19

No. If we view the Schrödinger equation simply as a second order differential equation, then no, it doesn't have a general solution that we can find. Let me add that, from the mathematical standpoint, the equation alone is not even sufficient to have a well posed problem; it needs to be supplemented by the boundary conditions. The solutions for the same type ...


12

For a few potential functions $V(x)$, there are exact analytical solutions. For a general potential, there aren’t, so one must use either numerical techniques or analytic approximations.


11

I don't know why (or if) people originally expected Hartree-Fock to work as well as it does, but after thinking about it for a while, I'm personally a little less surprised by it. It seems surprising at first because most wavefunctions are not Slater determinants, but in view of the constraints that I'll derive below, finding any wavefunction that does ...


10

The general approach is that for Schrödinger equations where the potential is separable (in the sense that $V(x,y,z) = V_1(x) + V_2(y) + V_3(z)$) then there exists a basis of hamiltonian eigenfunctions which are separable (in the sense that $\psi(x,y,z) = \phi(x)\chi(y)\xi(z)$). However, in general, there are also non-separable eigenfunctions of the ...


9

The Schrödinger equation is just another way of writing the conservation of energy, right? Wrong. The original motivation for Erwin Schrödinger can be read in his 1926 paper An Undulatory Theory of the Mechanics of Atoms and Molecules (behind paywall, but freely readable copies can easily be found on the web). Namely, the ansatz is the Hamilton-Jacobi ...


6

Why is Schrodinger equation taught while it does not describe an electron? This question is of the same order as asking: "Why is Newtonian gravity still taught everywhere since it has been seen that General Relativity is the underlying theory of gravity" and thus should explain everything? There are even observations in the solar system. Physics ...


5

There are two equations commonly called the Schrödinger equation. Most physicists I know would prefer to call*: $$i\hbar \frac{d}{dt}|\psi\rangle = \hat{H}|\psi\rangle$$ 'the' Schrödinger equation. This is also called the 'time dependent' Schrödinger equation. It is the quantum mechanical replacement for Newton's second law: $$ m \frac{d^2}{dt^2}\vec{r} = F(\...


5

The answer is no. You are mixing up domains of your functions, which is why you are getting such a result. There is a very large difference between a three-dimensional potential that depends only on $x$ and a proper one-dimensional potential. Recall that a function is defined by stating domains and then a rule. The rule may be the same, but the domains ...


4

There is no general solution. But this is not only due to different possible potential functions, but also due to the boundary conditions. No differential equation can be solved without boundary conditions, and those can vary according to the problem. Moreover the case distinction $E<0$ and $E>0$ leads to very different solutions. The first leads to ...


4

I think you are confusing the treatment of relativistic (spin-$\frac{1}{2}$ particles) electrons as compared to the non-relativistic case. The Schrodinger equation can perfectly describe the properties of the non-relativistic electron. The Dirac equation describes the interactions of relativistic electrons (and other spin-$\frac{1}{2}$ particles).


4

Schrodinger's equation in general is not an equation about energy conservation. The equation you have on the board is specifically the time independent Schrodinger equation, which already assumes stationary states whose energy do not change. In general, we actually have (in the position basis) $$-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf r,t)+V(\mathbf r,t)\Psi(...


4

Well in QM conservation of energy is violated right?(nuclear explosions, nuclear energy, particle accelerators, the sun, other stars) None of these things violate the conservation of energy. They merely change energy from one form (e.g. mass) to another form (e.g. electromagnetic). In all these cases the energy is the same before and after the interaction. ...


4

To show this, we just need to show that $\partial\psi/\partial y$ and $\partial\psi/\partial z$ are zero. No, this is wrong. These don't need to be zero. Instead, you can solve the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}(x,y,z,t)= -\frac{\hbar^2}{2m}\nabla^2\psi(x,y,z,t)+\mathcal{V}(x)\psi(x,y,z,t).$$ by separation of variables with ...


3

The time-independent Schrodinger equation is in the class of real second-order linear ordinary differential equations covered by 19th-century Sturm-Liouville theory. The eigenfunctions of such equations are orthogonal with respect to an appropriate inner product on the function space. So one does not have to prove orthogonality for each potential one is ...


3

The eigenstates $\Psi_n$ of the Harmonic Oscillator are the Hermite polynomials (with a Gaussian weight) which form a complete set, and so you can write any initial state $$\Psi(x,0) = \sum_{n=0}^\infty c_n \Psi_n(x),$$ and solve for the $c_n$s. Once you've done this, you just need to tack on the time dependence for each eigenstate (of the form $e^{-iE_n/\...


2

Or should I use... No! You should not use flakey wrong expressions simply because they might get you to the right known answer by cancelling errors. You should do the FT correctly, since you know it's wrong, as, for t =0, its inverse FT you start with $$\Psi(x,0) = \frac{1}{\sqrt {2\pi\hbar}} \int_{-\infty}^{\infty} \Phi(p) ~ e^{i\frac{px}{\hbar}} \,dp$$ ...


2

Essential knowledge about the Schrödinger-equation SE (I limit my post to the time-independent equation) is that it is an eigen-value equation. I assume that you know the basics of linear algebra so if you take a, for instance a 2x2 matrix $A$, an eigen-value problem can be set up: Let's assume $$A= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\...


2

Sometimes the expansions are not obvious. For example The harmonic oscillator time-dependent Schr"odinger equation $$ i\partial_t \psi = -\frac 12 \partial^2_x \psi +\frac 12 \omega^2 x^2 \psi $$ has a ``breathing'' solution $$ \psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \...


2

It seems that it is quite useful to notice that the last term in the Hamiltonian reduces, $$K(pq+qp)=K\left[pq+(i+pq)\right]=2Kpq+iK,$$ with help of $[q,p]=i$ with $\hbar=1$. Then, we have $$H=\frac{1}{2}(p^2+q^2)+2Kpq+iK.$$ The first terms can be rewritten with help of ladder operators, $$p=i\sqrt{\frac{\omega}{2}}(a^{\dagger}-a),\quad q=\sqrt{\frac{1}{2\...


2

The reason for the traditional method of pulling out the factor of $e^{-x^2/2}$ and only then seeking a series solution is that, just given a power series in $x$, it is hard to determine how its sum behaves as $|x|\to \infty$. When the series terminates, as it does for the Hermite polynomials, then it is easy to see that the wavefunction goes to zero at ...


1

It should be noted that the most famous analytic solution can be acquired for $V(x) \propto x^2$, which is the harmonic oscillator. For $V(x) \propto x^3$ you already need perturbation theory.


1

On place to look for information is by reading about Rabi oscillations, which arise when a two-level system is driven by a near-resonant electromagnetic field. For some situations one could find solutions in terms of wave functions that are exact or nearly exact. More generally, the way to approach this problem is by considering a joint wave function of the ...


1

Schrödinger’s equation is a differential equation which describes the evolution of the quantum state vector of a system over time. So one of the real axes of the graph is probably time. The other two axes are probably a representation of one component of the state vector in the complex plane. Each component of a quantum state vector is a complex number, so ...


1

A Bogoliubov transformation to new operators $\hat b$ and $\hat b^\dagger$ \begin{align} \hat b= u\hat a+v\hat a^\dagger\, ,\qquad \hat b^\dagger u^*\hat a^\dagger +v^*\hat a \end{align} with $\vert u\vert^2-\vert v\vert^2=1$ and suitably chosen $u$ and $v$ will bring $H$ to a diagonal form. Alternatively, this Hamiltonian is expressible in terms of ...


1

This is not a solution, but multiple suggestions for attacking the problem: Brutal force The equation is likely reducible to a hypergeometric equation, perhaps even something simpler, like parabolic cylinder functions or confluent hypergeometric. The way to start is to try to reduce it to one of the canonical forms by writing ψ(q)=f(q)ϕ(q) and choosing f(q) ...


1

In fact, it is the other way around. It is true that the behavior of matter, be it massive or massless, is governed by wave equations, such as Maxwell, Schroedinger, Klein-Gordon and Dirac. However, these equations describe the statistical behavior of matter, while matter itself consists of discrete particles.


1

It won't affect the measurement, due to the fact, that identity operator commutes with everything. In Schroedinger picture: $$ \langle \psi(t) | A | \psi(t) \rangle = \langle \psi | e^{-i (H - \lambda I) t} A e^{i (H - \lambda I) t} | \psi \rangle = $$ $$ = e^{-i \lambda t }\langle \psi | e^{-i H t} A e^{i H t} | \psi \rangle e^{i \lambda t} = \langle \...


1

Let $|\Psi \rangle$ be some state in which the system is. Then the energy expectation value is: $$\langle \Psi|(H-\lambda I)|\Psi \rangle = \langle \Psi|H|\Psi \rangle - \lambda.$$ Note that $|\Psi \rangle$ is arbitrary so we see that this just corresponds to an overall shift of the energy. For comparing energy levels this does not matter but for energy ...


1

Yes. Provided that the resulting wavefunction is normalizable. The point is that that, in polar coordinates the radial part of the Laplacian is a singular point of the equation at $r=0$. Depending on the form of the potential, such singular points can be in Weyl's limit point or limit circle case. In the latter case there is a one-parameter family of ...


1

So a complex number $a+bi$ acts upon the 2D space $(x,y)\leftrightarrow x+yi$ by a matrix we could write as$$\begin{bmatrix}ax-by\\bx+ay\end{bmatrix}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix},$$ and in fact all of the strange operations of complex numbers are actually very straightforward operations of matrix addition and ...


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