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1

For the time-dependent Schrödinger equation there might be one. The spatial derivative of the wave function is connected to a "flow of probability" associated with the squared absolute value of the wave function which gives a probability density. You can view this probability density as a fluid with mass conservation (probability conservation). A ...


-2

It started out smooth, and it can't go from smooth to not smooth.


0

I'd say that a good physical reason for the first derivative being continuous is for the probability current to be continuous too, since it's constructed from $\psi(x,t)$ and $ \nabla \psi(x,t)$.


2

It comes from the mathematical framework of Quantum Mechanics. A general expectation value is the result of computing a state $\omega$ over some observable $A$. Mathematically speaking $A$ is a self-adjoint operator from the C*-algebra $\mathfrak A$ of observables and $\omega$ is a state over $\mathfrak A$, i.e. a normalised positive linear functional on ...


4

$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\bra}[1]{\langle #1 \rvert}$As a special instance of the Born rule stating that, given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$, it is an axiom in the standard formulations of quantum mechanics that ...


0

This is a very interesting question. I don't know if there is a general and definitive answer, but I'll try to make some comments. I apologize if this ends up rambling; I'm finding this out as I write this answer. Operators have dimensions, since their eigenvalues are physical quantities. For bras and kets it gets more complicated. First, you cannot in ...


0

Really good question. Measurements have unit but in quantum mechanics, a measurement is the "evaluation" of an observable on a state (or a state on an observable) something like $$ \left\langle \psi | A | \psi \right\rangle ,\quad \psi\in\mathcal{H},\ A\in\mathcal{B}(\mathcal{H}) \ \text{self-adjoint}$$ A priori, there seem to be an arbitrariness in the ...


0

To find the solution to your problem: Area 1: U1,E>U1 The Schrodinger equation and it's solution is: $$ψ'' + {2m \over \hbar ^2}(E-U1)ψ=0 ---> ψ_A=e^{ik_1 x} + c_1e^{-i k_1 x} $$ where $k_1 ^2 = {2m \over \hbar ^2}(E - U1) $. Area 2: U2, E As in the above: $$ψ'' - {2m \over \hbar ^2}(U2-E)ψ=0--> Ψ_Β = c_2 e^{k_2 x} + c_3 e^{-k_2 x} $$ where $k_2 ^2 ...


2

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


1

SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason. Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation ...


2

A slight expansion on danimal's comment: you can generally get the state $\psi(x,t)$ from the $\psi(x,0)$ you provided by operating on it with the unitary time evolution operator $\exp(-i \hat{H} t/\hbar)$. Since you know the eigenstates, you can write the Hamiltonian in a diagonal basis and this operator will appear to multiply $\psi_n$ simply by $e^{-i E_n ...


1

Since $V(x)$ is bounded from above we have three possibilities. Either it oscillates at infinity with an upper bound, or it asymptotes to a constant $<E$ or it diverges to $-\infty$. Since we are interested in $x\rightarrow\infty$ we may average the oscillation in the first case to the mean, and if it diverges then we concern our selves with the leading ...


3

I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...


1

I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...


0

While $[2]$ is "legal", $[1]$ definitely isn't. We are able to write: $$\int dx\ \psi^\ast H\psi = \int dx\ \psi^\ast E\psi = E$$ only if $H\psi = E\psi$ i.e. when $\psi$ is an eigenfunction of $H$. In the general case, it is not, and we generally have $\psi = \sum_n c_n \psi_n$, where $\psi_n$ are the eigenstates of $H$, with $H\psi_n = E_n\psi_n$. The ...


0

Your step $[1]$ is not OK. Just because $\hat H$ has eigenvectors whose eigenvalue is $E$ does not mean that any vector will be an eigenvector of $\hat H$ with eigenvalue $E$. In particular, the position eigenstates are emphatically not eigenstates of the hamiltonian. To go beyond $$⟨\hat{H}⟩ = \int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩$$ you need to ...


1

It seems to me that you are making some confusion. The problem with the passage [1] (and [2]) that you outline is that you are not allowed to do that (on a rigorous level) if the operator has continuous spectrum, for there are no corresponding eigenvectors on the Hilbert space (and it is wrong also on a non-rigorous level as pointed out by others). Anyways, ...


1

For a normalised linear combination of (orthogonal) states like this, the probability of measuring one of them is the absolute square of the coefficient in the combination: If $$\Psi = a_1\psi_1+a_2\psi_2+...$$ where $|a_1|^2+|a_2|^2+... = 1$ then $$P(\psi_1) = \left|a_1\right|^2, P(\psi_2) = \left|a_2\right|^2$$etc. Slightly more generally, if you know ...


0

Assuming you've done the algebra correctly, these equations can be solved for a relationship between $k$ and $K$, which should lead to the quantization of energy levels in terms of $a$, $b$, and $V_o$. Then you solve for $C$ in terms of $A$ from either equation (you MUST get the same result with either) and then normalize.


1

The two questions are slightly different. Each individual measurement of $L^2$ or $L_z$ will return an eigenvalue. In this case, you have only one possible measurement for $L^2$ (corresponding to $l=1$), but you have two possible measurements for $L_z$; 2/3 of the time you'll get $m=1$, and 1/3 of the time you'll get $m=0$. The expectation value, on the ...


3

Angular momentum is that which is conserved under rotations. Equivalently, the angular momentum operators are the generators of rotations. This holds both classically and quantumly by (versions of) Noether's theorem. Defining "angular momentum" as $\vec x \times \vec p$ classically and then showing that it is conserved is doing it the wrong way around from ...


2

Suppose $\psi$ satisfies the (dimensionless) time-dependent Schrödinger equation: $$ i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi $$ It will also satisfy the conjugate equation: $$ -i\frac{\partial\psi^*}{\partial t}=-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^* $$ Now consider how the normalisation changes over time: ...


1

The wavefunction $\psi(x_0, x_1, t)$ gives the probability of finding one electron between $x_0$ and $x_0+dx_0$ and the other between $x_1+dx_1$: $$P(x_0, x_0+dx_0; x_1, x_1+dx_1; t) = \vert\psi(x_0, x_1, t)\vert^2dx_0dx_1$$ We expect the brightness of the scintillation at $y$ to be given by the probability (density) that either electron is found at $y$: ...


1

For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ...



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