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Ján Lalinský
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This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

The function $\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$ is of similar status as the "delta function" $\psi(\mathbf r) = \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)$ is. They are useful as tools to ease the work with the normalizable functions, but they can never be used as functions describing actual state.

In order to have existence of some proper Hamiltonian eigenfunctions (so they are members of some Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

In order to have existence of some proper Hamiltonian eigenfunctions (so they are members of some Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

The function $\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$ is of similar status as the "delta function" $\psi(\mathbf r) = \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)$ is. They are useful as tools to ease the work with the normalizable functions, but they can never be used as functions describing actual state.

In order to have existence of some proper Hamiltonian eigenfunctions (so they are members of some Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

added 369 characters in body
Source Link
Ján Lalinský
  • 41.3k
  • 1
  • 34
  • 98

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

In order to have existence of some proper Hamiltonian eigenfunctions (membersso they are members of ansome Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

In order to have existence of some Hamiltonian eigenfunctions (members of an Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

In order to have existence of some proper Hamiltonian eigenfunctions (so they are members of some Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

Source Link
Ján Lalinský
  • 41.3k
  • 1
  • 34
  • 98

This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

In order to have existence of some Hamiltonian eigenfunctions (members of an Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.