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Well, the problem is easy from the standpoint of chemistry. Yes, there are multiple isomers. So why are there different structures and different energy eigenvalues? The function you describe is high-dimensional. For even something "small" like water, we're talking about 3 atoms and 10 electrons. If we restrict ourselves to a Born-Oppenheimer picture and ...


1

I see you have some great answers already, but here is a different way of looking at it which might help you with regard to your comment above. This is linear variational theory, meaning you have to keep all non-linear parameters constant throughout the calculation. We wish to find values of the linear parameters $C_i$ that minimise the variational integral: ...


0

The simple product $\psi_1(x_1)\psi_1(x_2)$ is already symmetric under $x_1 \leftrightarrow x_2$. So it is not necessary to symmetrize the wave function by adding in the term you get when you switch $x_1$ and $x_2$.


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Let's run through the variational principle very quickly. The idea is that an arbitrary state $\psi$ can be decomposed into a sum of orthogonal energy eigenstates: $\left|\psi\right> = \sum c_n \left|\psi_n\right>$ where $\sum |c_n|^2 = 1$ and $H\left|\psi_n\right> = E_n \left|\psi_n\right>$ Then the expectation of the energy ...


3

I don't find this proof a good one, since the notation is messy and not very clear (not to say wrong). One proof can be given in a similar way to the one you posted in the link. Suppose the spectrum of $H$ is discrete and the set of eigenstates $\{|\phi_n\rangle\}$ constitutes an orthonormal basis with eigenvalues $E_n$, such that $E_0\leq E_1\leq ...


2

Let me first quickly answer what I believe will be a misunderstanding of your question: a pure quantum state has no Shannon entropy, in the sense it can be treated as a known point in the spectrum of an observable: you can think of this spectrum as an alphabet of symbols and knowing the pure quantum state is tantamount to knowing which symbol we have. A ...


2

While I am not versed in the wave functions of particles, I can add perhaps a little bit of intuition as to the approach you are taking to this process. What you are asking for sounds a lot like a common question in number theory and computer science: what is the Kolmogorov Complexity of a given piece of data? In other words, what is the most efficient way ...


1

The main point that although a pointwise convergent Fourier series of cosine modes is an even function $\psi(-x)=\psi(x)$, it does not have to be differentiable at $x=0$. A pointwise convergent infinite sum of differentiable functions is not necessarily a differentiable function. More generally, as OP already mentions, the wave function $\psi(x)$ is not ...


3

Once you consider all three fermions, their wave function must be completely antisymmetric which, in your case, means that it must be a multiple of $$ \frac{ |123\rangle+ |231\rangle+ |312\rangle- |132\rangle- |321\rangle- |213\rangle }{\sqrt{6}} $$ You may imagine that at the very beginning when you only know about the single fermion, it has the wave ...


2

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics. More generally, the dispersion relation tells us about the phase speed ...


0

Note that I didn't check your work leading up to the last integral. Assuming it is correct, first remove the $\cos(\omega_2-\omega_1)t$ term from the integral since it doesn't depend on $x$. You are left with evaluating $$ \int_{0}^{L/2} dx \ \sin^2\left(\frac{\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) = \frac{L}{\pi} ...


-1

What you want is that $Pr(right)-Pr(left)>0$. But $$Pr(right)-Pr(left)=-2\times\frac{96}{41\pi}\int^{L/2}_{0}\sin^2(\frac{\pi x}{L})\cos(\frac{\pi x}{L})\cos(\omega_2-\omega_1)tdx$$ Realize, that $\sin^2(\frac{\pi x}{L})\cos(\frac{\pi x}{L})>0$, for $x\in[0,L/2]$, and so is the integral. Hence what you want is simply the smallest $t$ such that ...


0

We are told that $\left|1\right>$ and $\left|2\right>$ are energy eigenstates, meaning they are eigenstates of the Hamiltonian $\hat{H}$, a Hermitian operator: ${\hat H}^\dagger = H$. Eigenstates of a Hermitian operator with different eigenvalues are orthogonal (see this): $\left<m|n\right> = \delta_{mn}$ (using Kronecker delta), assuming the ...


0

The choice of the constants $A$ and $B$ depends on the form of the solution. You could have denote one pair of constanst by $A$ and $B$, and the other by $C$ and $D$. A possible solution is $\psi (x)=A\sin(kx)$. In complex form, the sine is: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ If you've never proved this formula, try it using $e^{ix}=\cos x+i\sin x$. ...


1

I also don't quite understand what you mean by "imaginary $\psi$", but let me give you some general, more mathematically correct view. In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we ...


0

If $E < V\left(-\infty\right)$, $E<V\left(+\infty\right)$, and $E > V_{min}$ (necessary for $\Psi$ to be normalizable), then it is a bound state, and the spectrum will be discrete: $$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$ Otherwise -- if $E > V\left(-\infty\right)$ or $E > V\left(+\infty\right)$ -- it is a scattering ...


2

This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potential $$V(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}$$ It is not hard to check that $V(r)$ is a bounded ...


0

This is an interesting question because you have to start dealing with the atom surrounded by its environment. Atoms with electrons in states other than the lowest possible filled-up states, have lots of opportunities. They can emit photons (and thereby go to a lower energy state) in many possible directions, and that photon can interact with any number of ...


1

The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions $$ \begin{align} \psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\ k_n&=\frac{n\pi}{L}\;,\\ E_n&=\hbar \omega_n\;,\\ \omega_n&=\frac{\pi h n^2}{4L^2m}\;. \end{align} $$ Here, $|\alpha \rangle$ represents the spin state. The global fermionic ...


1

We all know that $E=k_E + V(x)$; $E$ = total energy, $k_E$ = kinetic energy, $V(x)$ = potential One reason why you may be confused by this is that the equation $$ E = \frac{1}{2}m\dot{x}^2 + V(x) $$ comes from classical mechanics. When we are solving the Schroedinger equation $$ H\Phi = E\Phi $$ with Hamiltonian operator $H$, the meaning of the ...


1

Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0. In other words when you measure this operator you will always get results which are larger or greater than zero. This "contradiction" is resolved by the fact that the potential is a ...


1

In essence you are mixing the notion of electron as a point particle, and quantum physics. While classical mech have us believe that electron are point-wise particles, Quantum mech actually forbids it, by the same Uncertainty principle you are invoking. Thus the electron has some dimensionality in the z axis as well as the y axis, I'll go ahead ad ...


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Why do electrons in an atom occupy only the stationary states? This isn't true. An electron in an atom can be in any superposition of states. This is one of the basic postulates of quantum mechanics: linearity. For example, say an atom has a ground state 1 and an excited state 2, and let's say we're able to prepare it in a pure state 2. It will decay ...


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One reason we focus on energy eigenstates is that atoms spend almost all of their time in an energy eigenstate, and their spectrum is a result of transitions between them. Another reason is pedagogical: to peel back the onion one layer at a time. But before too long, many courses do include examples of systems that are not in an energy eigenstate. One ...


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Answer to part 1: This is a common method of solving differential equations, employed by physicists to quickly extract a solution without having to make slow progress using more rigorous methods. The first step is to look for an asymptotic solution (i.e. the limit of large $\xi$ in this case), where the equation is easily solvable. Now, we know that this ...


3

The issue is with short-hand notation. The term $\psi^{n+1/2}_{j,k}$ being operated on by $D_x^2$ really means \begin{align} D_x^2\psi^{n+1/2}_{j,k}&=D_x^2\left[\frac12\left(\psi^{n+1}_{j,k}+\psi^n_{j,k}\right)\right]\\ &=\frac12D_x^2\psi^{n+1}_{j,k}+\frac12D_x^2\psi^{n_{j,k}} \\ &=\frac1{2\Delta ...


0

edit after comments: We have the experimental observation of fixed spectra coming from specific atoms. The same for nuclei, and both are stable in their ground state ( unless energetically disturbed or are unstable isotopes). Quantum mechanical solutions reflect these experimental observations and the spectra of atoms and nuclei have been fitted with ...


1

The paper addresses the case of a circular, finite quantum well with different electron effective masses inside and outside the well. The equation you point to is the result of a derivation from the time-independent Schrodinger equation, effective mass approximation and BenDaniel-Duke boundary conditions. It is also, however, dependent on the statement ...


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Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get $$\begin{align} \Psi(\theta) &= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\ &= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 ...


4

To explicitly verify this, one solves the problem for a box of finite depth $V_0$. If you additionally assume the wavefunction and its first derivative to be continuous across the potential step, the solution becomes a matter of Solve the Schrödinger equation in the distinct regions in- and outside of the box. Match $\phi$ and $\phi'$ at the potential ...


2

The 2nd way $$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$ will produce a complex result in general (in the example above it is will simply be zero), not having a physical measurement analog. The operator operates on some vector (either $\Psi$ or $\bar{\Psi}$), whereas the $|\Psi|^2$ is a simple real number.


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So I believe it's standard to place the operator inbetween the conjugate of the wavefunction and the wavefunction itself. For instance, $$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$ Yes, that is correct, and Is it wrong to do this? $$\langle p\rangle = ...


4

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob ...


5

It doesn't matter what sign you choose. Notice that since $|A|^2 = \frac{2}{a}$, you could even pick $A = \sqrt{\frac{2}{a}} e^{i\phi}$, so $A$ doesn't have to be real. The reason is that a wavefunction is only defined up to a global phase. The reason is that we calculate probabilites with $|\psi|^2$ and mean values of operators with $\int \psi^* \hat{O} ...


0

A Gaussian wave packet $\psi(r)=e^{-r^2/2a}$ actually satisfies $\triangle x \triangle p_x = \frac{\hbar}{2}$. As @luming points out, this is the ground state of the harmonic oscillator. In general a Gaussian isn't going to be an eigenstate of the Hamiltonian, but the lowest energy eigenstate should have the fewest number of oscillations, and so has the ...


0

Take it as you want, but this is the way I interpret the necessary existence of a ground state (at finite energy) for any bound system in quantum mechanics. The idea, in my view, consists in finding the minimum energy value of an hamiltonian of the form $H = \frac{\vec{p}^2}{2m} + V(\vec{x})$ under the statistical constraint that, say, $\Delta x \Delta p_x ...


2

The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ...


0

The ground state of a system is by definition the state of minimal energy, i.e. the system is located at the minimum point of the potential. Now, if we were in classical mechanics, this would mean that the system is at a stable fixed point. Of course in QM that is not possible since we have to satisfy the Heisenberg uncertainty. And so, I would say yes, ...



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