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In Quantum Mechanics textbooks, the equation for a electron tunneling through a barrier is $$-\frac{\hbar ^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi \left( x\right) +U\psi \left(x\right) =E\psi \left( x\right).$$ In numerical computations, we choose $E_{0}=\hbar ^{2}/2ma^{2}$ as a reduced energy, then these parameters change into: $\overline{x}\rightarrow x/a$, $\overline{E}\rightarrow E/E_{0}$, $\overline{U}\rightarrow U/E_{0}$, and the corresponding equation becomes $$-\frac{d^{2}}{d\overline{x}^{2}}\psi \left( \overline{x}\right) +\overline{U}\psi \left( \overline{x}\right) =\overline{E}\psi \left( \overline{x}\right).$$ After we obtain the wave function $\psi \left(x \right)$, the current density can be calculated by $$J=\frac{-i\hbar }{2m}\left[ \psi ^{\ast }\nabla \psi -\psi \nabla \psi^{\ast }\right].$$ If we plot the curve of current density and $x$, what is the unit of the current density?

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  • $\begingroup$ AFAIK current is measured in Ampères, or mA etc $\endgroup$ – user207455 Jul 9 at 11:34
  • $\begingroup$ In experiments that is exactly the truth. What about theoritical models? $\endgroup$ – longquan Jul 9 at 11:55
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    $\begingroup$ Welcome to Physics.SE! Because of the question-and-answer format of the board, it's best to avoid asking multiple questions in the same post. I would recommend that you split out your question about the bias voltage and ask it as its own separate question. $\endgroup$ – Michael Seifert Jul 9 at 11:56
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In standard units, the quantity $\hat{\mathbf{J}} = \frac{1}{2m} (\psi^* \hat{\mathbf{p}} \psi - \psi \hat{\mathbf{p}} \psi^*)$ is a probability current, which has the same units as a number flux. Its units depend on the number of dimensions you're working in, since you need to integrate it over a closed surface to obtain something with the units of $\text{s}^{-1}$ (i.e., particles per unit of time.) In general, if you're working in a $D$-dimensional problem, then the units will be $\text{s}^{-1} \text{m}^{-(D-1)}$.

Note that if you multiply the probability/number current by the charge on each carrier, you will obtain the charge current density. If the carrier charge is expressed in coulombs, then this will have units of $\text{C/(s} \cdot \text{m}^{D-1}) = \text{A/m}^{D-1}$, as expected.

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  • $\begingroup$ Thank you for your answer! I get your point. Basically the units of current and voltage are Ampere $\mathbf{A}$ and Volt $V$, respectively, and the magnitudes differ under different conditions. But in a theoretical model or a numerical simulation, if one wants to plot the curve of current-voltage characteristics, the data may be handled, and the units should be stated. So, how can we write down the explicit expression of the units based on the above equation with different reduced quantities? I mean, how can we know the specific dependence of the current on a certain reduced quantity? $\endgroup$ – longquan Jul 9 at 13:19

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