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New answers tagged schroedinger-equation

1

I personally (maybe wrongly) see this feature as an early sign of the $CT$ symmetry where $C$ is the charge conjugate symmetry operation and $T$ is the time-reversal symmetry operation. Having no explicit charge in your equation, the charge conjugate symmetry operation would be simply taking the complex conjugate of the wave function while the $T$ operation ...

2

It is the definition of complex number. Let's say $z=x+iy\quad \Rightarrow z^*=x-iy$ $z=x-iy\quad \Rightarrow z^*=x+iy$ In simple words, you just have to change the sign of the Imaginary part. The thing is that $\psi(x)$ it's a imaginary number, so it's conjugate it's just $\psi^*(x)$. If you have the $\psi(x)$ function, then you can change $i\to -i$ or ...

5

Schrodinger's equation cannot be derived. It was thought up using logical arguments and so far it has seemed to work experimentally. The equations is essentially a re-write up for energy conservation: $$E = T + V$$ Where $T$ is the Kinetic Energy and $V$ is the potential. However, to be more explicit we must work with operators (if you are unsure what ...

1

The time dependent Schrodinger equation is one of 5 (or 6) postulates of quantum mechanics. It is not proper to say that it is derived, unless you have a different set of postulates. for example, in the references below, the time dependent Schrodinger equation is the 5th postulate. http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html ...

3

From Feynman's lectures :) Where did we get that from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger. ADDITION Though a postulate such as Schrödinger equation cannot be proven, one can notice that in QM a state vector $\Psi$ is said to give the most complete description of a state of a system. So it is ...

4

You might be interested in this "elementary" derivation of the free particle Schroedinger equation from Maxwell's equations. It seems to be in the same spirit as Schroedinger's original reasoning. The niceness of this approach is that if you also include special relativity, it nets you both the free particle Schroedinger equation and its relativistic ...

1

Heisenberg's uncertainty principle is $$\Delta x \Delta p \geq \hbar/2.$$ Since the well is of width $L$, you have a measure for the uncertainty on the position $\Delta x$. Then assume the lowest possible value for $\Delta p$, i.e. the one for which the above inequality becomes an equality. Lastly, use $E = \dfrac{p^2}{2m}$ to find an expression for $E$. ...

1

In Mathematica, using the formula $\sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases}$ where $u_0=mL^2V_0/(2\hbar^2)$, we can first compute $u_0$: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since $u_0$ is typically on the order of an ...

2

First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such $\Psi(r,t)$, that fulfill this equation. The finite square well 1.0 fm wide means you have a potential $V(r)$ which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting $\Psi(r,t)$ ...

1

Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to ...

0

In general every linear combinations of a separable solution is still a solution (superposition principle), so you can take the simplest separable solution, put it in a linear combination with arbitrary coefficient (complex numbers, phases) and you obtained a solution of Sc. equation NON-separable.

0

You are deal with action variables so the integrations extrem are always give by the classical trajectory. So for the hidrogen atom in bound state they are 2 times the distance between R1 and R2.

0

How do I transform equation (1) to equation (2) plug $R(r)=u(r)/r$ into (1), you'll get (2) immediately, where $k(x)$ would the expression before $R(r)$ in the second term of (2) and what do I use for the bounds of integration in equation (4) to get the energy eigenvalue? $\int_0^\infty$

0

The bounds for r should still be the classical turning points, as you mentioned for the harmonic oscillator. Presumably you're in a bound state of Hydrogen, i.e. have an energy of the form $\frac{-13.6 eV}{n^2}$ for some integer n. The problem then reduces to finding the zeros of the equation $$\frac{-13.6 eV}{n^2} = -\frac{e}{r^2} - \frac{l(l+1) ... 0 The only thing that's really important is the differential eqaution. The situation is, outside the well, in both cases: \dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi Now it's foundamental notice that for bound states E<0 so we can write: E=-|E| and Sc. equation become: \dfrac{d^2 \psi}{dx^2}= + \frac{2m|E|}{\hbar^2} \psi So the usual way ... 1 First, note that \hat H_f has degenerate spectrum: it has equal eigenvalues for states with \left|k\right\rangle and for \left|-k\right\rangle. This in turn means that, in particular, the state \frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right) is also an eigenvector of \hat H_f. But in position representation it will look ... 0 Firstly, notice that if you take linear combinations of \psi_f you get \psi_b -- so the basis for both Hilbert spaces is the same. Though I'm not sure how rigorous the following statement is, I'm tempted to say that: The wavefunctions do not sense any region outside the potential well, where we have infinite potential. So as far of the states of our ... 0 Urgje gave you the answer. In its basic form (Schrödinger), the Hamiltonian is time-independent, therefore the general theory will tell you how to write the general solution of the Schrödinger equation as the sum/integral of the solutions of the spectral equation weighed by time-dependent exponentials. 8 Short version: In the infinite potential well, E \geq 0 (because V_{min}=0, and E \geq V_{min}). In your finite potential well, it sounds like you are looking for bound states, in which case E < 0, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ... 0 There are a variety of methods for determining the spectrum of a Hamiltonian system in one or more dimensions. For systems with small numbers of degrees of freedom, a direct matrix approach can be taken, often using some variant of the Sinc Discrete Variable Representation of the Hamiltonian. Depending on the size of the Sinc-DVR matrix, either direct ... 1 What you are proposing will work, it is essentially what is known as the shooting method for solving the eigenvalue problem. Note that the eigenfunction is defined up to a multiplicative constant so you can just set \epsilon=1 and there is only one parameter \epsilon' to vary in order to achieve a properly decaying solution at infinity. The shooting ... 0 Well, sure! The wavefunction, in polar coordinates, for the N^\text{th} Landau level is:$$\psi(r,\theta)=\frac{1}{\sqrt{\pi N!}}\Big(\frac{eB}{2\hbar}\Big)^{(N+1 )/2} r^N (e^{i\theta})^N \exp\big(-\frac{eB}{4\hbar}r^2\big)

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(a) It depends on what you mean by "variational methods". The stationary S. equation it is nothing but an elliptic equation, so you can use all the standard method appropriate for elliptic equations, existence of weak solutions and elliptic regularity (Weyl, Sobolev, Hoermander...). The temporal equation, to some extent is similar to the heat equation. At ...

-1

Energy would exhibit both positive as well as negative energy if it were a living entity. So first one must answer is time alive? To solve any equasion shouldn't you know the values of all propertys within it?Idetify the propertys first. Only then could you solve it.

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