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The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

How do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-\frac{1}{2}}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

How do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-\frac{1}{2}}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

My question is how do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units. Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-1}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

How do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-\frac{1}{2}}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

My question is how do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units. Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-1}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

My question is how do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units. Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-1}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match