# Tag Info

## Hot answers tagged schrodinger-equation

22

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". The following extensive exposition by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this ...

15

Good question! First you need to know that parity refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two "kinds" of parity: If $f(x) = f(-x)$, we say the function $f$ has even parity If $f(x) = -f(-x)$, we say the function $f$ has odd parity Of course, for most ...

12

Maybe in principle, not really in practice. Most people talking casually will use the interchangeably. If you want a rule that makes sense though, "states" are what we call the objects that live in a hilbert space (we also call them vectors, thus eigenvectors!) so the generic, "coordinate-independent" ket should be the eigenstate. It's projection onto a ...

10

1.) No. All the calculations one does in elementary quantum mechanics courses are at zero temperature. If they were at a finite temperature, you could never reliably say what quantum mechanical state your system is in; it would always be in an ensemble of different states. Since the ground-state wavefunction and ground-state density is not a 2d surface, you ...

10

Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator: $$\hat{H} \psi = E \psi.$$ As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then $$\hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ... 9 The issue is that the assumptions are fluid, so there aren't axioms that are agreed upon. Of course Schrodinger didn't just wake up with the Schrodinger equation in his head, he had a reasoning, but the assumptions in that reasoning were the old quantum theory and the deBroglie relation, along with Hamiltonian idea that mechanics is the limit of wave-motion. ... 8 This is fundamentally no more difficult than understanding how quantum mechanics describes particle motion using plane waves. If you have a delocalized wavefunction \exp(ipx) it describes a particle moving to the right with velocity p/m. But such a particle is already everywhere at once, and only superpositions of such states are actually moving in time. ... 8 The Harmonic oscillator has the same spectrum as a weaker harmonic oscillator with a hard wall at x=0. LATER EDIT: I see that I have to be more explicit--- the potentials V(x)= 2x^2 - 2 (x>0) V(x)= x^2 - 3 and (x<0) V(x)= \infty have the exact same spectrum. 8 The difference is due to the fact that solid harmonics are not spherical harmonic. So, equation (2) and the more conventional equation from Griffith are equations for different functions \phi. The Schrodinger eq. (1)$$-\frac{1}{2r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r}\psi\right) + \frac{\hat{L}^2}{2r^2}\psi + V\psi ~=~ E\psi ...

7

The solid harmonics were explained by Misha, so let me fill in the rest of the details. Laplacian operator is given by $\Delta = \partial_x^2 + \partial_y^2 + \partial_z^2$. First suppose there we are only interested in the radial part. Using chain rule (and letting $D \equiv \partial_r$, as in your references), we can write $$\partial_x = r_{,x} D, \quad ... 7 First let me start by saying that the N-body problem in classical mechanics is not computationally difficult to approximate a solution to. It is simply that in general there is not a closed form analytic solution, which is why we must rely on numerics. For quantum mechanics, however, the problem is much harder. This is because in quantum mechanics, the ... 7 The most general Schrödinger equation has total derivatives$$ i\hbar \frac{d}{dt}|\psi\rangle = \hat H |\psi\rangle $$because the state vector |\psi\rangle only depends on one variable, t. It's a complicated object that knows about the probability of anything in the given state, but this is hidden "inside" the state vector. However, if you rewrite the ... 7 The infinitesimal probability for the electron to be in the volume dV around a point (r,\theta,\phi)\leftrightarrow (x,y,z) is given by$$ dP = dV\cdot |\psi(x,y,z)|^2 = dV\cdot |R(r)|^2\cdot |Y_{lm}(\theta,\phi)|^2 =\dots$$as you can see if you substitute your Ansatz for the wave function. However, the infinitesimal volume dV=dx\cdot dy\cdot dz may ... 6 I suppose the walls of your well are located at x=0 and x=a. It is easier to shift the coordianate to y=x-a/2 so that the walls are located at y=\pm a/2. Your wave function is then$$ \psi(y)=\frac6{a^3}\Bigl(\frac{a^2}4-y^2\Bigr). $$Since the wave function is even in y, it is clear why the probability is zero for all even n, since the ... 6 You certainly couldn't recover quantum effects with a classical treatment of that Lagrangian. If you wanted to recover quantum mechanics from the field Lagrangian you've written, you could either restrict your focus to the single particle sector of Fock space or consider a worldline treatment. To read more about the latter, look up Siegel's online QFT book ... 6 One of my favorite non-trivial, exact many-body ground states is the solution of a very specific spin-1 magnetic insulator in 1D, with a hamiltonian$$H_{AKLT}=\sum_{\langle ij\rangle}\vec{S}_i \cdot \vec{S}_j + \frac{1}{3}(\vec{S}_i \cdot \vec{S}_j)^2$$It turns out that you can construct the ground state by looking at the spin-1 operators as a projection ... 6 The random potential is a model for small fluctuations in the local energy of an electron, when there are defects or impurities that raise the energy of an electron at certain spots by a little bit, and lower it at other spots. This is a good model, although it doesn't look like one at first, because the quantum mechanical solution is not sensitive to the ... 6 It's because when$$V(x,y,z) = V_x(x) + V_y(y) + V_z(z),$$(I guess that your extra identity V(x,y,z)=V(z) is a mistake), we also have$$ H = H_x + H_y + H_z$$because H = (\vec p)^2 / 2m + V(x,y,z)  and (\vec p)^2 = p_x^2+p_y^2+p_z^2 decomposes to three pieces as well. One may also see that the terms such as H_x\equiv p_x^2/2m+V_x(x) commute with ... 6 I) OP's potential$$V(x)~=~a^x~=~e^{bx}, \qquad b~:=~\ln a ~\in~ \mathbb{R}, $$is the so-called Liouville potential. There are no (discrete) bound states. In scattering theory, an incoming wave at x=-\infty gets reflected by the so-called "Liouville wall", and returns to x=-\infty. This potential is used in e.g. Liouville theory, which is ... 6 In mathematics, there is a complete symmetry between +i and -i. Both the imaginary unit and the minus imaginary unit obey$$ i^2 = (-i)^2 = -1 $$The exchange of i and -i is known as the {\mathbb Z}_2 automorphism group of the complex numbers {\mathbb C}. When you introduce the complex numbers for the first time, it's a complete convention ... 6 What you write is the time-dependent Schrödinger equation. This is not the equation of a true wave. He postulated the equation using a heuristic approach and some ideas/analogies from optics, and he believed on the existence of a true wave. However, the correct interpretation of \Psi was given by Born: \Psi is an unobservable function, whose complex ... 5 What you've written down is the spatial part of the electron wavefunction. The spin state is not included. The full wavefunction of the electron involves both the spatial part and the spin part. Sometimes in quantum mechanics books the full electron wavefunction is written as the tensor product of the spatial and spinor parts, sometimes you'll just see it ... 5 The answer is fairly simple -- classical N-body problem has its solution in 6N 1D functions of time, quantum N-body problem has its solution in one complex function, but 3N-dimensional (not counting spin and similar stuff). Then, there is no wonder why one can find analytical solutions only for trivial problems or at least make N huge and escape into ... 5 the wave function describing an electron in a p-state (L=1) with m=0 indeed vanishes in the z=0 plane because the spherical harmonic Y_{10} is proportional to \cos(\theta) which vanishes in that plane - even at the very origin. This wave function's squared absolute value describes the probability density that the electron is at a given point. This ... 5 Are you looking for a proof? If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors): We start by differentiating the definition of the probability with respect to time only:$$ \frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ ...

5

The physical observable is not the wavefunction, but its integral over a finite area. In spherical coordinates, this is: $P({\vec x})=\int dr\, d\theta\, d\phi r^{2}\sin\theta \psi^{*}\psi$ This integrand is manifestly finite at $r=0$, even if $R(r)$ has a $\frac{1}{r}$ divergance.

5

I like the previous answer, but I thought you might want something less abstract. To simulate the Schrodinger equation, replace space by a grid of points, and the wavefunction values with a wavefunction value on each point of the grid. Then replace: $$\nabla^2 \psi$$ in the Hamiltonian with $$(\sum_e \psi(x+e) ) - N \psi(x)$$ Where e runs over the ...

5

Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form $|\Psi \rangle$ only depends on time perhaps it is more correct to use the last one, but it is matter of taste. In order to get your first equation (a wave equation), you must project on $\langle x|$: H(P=-i\hbar ...

5

Relativistic effects are those that disappear in the non-relativistic approximation $1/c\to 0$, usually small corrections to the non-relativistic approximate results that are proportional to $1/c^2$ or higher powers of the inverse speed of light. Let me correct a typo: "cannot account for GR" should have read "cannot account for the special theory of ...

5

No, you are wrong. Particular for the following statement: If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius x? The probability density at any points is given by $|\psi(r,\theta,\phi)|^2$. Certainly, the probability is for any region $V$ ...

Only top voted, non community-wiki answers of a minimum length are eligible