# Tag Info

9

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) ...

5

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 ...

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 ...

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 ...

3

The wavefunction: $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$ is an infinite plane wave. So it describes a particle that has an infinite extent in both time and space. That is, it exists for $-\infty \le x \le \infty$ and for $-\infty \le t \le \infty$. Unsurprisingly, if the particle has an infinite extent then it's amplitude is everywhere zero and ...

2

You write that you do not like the wave-particle duality explanation of the Young experiment, and therefore turn to QFT. Before going further I would like to point out that the double slit experiment is a one-particle effect. That means you only need consider one particle at a time to explain what is happening. Because of this QFT will not buy you much as ...

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 $\partial_t\psi(x,t) = 0$. ...

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 ...

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. ...

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 ...

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 ...

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 $L^2$ ...

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 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

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 ...

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 $\left< n'l \right\| ... 1 First, you are not talking about a phase space, but configuration space. Now, the space of wavefunctions of a single particle in 3D space is the space of Lebesgue square-integrable functions$L^2(\mathbb{R}^3)$on the configuration space$\mathbb{R}^3$. So already for a single particle, there is no generic requirement for the wavefunction to be smooth, ... 1 According to this paper, an experiment was performed that measured the single electron's physical wave function by causing it to interfere with itself. The interference pattern matched the predictions of the Schrodinger equation. So, apparently this was a direct measurement of an electron's wave. Hydrogen Atoms under Magnification: Direct Observation of ... 1 In this link you will see the radial hydrogen wavefunctions. It is only the l=0 states, S states, that have a value different than zero at r=0. The other angular momentum states get a very small contribution to the probabilities from r>0 to r=1 fermi ( the charged radius of the proton) as 1 fermi is of order 10^-15meters, and the probability is the ... 1 One calculates the probability that the electron is inside the nucleus by integrating$\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$is ... 1 You can have a repeatable measurement process (i.e. a measurement process that, roughly speaking, gives the same result if done twice in a row) only for discrete observables. A discrete observable is an observable whose spectrum is purely discrete. So with the Hamiltonian it is possible to have repeated measurements, provided it is a system with purely ... 1 The wavefunction is always antisymmetric. It does not matter if they have similar, different, or even identical numbers for$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$whereas for two spin ... 1 The wave function is "reduced", meaning that there is a reduction in size of the continuous range of states (positions) that have non-zero probability. However, it never goes to being a single eigenstate, due to the quantum uncertainty of the probe used in the measurement or of the measurement apparatus, itself. That uncertainty can never go to exactly zero. ... 1 You should also match the derivatives at$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)$where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ... 1 First, your explanation is...sort-of-right. What's travelling is a quantum object, not a particle, not a wave. The probability of detecting a particle-like localized blip with some sort of detector is given by a probability density$\rho$, which is the "sqaured amplitude" of a "wavefunction"$\psi$. For free particles, the Schrödinger equation that$\psi\$ ...

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