7
You are not wrong, but it is worth noting that the same thing is true of any momentum eigenstate (or closely related unbound eigenstate of a Hamiltonian with a potential well in it). Explicitly
$$
-i\hbar\frac{\partial}{\partial x} \psi(x) = p\,\psi(x)
$$
then
$$
\psi = A e^{i \frac{p}{\hbar}x}
$$
which is not normalisable either. This means that we can ...
6
Well, for continuous spectra, say the free particle, $E=\frac{\hbar^2k^2}{2m}$, you do it al the time, in transforming integral measures, $dE=2\frac{\hbar^2}{2m} kdk $.
You are basically asking: "when can I analytically continue discrete spectra in-between discrete (~integral) indices"? The answer would, of course, depend on the context, but most ...
6
If you want to be formal, the function $\psi : \mathbb{R}\to \mathcal{H}, t\mapsto \lvert\psi(t)\rangle$ needs to be understood as a function between Banach spaces (every Hilbert space is in particular a Banach space). The correct notion of derivative is then the Fréchet derivative.
Note that this vector-valued function is much easier to differentiate ...
4
As the electronic Hartree potential is accounting for the average electrostatic repulsion between the electrons, it is indeed completely spurious for a one electron system, as there should be no electron-electron interaction whatsoever.
And yes, using an exchange-correlation potential that exactly cancels the Hartree potential is the perfect choice for a ...
4
I think that the problem is that the square root of the Laplacian is a non-local opertor and non-locality is usully regarded as a bad thing in physics. The long range nature shows up in the general expression
$$
(-\nabla^2)^s f(x)\equiv \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \int_{{\mathbb R}^n} d^ny \frac{\left\{f(x)-f(y)\right\}}{|x-y|^{...
3
1) We are in the Schrodinger picture. Physical states are labelled by a state vector $|\psi(t)\rangle$, which evolves in time as the state evolves in time. The time $t$ is a parameter-it labels different state configurations. This is a 'ket'-it stores all possible information about the system, at the time $t$.
2) We have, by definition, $$\frac{d}{dt}|\psi(...
3
You are dividing the $x$-coordinate into three intervals. Within each interval the function is linear, and so it is trivially continuous there. The only places where discontinuities may arise is at the boundaries where the function changes form. To finish checking continuity you need to individually look at these boundaries. For example, if the function is $...
3
You have $A sin(ra) \pm B cos(ra) = 0$ ; This implies that $Asin(ra) = 0$ and $Bcos(ra) = 0$. The trivial solution to these two equations is $A=B=0$.
For a non-trivial solution at least one of $A$ and $B$ have to be non-zero. Hence, we can choose : $A = 0$ , $cos(ra) = 0$ , or , $B=0$ , $sin(ra) = 0$ .
If we want both $A\neq 0$ and $B\neq 0$, ...
2
The resonances' wavenumbers $k_n$ should obey the Fabry-Perot condition
$$
k_n = \frac{2\pi n}{2L}, \quad n\in {\mathbb Z}.
$$
The widths of the resonances will depend on $k$ because the reflectivity of the individual $\delta$ potentials are $k$ dependent --- but the change with $k$ should be smooth. Your irregularity suggest that you have made an error. ...
2
I would like to add two points to the accepted answer:
If you use the periodic boundary conditions trick to normalize the momentum eigenstates, then all the momentum eigenstates become normalizable, including the zero momentum eigenstate.
The discussion of tunneling in QM books is usually within the quasi-classical approximation, which breaks down when the ...
2
The Euler-Lagrange (EL) equations are not affected by total derivative terms, cf. e.g. this Phys.SE post.
In OP's concrete example the Lagrangian density (2) is preferred as it is manifestly real. See also this related Phys.SE post.
2
You don't use $\psi(s) = \psi(s + n L)$ for $n$ integer, because the ring always has length $L$, its length has nothing to do with $n$. Instead you impose the condition $\psi(s) = \psi(s + L)$, which implies that
$$\phi(L) - \phi(0) = 2 \pi n$$
which $n$ is the energy level. Then you get
$$n h = \int_0^L \sqrt{2 m (E_n - V(x))} \, dx$$
which is a typical ...
2
A Hilbert space is, by definition, an inner product space (meaning that it is also a normed space) and it is Cauchy Complete. Completeness means that all the normal definitions of limiting procedures go through as usual (with no more than minor notational changes).
2
Just integrate by parts a few times to get it in the desired form. For the first term in the identity,
$$\int d\mathbf{r} \, (\nabla^2 \psi^*) \psi = - \int d\mathbf{r} \, (\nabla \psi^*) \cdot (\nabla \psi) = \int d\mathbf{r} \, \psi^* \nabla^2 \psi.$$
For the third term in the identity,
$$- 2 \int d \mathbf{r} (\nabla \psi^*) \cdot (\nabla \psi) = 2 \int ...
1
The state of the particle, which was previously a single energy eigenstate of the old box, is now equal to some superposition of the different energy eigenstates of the new box. If you now measure the energy, you can get various values for the energy, with various probabilities, as determined by that superposition.
1
If you are okay with approximations and some nasty follow-up calculations, you can do the following not-so-delightful recipe:
1- Expand the $V(r)$ in its analytical form using Maclaurin Seires. (Hint, use $r=\zeta + 1$)
2- Stop when you reach the order that you feel appropriate for you. (e.g. 3)
3- Solve the Schrödinger Equation for each term after ...
1
The difference is in the sign of $E$.
The definition $k_2^2=-2mE/\hbar^2$ with $E>0$ implies that $k$ is pure imaginary $k_2$, i.e $k_2=i k_1$ with
$k_1^2=+2mE/\hbar^2>0$. Then $e^{k_2 x}= e^{ik_1 x}$.
On the other hand the definition $k_1^2=+2mE/\hbar^2$ gives $k_1$ real so again
$e^{i k_1x}$.
1
First, your Shrodinger equation seems to have a some small problems.
From
$$ - \frac{\hbar^2}{2m}\Psi'' +V(x)\Psi = E\Psi$$
one get:
$$ \Psi'' + \frac{\sqrt{2m}}{\hbar} (E-V(x)) \Psi =0.
$$
For your barrier, and for $0<E<V_0$,
One has in the barrier ;
$$ \Psi(x) = A e^{\kappa x} +
B e^{-\kappa x},$$
where $\kappa=\sqrt{\frac{2m (V_0-E)}{\hbar^2}}$...
1
Let us for simplicity assume that the lower bound $x_1=0$ is zero. (It can of course be generalized to any $x_1\in\mathbb{R}$.) Then
$$\phi(x+y)~=~\phi(x)+\phi(y).\tag{1}$$
From the single-valueness of the wavefunction, we get
$$ \psi(x)=\psi(x+L),\tag{2}$$
or equivalently,
$$ \sum_{\pm} C_{\pm}e^{\pm i\phi(x)}~\stackrel{(2)}{=}~ \sum_{\pm} C_{\pm}e^{\pm ...
1
There are several points: how do you define the derivative? No problem: if $\psi_1$ and $\psi_2$ are two functions in $\cal H$, there is by definition a scalar product $(\psi_1,\psi_2)$. If the 2 are functions on $\mathbb R$, the scalar product is often defined as
$$
(\psi_1,\psi_2)=\int_{-\infty}^\infty dx\,\psi_1^*(x)\psi_2(x)
$$
But a scalar product is ...
1
Rearranging the radial equation, we can write :
$\frac{d^2}{dt^2}u-[{\frac{l(l+1)}{r^2}}-\frac{\alpha}{r}]u -\kappa^2u = 0$ , where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}}$ , and $\alpha = \frac{2me^2}{4\pi\epsilon_0\hbar^2}$
As $r \to \infty$ , The the effect of the term in the $[$ $]$ (which is the "effective Potential" , $V_{eff}$) becomes ...
1
Roughly speaking, the one-dimensional equation of continuity says that if the probability current going in is different from the probability current going out, then the probability density will change.
So, for instance, if there is more (of whatever) going in than going out, the probability will increase at that position.
1
Since the Lagrangian density (which is confusingly also referred as a Lagrangian) is defined as a function which is integrated on, we may always think the $\mathcal{L}$ in inside a 4D integral, since the action is defined via
$$
S = \exp \left( \int d^4x \mathcal{L} \right).
$$
Now, we may decompose the term to two identical parts and integrating one of ...
1
$\Psi^*\Psi=|\Psi(x)|^2$ according to the Born Rule is the probability density function $P(x)$, not the 'point probability' in $x$.
The probability of finding the particle in a part $\delta x$ of the particle's domain is given by (for a normalised wave function):
$$P(x, \delta x)=\int_x^{x+\delta x}\text{d}x|\Psi(x)|^2$$
It is obvious that for $\delta x=...
1
Both $\frac{Ax}{a}$ and $\frac{A(b-x)}{(b-a)}$ are polynomial functions which when graphed are functions that have no holes, no jumps or asymptotes and behave well with limits, etc. They are continuous functions on the real line. And for $0$ the function obviously only has one value over it’s domain so it is continuous as well.
To find the derivative of ...
1
Let me make some complements. What James Johns said about Helium atom is a quite a good
example.
Basically, if we are capable of solving the many body Schrodinger equation directly, then no non-local potential appears. We have no non-local interaction in the nature. However, if we want some effective models or make some approximations, sometimes we have ...
1
After having thought about this question for a long time, I have arrived at the conclusion that what Bohm really meant must be far simpler than I had conjectured in my previous speculations.
I already remarked in my NOTE (2) to the post that the impossibility result stated by Bohm is trivial if we assume Property (4) in the form given there. Actually, I ...
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