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It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized, even though zero is its average position. But it has an infinite variance in position--- if you measure the position again and again, you will not have an average deviation. This wavefunction is infinitely spread out, so it is not really useful for describing a particle which is somewhere. It is an idealization, like a plane-wave state.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized, even though zero is its average position. But it has an infinite variance in position--- if you measure the position again and again, you will not have an average deviation. This wavefunction is infinitely spread out, so it is not really useful for describing a particle which is somewhere. It is an idealization, like a plane-wave state.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized as a Gaussian because although it has a peak at zero, zero is not its average position: it does not even have an average position since, as we saw for the other function, the expectation or average position is infinity, so it is not really useful for describing a particle which is somewhere.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized, even though zero is its average position. But it has an infinite variance in position--- if you measure the position again and again, you will not have an average deviation. This wavefunction is infinitely spread out, so it is not really useful for describing a particle which is somewhere. It is an idealization, like a plane-wave state.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

It usually just means that the state has infinite mean energy and so is not realizable by time evolution and measurements, when you have a finite energy at your disposal, so that th physical Hilbert space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is square normalizable, but multiplying by P takes you out of square normalizable states. The Hamiltonian has a P^2 term, so that this state has infinite kinetic energy, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x is defined by replacing p by x above. Such a state is not localized at all, so it is not really useful for describing a particle which is somewhere.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized as a Gaussian because although it has a peak at zero, zero is not its average position: it does not even have an average position since, as we saw for the other function, the expectation or average position is infinity, so it is not really useful for describing a particle which is somewhere.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants