# Tag Info

1

Now, to me these boundary conditions simply mean that at $a$ the wavelength must be zero. It can be "after" a full length or a half length. My question is, why can't there be a half integer (say 2,5; 4,5) number of wave lengths in any of the box dimensions? You really need to solve the time-independent SchrÃ¶dinger equation for that: $$\Big[\frac{-\hbar^2}{... 2 Aside from the fact that everything about the electron in an atom must be understood in the sense of quantum statistics, as already pointed out in the other answer and the comments, there is still a definite sense in which electrons do "circle around the nucleus" in (stationary) eigenstates of well-defined angular momentum. Think of electron eigenfunctions ... 3 This is one of the mysteries of quantum mechanics. If you could measure the velocity of an electron, you would get a non-zero value. But what you can't do is use that velocity to predict where to find it next. The act of measurement disturbs the electron in an essential way. One popular interpretation of quantum mechanics is a statistical one. This says ... 1 Someone might give a more precise answer in terms of group theory but I'll give it a go anyway ; feel free to edit my post. Instead of considering the case of an odd number of fermions, one can consider just a single spin 1/2 - fermion to discuss 2 \pi rotations. Spin 1/2 is a representation of dimension 2 of the rotation group, which is called the ... 2 The radius of the wavepacket is unrelated to the radius of the particle. For instance you could have a particle of zero size (as the electron is suspected) but whose probability of being found somewhere has a certain size. For instance a wave packet: The wave packet only gives you the probability of finding a particle at that place (that is, more likely ... 2 For question 1, it comes down to probability. I have two distinguishable particles, a and b. The probability to find particle a at x_1 is$$P_a (x_1)=\int dx_1 \Psi_a(x_1) \Psi_a^*(x_1),$$and we have a similar expression for particle b at x_2. The probability to find particle a at x_1 and particle b at x_2 is just the product of ... 1 If the state of two particles is the tensor product of the two single particle states, then the wave function of the two particles is the product of the two single particle wave functions. For indistinguishable particles it is an experimental fact that the final state must be either symmetric or antisymmetric with respect to the exchange of the two ... 4 There's no analytic proof, but numerical evidence suggests that if you know that the Hamiltonian is local, and it satisfies the Eigenstate Thermalization Hypothesis (which most local Hamiltonians do), then you can extract the entire Hamiltonian from a single excited eigenstate, though not from the ground state: https://arxiv.org/abs/1503.00729. 19 IF you know that your Hamiltonian is of the form$$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state \Psi_0(\mathbf r) obeys$$ \hat H\Psi_0(\mathbf r) =...

3

If unknown part of Hamiltonian is potential $V({\bf{r}})$, then you can write a stationary Schrodinger equation and figure out what the potential should be.

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Assume for simplicity that all the operators are bounded. If you know the wave function $\psi$ associated with the ground state of the unknown Hamiltonian $H$, then $H$ has the form $$H = E_0|\psi\rangle\langle\psi| \oplus K$$ where $K$ is another Hamiltonian defined on a subspace of the original Hilbert space of co-dimension 1, and $E_0$ is the energy of ...

1

First of all, physical descriptions usually cannot be proven in a mathematically sense. Of course they should make sense and therefore mathematically sound. In particular when it comes to particle and antiparticles which is a concept of relativistic quantum field theory (QFT), actually a wavefunction as it exists in quantum mechanics no longer makes sense. ...

2

$$A\sin(k_iL)=De^{-qL}$$ $$Ak_i\cos(k_iL)=-Dqe^{-qL}$$ $$k_i\cot(k_iL)=-q$$ Insert the values for $k_i$ and $q$: $$[2m(V_1-E_i)/\hbar^2]^{1/2}\cot[2m(V_1-E_i)/\hbar]^{1/2}L=-[2mE_i/\hbar^2]^{1/2}$$ The allowed energy levels ($E_i<V_2$) for bound states can be determined by numerical solution of that equation. But particles with energy $E>V_2$ can ...

2

Qualitatively, the wave functions of the bound states in a triangular potential well like the one you described, look like this: For $x<-a$, $\psi=0$ because of the infinite potential in that region. Where the wave function crosses the potential line, quantum tunnelling occurs and $\psi \to 0$. For particle energies above $V_0$, no bound states can ...

36

An animation is worth a million words:

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This will be a purely mathematical treatment. It needs to be combined with some practical playing around to really "get" it. Traveling wave Let's start with the description of a harmonic traveling wave in one-dimension. Here "harmonic" just means the mathematical form of the wave is sinusiodal in both time and space. For concreteness we'll using talk ...

5

how is a standing wave related to the atomic orbit (It is my understanding that the atomic orbit is both a mathematical function that describes the probability of an electron being at a certain place, but it is also the image of this function in terms of real space, i.e. the actual 3 dimensional volume around the nucleus that a particular electron calls "...

2

A standing wave is basically two opposing waves of equal amplitude, as shown in the diagram below (where n is a positive integer): You can see this more clearly if you look at the top line where n=3, and follow it as it goes down, up, down. That's wave one. Then, if you look at the bottom line in the same case as it goes up, down, up, that is the opposing ...

0

Atoms. We sometimes use somewhat loose language when we speak of the wave function of the electron. We typically either have chosen the center of mass frame, or pretended that the nuclei are fixed (as you point out). It's the atom that has energy levels. We can get away with the loose language because if we really were able to fix the positions of ...

2

The time-independent Schroedinger equation for the hydrogen atom is $$H \ \Psi(\vec r_N, \vec r_e) = E \ \Psi(\vec r_N, \vec r_e)$$ where $\vec r_N$ is the position of the nucleus, $\vec r_e$ the position of the electron and $$H = \frac{p_N^2}{2 m_N} + \frac{p_e^2}{2 m_e} - \frac{e^2}{4 \pi \epsilon_0 \mid \vec r_N - \vec r_e\mid} \, .$$ We usually use ...

1

Let's say you measured the position of the electron exactly. While you might think that the electron has lost all it's quantum properties since it is localized in space and the wavefunction has collapsed, this is not true. This is because the collapse of the wavefunction depends intrinsically on the measurement that you are performing. Let me clarify that ...

1

The most probable radius is found by maximising probability per unit radius, whereas $|\psi^2|$ gives us probability per unit volume. To find the conversion factor from one to the other, we need to ask how much volume is there per unit radius near a radius $r$? The answer to this is $4 \pi r^2$.

1

You can think of an MPS as being built up by objects with three indices. How to easily represent such an object? We can think of this as a matrix where each entry is a vector (in particular the vector will be a vector in the on-site Hilbert space, so e.g. for a spin 1/2 system it will be of the form $\alpha |\uparrow \rangle + \beta |\downarrow \rangle$). ...

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Preliminaries: How do we define 'localized?' For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^... 2 In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual ... 3 To solve your problem exactly, you would have to solve the SchrÃ¶dinger equation $$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$ where$\Psi (\vec r_1 \dots \vec r_N,t)$is the wave function of the$N$particles and $$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$ where$u_{ij}...

0

There is a difference between pressure and displacement: pressure is "absolute", displacement has a direction. With speakers on the left and right, pointing to each other, the pressure wave from the right will be the mirror image of the one from the left, the displacement wave is not only mirrored but also changes sign (mirrored by the x-axis). Taking that ...

0

I suppose you mean that the gas is contained in a magic box. Otherwise the walls become part of the system, exchanging momentum/energy with the 'particles'. I have no answer for you; I don't know. What I do know is that none of the particle-particle collisions can be characterized other than by using a probability distribution. Common sense demands that ...

2

Quantum effects appear if the concentration of particles satisfies, $$\frac{N}{V} \ge n_q$$ where $N$ is the number of particles, $V$ is the volume, and $n_q$ is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum ...

3

It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $\Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then ...

0

The behaviour of the molecules in your gedanken experiment can be approached by using decoherence. But I do not believe you can get a definitive answer until somebody makes a full scale simulation (or until some expert's answer can make a formal proof of what really happens, but I am not skilled to do that). The decoherence effects can be argued ...

0

It doesn't make sense to apply periodic boundary conditions to the infinite square well, since the solutions in each well will be independent from each other (it is not possible for a particle to tunnel between different wells). Electrons will be strictly confined within the infinite square well of width $L$. When you consider the finite well, there is a ...

0

To make a comment a bit more formal, a state $\Psi(x,t)=\psi(x)e^{-iEt}$ with a nonzero imaginary component of energy $E=E_0+i\Gamma$ will tend to either decay or grow exponentially in norm. This can be seen from the explicit probability density, $$|\Psi(x,t)|^2=|\psi(x)|^2e^{2\Gamma t},$$ or from its evolution equation, \begin{align} i\frac{\partial}{\...

1

Let $\:\mathbf{L}\:$ the vector angular momentum operator $$\mathbf{L} =\left( L_{x},L_{y},L_{z} \right) \tag{01}$$ The eigenfunction $| \psi_{n\ell m}\rangle\:$ is a common eigenstate of $\:H,L^{2},L_{z}$ and more precisely for our case is an eigenstate of $\:L_{z}\:$ with eigenvalue $\:m\:$, so  L_{z}| \psi_{n\ell m}\rangle =m\hbar|\psi_{n\ell ...

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