# Tag Info

1

You should also match the derivatives at $x=0$ so that they took into account the $\delta$-function. If you take smooth well $V_\epsilon(x)$ and consider small region near zero $(-\epsilon,\epsilon)$ where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ...

3

Almost. You do know its state. It's $(c_1\Psi_1+c_2\Psi_2)$ (apart from a normalization constant). Remember that your choice of basis vectors to represent the degenerate subspace is arbitrary. There's nothing in the physics distinguishing the two you happened to choose. That superposition state is just as good as any of the other states in that degenerate ...

0

As pointed out above, this is a standard ODE in which u are normally to assume a solution to the DE as Psi(x)=Aexp(ax) [where A and a are both constants] If u then substitute for the assumed solution in the Schrodinger equation, u have; a^2+k^2=0 or a^2=-(k^2) or a=+/-(ik) since u now have two values for a (ik and -ik), it means you should have two ...

1

The wavefunction is always antisymmetric. It does not matter if they have similar, different, or even identical numbers for $n$ and $l$. But let's be clear. The wavefunction is a function from the configuration space into the joint spin state. So for a spin 1/2 particle the wavefunction is a function from $\mathbb R^3$ to $\mathbb C^2$ whereas for two spin ...

1

You can have a repeatable measurement process (i.e. a measurement process that, roughly speaking, gives the same result if done twice in a row) only for discrete observables. A discrete observable is an observable whose spectrum is purely discrete. So with the Hamiltonian it is possible to have repeated measurements, provided it is a system with purely ...

3

Energy is a bit of a special case because the eigenfunctions of the Hamiltonian are time independent (assuming a time independent Hamiltonian). So when you make an energy measurement and collapse the system to an eigenfunction of the Hamiltonian it stays there. However the position operator does not commute with the Hamiltonian so when you measure the ...

1

The wave function is "reduced", meaning that there is a reduction in size of the continuous range of states (positions) that have non-zero probability. However, it never goes to being a single eigenstate, due to the quantum uncertainty of the probe used in the measurement or of the measurement apparatus, itself. That uncertainty can never go to exactly zero. ...

8

The Born rule (and hence any discussion of collapse in the sense of the Copenhagen interpretation) is relevant only when an observer has made a distinction between a (tiny, observed) system and its (huge, observing) environment (= everything else, containing in particular the measurement equipment). This distinction (not present in relativistic QFT itself) ...

4

I take a minimal interpretation of QFT in a Copenhagen style to seek to make a connection between a classical description of/model for an experimental apparatus and classical records of its measurement results and a QFT model for the same apparatus. Classically, a modern measurement device is most often a thermodynamically metastable system that we engineer ...

4

It is your Eq 1 and 2 that are mathematically inconsistent -- just take complex conjugate on both sides of Eq 1 and you will see. I think you confused a state being an element of the Hilbert space and a state satisfying the schrodinger equation. I can write down plenty of elements of the Hilbert space that does not satisfy the Schrodinger equation, for ...

1

One calculates the probability that the electron is inside the nucleus by integrating $\psi^*\psi$ over the volume of the nucleus. For example, the radial part of the hydrogen ground state wavefunction is $\psi=\frac{e^{-\frac{r}{a_0}}}{\sqrt{\pi a_0^3}}$, so the integral is $\frac{1}{\pi a_0^3}\int_0^b e^{-2r/a_0} 4\pi r^2 dr$. In the above, $a_0$ is ...

1

In this link you will see the radial hydrogen wavefunctions. It is only the l=0 states, S states, that have a value different than zero at r=0. The other angular momentum states get a very small contribution to the probabilities from r>0 to r=1 fermi ( the charged radius of the proton) as 1 fermi is of order 10^-15meters, and the probability is the ...

0

Without any attempt for rigorous arguments: If the wave function is not continuous across the delta potential, then evaluating the second derivative of it (because of the kinetic term in the Hamiltonian) would give you the derivative of a delta function, which is not a delta function, and the Schroedinger equation is not satisfied at the position of the ...

-1

In any quantum mechanical setting, regardless of the type of the potential, one imposes the requirement that the wave-function $\Psi(x)$ (i.e., the solution to the Schrödinger equation) be $C^1$-smooth (continuous with well-defined 1st derivative) or at least, continuous, $C^0$ [see here for a definition], in position. Notice that a discontinuous ...

0

To understand what is going on you have you to distinguish a definition from an equation. As an example you could consider the heat equation $\partial_{xx}u=k\partial_t u.$ Both sides have their own meaning, and the equation says that for the solution to the heat equation, the two things are equal. You have to first be able to compute the left hand side ...

2

The "independent" in "time-independent Schrödinger equation" doesn't mean that the wavefunction $\psi(x,t)$ is independent of time, but that the quantum state it defines doesn't change with time. Since $\psi(x)$ and $\mathrm{e}^{\mathrm{i}\phi}\psi(x)$ for any $\phi\in\mathbb{R}$ define the same quantum state, this does not imply $\partial_t\psi(x,t) = 0$. ...

1

A general wavefunction for a system can be expanded as a sum of the eigenfunctions, $\psi_i$: $$\Psi = a_0\psi_0 + a_1\psi_1 + a_2\psi_2 + \,...$$ The coefficients $a_i$ are calculated using: $$a_i = \langle\psi_i | \Psi\rangle$$ If you do a measurement just once you will get one of the eigenvalues $E_i$, and the probability of getting each ...

1

One has to keep in mind 1) that it is the complex conjugate square of the eigenfunction that gives the probability of finding the electron with energy E at a specific radius. 2) There are no fixed orbits in the quantum mechanical solution, only a locus of probability called orbital 3)orbitals overlap in space, it is the energy that is keeping the electron ...

1

A wave is a traveling disturbance of some kind. In a medium the disturbance is generally a displacement of some part of the medium from its resting position. Now, let's unpack that. If "some part" is displaced then it is not he same in all places, so there must be a spacial dependence. If it "travels" then it is not in the same place at all times, so ...

1

The expectation value is a single number, it is the sum of all the possible values with a weight based on how often you get the result. So $\langle S_x\rangle=P(+\frac \hbar 2)\frac{\hbar}{2}+P(-\frac \hbar 2)\frac{-\hbar}{2}$ And you can get the probabilities by projecting the original spinor onto the eigenspaces of the operator and comparing the $L^2$ ...

2

Your confusion stems from the fact that $\lvert x \rangle$ is not inside the Hilbert space of states. It cannot be because $\langle x \vert x \rangle = \delta(x-x) = \delta(0)$ is not an allowed value for an inner product in a Hilbert space to have. There are several things to say about $\lvert x \rangle$: If you want to make precise what kind of objects ...

2

Assuming everything is defined in the correct Hilbert spaces, project the decomposition $$|\psi\rangle = \sum_n {c_n |n\rangle}$$ onto the position kets ("states") $|x\rangle$ and obtain $$\psi(x) = \langle x |\psi\rangle = \sum_n {c_n \langle x |n\rangle} = \sum_n {c_n u_n(x)}$$ where the $u_n(x) = \langle x |n\rangle$ are the wavefunctions ...

0

I found the answer and wanted to share with you. First of all, we are obtaning $A(k)$ by Fourier Transformation, so like wave function we are allowed to normalize it. $\int\limits_{-\infty}^{\infty} |A(k)|^2dk = 1$ That's how we find constant $C$ . To obtain $\Delta P$ , we just need to calculate $\langle k \rangle$, because $p=\hbar k$ Remember we ...

2

I know this question was asked a long time ago, but since I thought very hard about the same question today and didn't find the other answer very helpful, I decided to write my own. The problem is that OP is not really asking about why the product form used in the BO approximation is valid, but how the given expansion (which is claimed to be exact, see e.g. ...

1

This answer gives an analytical approach for diagonal matrix elements. First of all, since $r^k$ is spherically symmetric you can immediately integrate the angular parts: $$\left< n' l' m' | r^k | n l m \right> = \left< n'l \right\| r^k \left\| nl \right> \delta_{l',l} \delta_{m',m},$$ where $\left< n'l \right\| ... 1 This looks to be a messy calculation for the most general case, and a direct attempt based on special-functions properties of the Laguerre wavefunctions is likely to simply falter and die in the not-quite-right forms of the integrals. These matrix elements are calculated in terms of recursion relations in Matrix-element calculations for hydrogenlike ... 0 Given that the eigenstates of the hydrogen atom can be separated into a tensor product of angular momentum eigenstates and a radial solution, the Wigner-Eckart theorem helps to calculate the matrix element of any tensor-like operator (a scalar as$r^k$is a trivial case) between states of the form$|lm\rangle$. The rest is an integral over the radial ... 2 We prove this by a reductio ad absurdum. We start by assuming that the wavefunction of a non-degenerate ground state is complex, then show this means the wavefunction must be degenerate. Suppose we have a complex ground state. Then we can write it as a sum of real and imagniary parts: $$\psi = \psi_r + i\psi_i \tag{1}$$ The ground state obeys ... 1 According to this paper, an experiment was performed that measured the single electron's physical wave function by causing it to interfere with itself. The interference pattern matched the predictions of the Schrodinger equation. So, apparently this was a direct measurement of an electron's wave. Hydrogen Atoms under Magnification: Direct Observation of ... 0 Continuity of a wave-function is not a necessary condition, as already pointed out by @ACuriousMind. For instance, consider a linearly dispersing particle in the presence of a$\delta$-function potential. The time-independent Schrödinger equation in this case is$-i \frac{\partial}{\partial x} \psi(x) + V \delta(x) \psi(x) = E \psi(x)$Typically, you have ... -2 How can quantum wavefunctions be smooth/continuous when particles are created/destroyed/changed? Because particle creation etc merely involves a change of configuration. There is no magic. In gamma-gamma pair production you start with two photons, and you end up with an electron and a positron. Each has a wave nature, as evidenced by things like ... 1 First, you are not talking about a phase space, but configuration space. Now, the space of wavefunctions of a single particle in 3D space is the space of Lebesgue square-integrable functions$L^2(\mathbb{R}^3)$on the configuration space$\mathbb{R}^3$. So already for a single particle, there is no generic requirement for the wavefunction to be smooth, ... 0 This is the wavefunction of a coherent state; it is a well-known solution of the quantum harmonic oscillator and it has a large number of nice properties. There is a good number of independent ways to derive it so it is pretty pointless to try and guess which one Griffiths was thinking of. 2 First note the following: If$\psi(x,0)$is any normalizable wave-function in your Hilbert space, then$\psi(x,t) = e^{-iHt} \psi(x,0)$(I've set$\hbar = 1$) satisfies the time-dependent Schrodinger equation (we call this "time evolution"). This can be seen by using the fact that the stationary states form a basis for your Hilbert space and so$\psi(x,0)$... 0 Look at the integral over$K$in your solution. It has the form $$\int dx' K(x,x';E)\phi(x') \;\;\text{~}\;\; \int dx'e^{i|x-x'|\sqrt{2mE}/\hbar}\phi(x') =\\ = \int_{-\infty}^x dx'e^{i(x-x')\sqrt{2mE}/\hbar}\phi(x') + \int^{\infty}_x dx'e^{-i(x-x')\sqrt{2mE}/\hbar}\phi(x')$$ If you take out the factors in$e^{\pm ix \sqrt{2mE}/\hbar}\$ in each of the last ...

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