $$\Psi[\phi] = \exp \left( -\int (\phi(x)-f(x))^2 dx^3 \right)$$ is maximum at $\Psi[f]$. Just like a normal wave function $\psi(x)= \exp(-(x-a)^2)$ can be sharply ...

0

Ok, So after alot of research I am attempting to answer my own question. Current state of the art research into this exact question was published in a paper by Daisy Williams, Xiaoyi Bao, and Liang Chen under the name: "Effects of polarization on stimulated Brillouin scattering in a birefringent optical fiber" Those who want to delve in, should with regard ...

1

This is quantum mechanics, my friend. The statement simply says that one hybridized orbital consists of many "pure" orbitals. In your first equation, one hybrid orbital has four pure orbitals $2s, 2p_x, 2p_y, 2p_z$. The coefficients in front of each term can be thought of as how much of one particular kind of pure orbital can be found in the final hybrid ...

3

Yes, hybridization is superposition. Assuming the basis functions are orthonormal, which we usually take to be the case, $\langle\phi_i\,|\,\phi_j\rangle = \delta_{i,j}$ the condition for normalization of $\psi$ is that the sum of the squares of the coefficients equal unity. More precisely, $$c_1^*c_1+c_2^*c_2+\cdots = 1$$ You should check the math to see ...

2

Yes, hybridization is just that the hybrid state $\psi$ is a superposition of the different orbital states $\phi_1,\dots,\phi_n$. Since the orbital states $\phi_i$ are assumed to be normalized ($\lvert\lvert \phi_i \rvert\rvert^2 = 1$) and orthogonal to each other, for the hybrid state to be normalized the squares of the coefficients must sum to $1$.

1

One thing that goes wrong here is that the Hamiltonian is no longer a self-adjoint operator acting on these singular wave-functions. Let's focus on the kinetic energy portion of the Hamiltonian: $$\hat K = - \partial_r^2 - {2 \over r} \partial_r = -{1 \over r^2} \partial_r r^2 \partial_r$$ To prove self-adjointness, consider the natural inner product on ...

0

I will elaborate on count_to_10's comment. Consider the following current $$j=\frac{i}{2m}(\Psi\frac{d}{dx}\Psi^*-\Psi^*\frac{d}{dx}\Psi)$$ where $m$ is the mass of the particle and I am working in units where $\hbar=1$. This is the famous probability current of QM. It is related to the probability density $\rho(x,t)=\psi\psi^*$ via ...

0

Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}\mathrm{d}x$$ show that if $\Psi(x,t)$ is normalized at time $t=0$, then the corresponding momentum space wave function $\Phi(p_x,t)$ is also normalized at time $t=0$. Good. By definition of the momentum space wave ...

0

To calculate $\langle\psi | \psi\rangle$ we have, using the expansion of the identity onto the momentum basis: $$\langle\psi | \psi\rangle = \int\textrm{d}p\,\langle\psi|p\rangle\langle p|\psi\rangle = \int\textrm{d}p |\psi(p)|^2$$ to be proven to be 1. Let us insert the identity as expanded onto $x$ and $x'$ $$... 0 In addition to what @DavePhD says, the Schrodinger model also calculates the angular momentum correctly and shows the angular momentum degeneracy of energy states. A similarity between the results is that the Bohr model orbital radii are equal to the mean radius, <\psi|r|\psi>, values of some of the angular momentum states. 2 It is true that the wave function \psi(x) is already not an observable, hence the statement about the group velocity is not needed to "falsify" the possible claims for \psi(x) to be physical. However, given a state |\psi\rangle \in \mathcal{H} there might still be a temptation to identify its wave function with a possible particle behaviour in terms of ... -1 I see here in the memoirs of Bryce De Witt that he says the Wheeler-de-Witt equation is WRONG and is only an approximation good for cosmology but not as a basis of quantum gravity. I will take that to mean the Wheeler-de-Witt equation is an approximation in the same way that the single-particle Shrodinger equation is an approximation to full quantum ... 1 Recall that states or wave-functions are only defined up to an overall phase, i.e. \psi(x) and e^{i \alpha(x)} \psi(x) are both wave-functions that describe the same state. The wave-function generically is a complex function of the form \psi = f(x) e^{i h(x)} where f(x) and h(x) are real functions. It is then often convenient to make a choice of ... 1 Your answer is perfectly fine. As you can see one can choose an abritrary phase \exp(i\phi) for c_n in the equation$$E_n + \frac{\hbar \omega}{2} = |c_n|^2$$and it will still hold. This relates to the fact that you can always choose an arbitrary phase for the eigenfunctions \psi_n. All physical observables (e.g. A_{nn} =\langle ... 2 There is a little confusion in your statements about the quantisation of the gravitational field, especially then at any time slice one can define the state of the universe as the probability of gravitons being at positions: is not correct. Let us start from the beginning instead. The standard way to quantise fields is to write down their path ... 1 My understanding of why Griffiths picked \frac{1}{\sqrt{ |x| }} as an upper bound for \Psi is from a dimensional analysis perspective. For example \Psi^*\Psi is a probability density. So \Psi^*\Psi for 1 dimensions must have units of \frac{1}{|distance|} in order for \int^\infty_{-\infty}\Psi^*\Psi\,dx=1 (for normalized \Psi). Thus \Psi must ... 0 Let's talk about 1d just to keep it as simple as possible. Consider the ground state of the simple harmonic oscillator. The energy eigenstates are each a Gaussian multiplied by a polynomial. The ground state is just a Gaussian. So each of those states goes to zero as x\rightarrow\pm\infty. And each of them is unit norm. And any finite linear combination ... 0 Let us first clarify the difference between a state and its wave function representation. A state |\psi\rangle is an element in a Hilbert (or equivalently) Fock space, whereas its wave function \psi(x)=\langle x |\psi\rangle is its representation onto the position (or momentum, respectively) basis. In quantum field theory things extend a little, ... 0 Plotting that probability distribution is what my wave function ψ(x) is. No the ψ(x) is a complex valued function. The probability density is given by ψ(x)ψ(x)^* , i.e. the complex conjugate squared with the original. For the second point, this P and its square roots are meaningless; there are no two independent probabilities; there is one ... 1 To me it seems you are a little confused on the basics, so I think going over some of the basics will do more to help you than trying to answer your question directly. As a result, first I am going to make a few statements about the basics of quantum mechanics, then I am going to comment on your two statements, and lastly I'll come to your question. OK, ... 1 To cut it short, the integral you need is (assuming \alpha>0):$$\underset{-\infty }{\overset{\infty }{\int }}{x}^{2}{e}^{-\alpha {x}^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi }{{\alpha }^{3}}}$$As suggested in the comments, it's one of the gaussian integrals. The mistake you made is a purely algebraic one, since you inserted -\infty into e^{-x^2} and got ... 1 Let's take an example. Suppose the wave function can be express as:$$\Psi=c_1\psi_1+c_2\psi_2$$If we add a global phase difference, say e^{i\varphi}:$$\Psi\to\Psi'=e^{i\varphi}(c_1\psi_1+c_2\psi_2)$$then, expectation value of x:$$\langle x\rangle _{\Psi'}=\langle \Psi'|x|\Psi'\rangle =e^{-i\varphi+i\varphi}\langle \Psi|x|\Psi\rangle =\langle ...

1

In nonrelativistic quantum mechanics the wavefunction is defined on configuration space. So for $n$ particles there is a $3n$ dimensional configuration space. And the quantum wavefunction is a function from $\mathbb R^{3n}$ into the complex numbers (or into a joint spin state if you have spin). There could be regions where the wave is zero or very small, ...

-3

As a quantum theorist, this area of physics perfectly intersects my area of knowledge. A simple Google search can lead you to my favourite site for philosophical inquiry. The Many-Worlds Interpretation (MWI) of quantum mechanics holds that there are many worlds which exist in parallel at the same space and time as our own. The existence of the other ...

3

You have the freedom to define the wavefunctions, such that one of the leading coefficients is one. Suppose you didn't do this, and instead defined: $$\psi(x') = C_1e^{ikx'}, x' > \frac{a}{2}$$ $$\psi(x') = C_2 e^{ikx'} + C_3 e^{-ikx'}, x' < \frac{a}{2}$$ Then, you could divide through by $C_1$ to give:  \psi(x') = e^{ikx'}, x' > ...

1

You could use a Wigner distribution which does generate positive numbers for regions of phase space that are large enough. But when operators don't commute it also means there is no experimental setup that can evolve them into a state that would give the same result repeatedly if performed again and again. So you've jumped over the whole fact that a so ...

1

Do you see an error in my steps? Yes, I see a conceptual error. So I'll talk about that. $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = \dfrac{\partial \Psi^*}{\partial t} \Psi + \Psi^* \dfrac{\partial \Psi}{\partial t}$ should be $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = ... 1 Now I just begin with double-slit experiment and then conclude that there is a probability wave for electron This is wrong. Firstly, the wave function is defined on configuration space, it isn't a wave in space. And secondly, it isn't using probability the way a mathematics textbook on probability uses the word probability. So it isn't a wave in space, ... 5 Well, there are many things you could do. You could: consider two gaussian beams (the linked article is for electrodynamics) apply some paraxial approximation (which would be more appropriate to treat electrons with a high forward momentum) do a cheap/cheesy symmetric point source approximation using Green's functions. I can do number three for you :) ... 1 How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves? Actually, it's better not to. In modern Quantum Physics the idea of electrons as standing waves is increasingly seen as no more than an analogy and not a very good one either. In some cases like this system it's a rather compelling one but even ... 0 How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves? It seems to me there is a confusion between a Bohr type model of atoms and molecules and the quantum mechanical framework with the orbitals. One can design an orbit of an electron as a standing wave classical solution and then one has to ... 3 No one thinks the wavefunction of nonrelativistic quantum mechanics is a wave in space and time. No one that wants to agree with observations that is. If someone is a realist about the wavefunction of nonrelativistic quantum mechanics then they start out by making a mathematical model by having the model include a wavefunction. Which when there are$n$... 0 Observables correspond to particular things you can do in the lab (or observe in nature). So let's first talk about something you can do in the lab. You can take a particle with spin and subject it to an inhomogeneous magnetic field. A particle with spin has a magnetic moment proportional to the spin and we know to Hamiltonian for a particle with a magnetic ... 3 To add to Yuggib's Answer, which I am in complete agreement with: I have never particularly liked the name "operator" for an "observable", because the former implies a mapping and, therefore, that the image$\hat{A}\,\psi$has a direct physical meaning. As in Yuggib's Answer, there is in general no direct physical meaning. Rather, an "observable", as I like ... 3 I do not think that the action$A\psi$has a direct physical meaning, when$A\$ is a generic observable. This is because the interpretation of a quantum system as a mathematical model yields the wavefunction and its corresponding Hilbert space as a sort of byproduct. In fact, the state may not always be a wavefunction: without entering too much into details, ...

Top 50 recent answers are included