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the solution of kroning-penny model inside the semiconductor crystall was something like this

enter image description here

where this solution can be approximated in the conduction band and valence band by a parabola where

\begin{gather*} E=E_{C}+\frac{(\hbar k)^2}{2m} \end{gather*}

where $E_{C}$ is the potential energy of the conduction band and $\frac{(\hbar k)^2}{2m}$ is the Kinetic energy

and also in valence band \begin{gather*} E=E_{V}-\frac{(\hbar k)^2}{2m} \end{gather*}

But the problem is : total energy in the valence band is the summation of kinetic and potential , but the kinetic energy in the valence band is Negative !

This doesn't makes sense because the \begin{gather*} K.E=\frac{p^2}{2m} \end{gather*} which can't be negative

so how this make sense at all?

Qmechanic
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Mans
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    The negative "kinetic energy" is precisely why valence band electrons close to the band edge have a negative effective mass, and why we often like to describe the valence band statistics and dynamics in terms of holes with positive effective mass. These answers don't address your question directly but they might still be helpful: holes, kinetic energy and group velocity of electrons. – Puk May 28 '23 at 20:40
  • @Puk, I don't really understand what you say ? Could you elaborate more ? – Mans May 28 '23 at 21:47
  • @Puk, secondly we could also model the valence band by the bottom of it and say the the total energy of valence band is equal to \begin{gather} E=E_{v.b} +K.E \end{gather} where the $E_{v.b} $ is the energy of the bottom of valence band ,like what we did with conduction band ,in this case the negative kinetic energy will not appear – Mans May 28 '23 at 21:51
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    Dealing with electrons with negative "kinetic" energy is somewhat counterintuitive. For example, as a consequence of this negative energy component, an electron at the top of the valence band accelerates in the direction of E-field rather than against it. Instead of electrons that act like they have negative mass, we often deal with holes that behave like ordinary positively charged particles. The bottom of the valence band is normally not relevant for charge or energy transport: the bottom is completely filled with electrons, the currents of electrons with wave vectors $\pm k$ cancel out. – Puk May 28 '23 at 22:11
  • @Puk, Are you saying the negative kinetic energy is because the particle has negative effective mass because \begin{gather} K.E=p^2/2m \end{gather} and because $m$ is negative so the K.E ? – Mans May 28 '23 at 22:37
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    Well, I would say it's the other way around: the fact that energy decreases (as it must) as you move away from the top of the valence band means the effective mass is negative at the top of the band, but maybe that's just semantics. Yes, it's no accident that they are both negative and $\text{KE}=p^2/2m$ is a valid explanation ($p$ here being the crystal momentum). – Puk May 28 '23 at 22:42
  • @Puk, but if we deal with the real mass , how would we explain it ? In case of effective mass the reason because effective mass was negative and so the kinetic, if we deal with the real mass what would be the case ? – Mans May 28 '23 at 22:49
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    An energy that decreases as momentum or group velocity increases doesn't break any laws, it just means the particles behave like they have negative mass (for a limited range of energies and wave vectors). I'm not sure if there is something more to explain, but perhaps someone else will chime in with a deeper insight. – Puk May 28 '23 at 22:58
  • @Puk,I mean the particle indeed behave as having negative effective mass, but I'm talking about if we dealt with the real mass case ,, we explained that $K.E=p^2/2m$ is negative because $m$ is the effective mass of the particle and it's indeed negative, if we swapped the negative $m$ by the $m_{r}$ which the real mass how could we possibly explain the negative of kinetic energy? – Mans May 28 '23 at 23:16
  • @Puk, There another thing that I forgot, the drawing of the energy with the wave vector is in the reduced form, the real form the energy increases as the wave vector increase – Mans May 28 '23 at 23:23
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    @Mans If you use the real mass as you propose you have changed the scenario. You are no longer talking about electron quasiparticles in the valence band. You are talking about free electrons. Those electrons will behave differently. – Matt May 29 '23 at 02:42

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