User:Maximilian Janisch/Quantum fluctuation

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In quantum physics, a quantum fluctuation (a special case of which is a vacuum state fluctuation or a vacuum fluctuation) is an informal name for the fact that multiple measurements of the same physical property of a quantum system,^{[note 1]} such as the position or the spin of a particle, may yield different results, even though the system was prepared in the same state.^[1] This is the case whenever the latter state is not an eigenvector of the operator corresponding to the observable that is measured.

It is incorrect to think of quantum fluctuations as dynamical processes in time or space^[2]

Energy conservation in quantum mechanics

Consider a quantum system whose states live in the separable complex Hilbert space ${\mathcal {H}}$ equipped with the inner product $\langle \cdot ,\cdot \rangle$ . Assume that the system was conditioned in the state $\psi (0)\in {\mathcal {H}}$ . In the Schrödinger picture, it is postulated that the evolution of this state is determined by a strongly continuous semigroup of unitary operators $(U(t))_{t\in [0,\infty [}$ , each operator $U(t)$ mapping ${\mathcal {H}}$ into itself, in the following way:

\psi (t)=U(t)\psi (0).

The Hamiltonian $H$ of the quantum system is defined as the infinitesimal generator of the before-mentioned semigroup. KABALLO

During unitary evolution (meaning that no measurement is performed), the energy expectation value in the state $\psi (t)$ , defined as^{[note 2]} $\langle \psi (t),H\psi (t)\rangle$ according to the third Dirac–von Neumann axiom, remains unchanged, since $U(t)$ is assumed to be unitary and since $U(t)$ commutes with the Hamiltonian (why?)

\langle \psi (t),H\psi (t)\rangle =\langle U(t)\psi (0),HU(t)\psi (0)\rangle =\langle U(t)\psi (0),U(t)H\psi (0)\rangle =\langle \psi (0),H\psi (0)\rangle .

Now, let $\phi \in {\mathcal {H}}$ be an eigenvector of $H$ with associated eigenvalue $\lambda \in \mathbb {R}$ (since $H$ is self-adjoint, its spectrum is a subset of the real numbers). Then ${\frac {\vert \langle \phi ,\psi (t)\rangle \vert ^{2}}{\lVert \phi \rVert ^{2}\lVert \psi \rVert ^{2}}}$ is also independent of $t$ .^{[note 3]} If $H$ has a discrete spectrum, then the latter expression is, by the Born rule postulate, the probability of measuring the energy level $\lambda$ when an energy measurement is performed on the quantum system in the state $\psi (t)$ .^[a]

Unitarity_(physics)#Hamiltonian_evolution

Footnotes

^ Also known as an observable.
^ The expression $\langle \psi (t),H\psi (t)\rangle$ is only well-defined when $H\psi (t)$ is well-defined, which is an assumption that physical systems have to obey.
^ This can be seen as follows: First, since $U(t)$ is unitary, we have $\lVert \psi (t)\rVert ^{2}=\langle U(t)\psi (0),U(t)\psi (0)\rangle =\lVert \psi (0)\rVert ^{2}$ . Second, $\vert \langle \phi ,\psi (t)\rangle \vert ^{2}=\left\vert \left\langle \exp \left({\frac {iHt}{\hbar }}\right)\phi ,\psi (0)\right\rangle \right\vert ^{2}$ . But $\phi$ is an eigenvector of $H$ with eigenvalue $\lambda$ , so the latter expression equals $\left\vert \left\langle \exp \left({\frac {i\lambda t}{\hbar }}\right)\phi ,\psi (0)\right\rangle \right\vert ^{2}$ . Since $\left\vert \exp \left({\frac {i\lambda t}{\hbar }}\right)\right\vert =1$ , this in turn equals $\vert \langle \phi ,\psi (0)\rangle \vert ^{2}$ .

TODO

Explain better, what the formal terms mean.
Explain what the Hamilton operator is (it is unbounded, densely defined, etc.)
Find rigorous justification for why H commutes with U. (Cf. Engel-Nagel.)
Heisenberg picture.
Turn brackets to footnotes.
Say that we are working with a time-independent Hamiltonian operator.
Siehe Kaballo Grundkurs, Seite 287
Es fehlt die Konstante in der Passage von U zu H.
Sei $Q$ das Spektralmass (cf. Kaballo Aufbaukurs, Theorem 16.6) von $H$ . Wie kann man beweisen, dass $Q(M)$ mit $H$ für alle Borel-Mengen M kommutiert?
Genauer sagen, was die semigroup ist (Exponential, Engel-Nagel).
Noether's Theorem
See also page 24, definition of the inner product on the Fock space, in the Lecture notes for Math 273, Stanford, Fall 2018 by Sourav Chatterjee, Michel Talagrand. The inner product of the vacuum with any state containing at least one particle is obviously 0.

^ "quantum fluctuation in nLab". ncatlab.org. Retrieved 2021-09-26.
^ Neumaier, Arnold (2016-03-28). "Learn the Physics of Virtual Particles in Quantum Mechanics". Physics Forums Insights. Retrieved 2021-09-26.

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[1] Also known as an observable.

[4] The expression $\langle \psi (t),H\psi (t)\rangle$ is only well-defined when $H\psi (t)$ is well-defined, which is an assumption that physical systems have to obey.

[5] This can be seen as follows: First, since $U(t)$ is unitary, we have $\lVert \psi (t)\rVert ^{2}=\langle U(t)\psi (0),U(t)\psi (0)\rangle =\lVert \psi (0)\rVert ^{2}$ . Second, $\vert \langle \phi ,\psi (t)\rangle \vert ^{2}=\left\vert \left\langle \exp \left({\frac {iHt}{\hbar }}\right)\phi ,\psi (0)\right\rangle \right\vert ^{2}$ . But $\phi$ is an eigenvector of $H$ with eigenvalue $\lambda$ , so the latter expression equals $\left\vert \left\langle \exp \left({\frac {i\lambda t}{\hbar }}\right)\phi ,\psi (0)\right\rangle \right\vert ^{2}$ . Since $\left\vert \exp \left({\frac {i\lambda t}{\hbar }}\right)\right\vert =1$ , this in turn equals $\vert \langle \phi ,\psi (0)\rangle \vert ^{2}$ .

[2] "quantum fluctuation in nLab". ncatlab.org. Retrieved 2021-09-26.

[3] Neumaier, Arnold (2016-03-28). "Learn the Physics of Virtual Particles in Quantum Mechanics". Physics Forums Insights. Retrieved 2021-09-26.

[note 1]

[1]

[2]

[note 2]

[note 3]

[a]