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Well, they "exist" in our (physicists') imagination, as a limit, but they are not normalizable.

Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you may readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III.20 of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero.

• Note it is translationally invariant, since the exponential of the momentum operator acting on it amounts to one.

The x-space wavefunction of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space; also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a "Gaussian" with infinite width.

But you may translate this state in momentum space by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and further define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...

$\require{cancel}$

Well, they "exist" in our (physicists') imagination, as a limit, but they are not normalizable.

Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you may readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III.20 of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. (Intelligent squeezed coherent states. )

• Note it is translationally invariant, since the exponential of the momentum operator acting on it amounts to one.

The x-space wavefunction of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space; also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a "Gaussian" with infinite width.

But you may translate this state in momentum space by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and further define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...

$\require{cancel}$

Well, they "exist" in our imagination, as a limit, but they are not normalizable. Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. You notice it is translationally invariant since the exponential of the momentum operator acting on it amounts to one.

The x-space wave function of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space, also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a Gaussian with infinite width.

But you may translate this state by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...

$\require{cancel}$

Well, they "exist" in our (physicists') imagination, as a limit, but they are not normalizable. 

Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you may readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III.20 of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero.

• Note it is translationally invariant, since the exponential of the momentum operator acting on it amounts to one.

The x-space wavefunction of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space; also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a "Gaussian" with infinite width.

But you may translate this state in momentum space by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and further define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...

Well, they "exist" in our imagination, as a limit, but they are not normalizable. Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. You notice it is translationally invariant since the exponential of the momentum operator acting on it amounts to one.

The x-space wave function of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space, also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a Gaussian with infinite width.

But you may translate this state by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } $. I didn't write this...

$\require{cancel}$

Well, they "exist" in our imagination, as a limit, but they are not normalizable. Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. You notice it is translationally invariant since the exponential of the momentum operator acting on it amounts to one.

The x-space wave function of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space, also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a Gaussian with infinite width.

But you may translate this state by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...