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For simplicity, let's take the 1D Schrödinger's equation for a single non-relativistic particle: $$-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x) \Psi(x,t) = i\hbar \frac{\partial\Psi(x,t)}{\partial t}$$

Applying separation of variables, $\Psi(x,t)=\psi (x)\phi (t)$, we get the time dependent solution. $$-\frac{\hbar^2}{2m} \frac{\psi'' (x)}{\psi (x)} + V(x)= i\hbar \frac{\phi' (t)}{\phi (t)}=C$$

$$\phi (t)=Ae^{-iCt/\hbar }$$

Here, the separation constant $C$ is taken as the energy of the particle, $E$. I see that this is convenient cause the exponent must be dimensionless. However, do we have further arguments for asserting it?

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This is essentially the definition of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically.

If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is defined to be the eigenvalue of the Hamiltonian. And we have $$H \Psi(x, t) = i \hbar \frac{\partial}{\partial t} \Psi(x, t) = i \frac{\partial}{\partial t} (\psi(x) \phi(t)) = i \hbar \left(- \frac{iC}{\hbar}\right) \Psi(x, t) = C \Psi(x, t).$$ So $C$ is the energy.

Of course, this just reduces the question to "why is the time-dependent Schrodinger equation true", and the answer is, again, that in quantum mechanics energy is $\hbar$ times the rate of change in phase.

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  • $\begingroup$ So it is just the case of us taking an "otherwise not particularly mathematically special" result and assigning it the physical property of energy? $\endgroup$ – Charlie Jan 22 at 17:27
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    $\begingroup$ @Charlie Perhaps, but there's nothing unprecedented about that. In classical mechanics, why is $mv^2/2$ the energy? Why not $mv^2/6$, or $m^2 v^7/38$, or perhaps $q^7 a^2 / c^3 m^{5/8}$? You need to make some definition of the word "energy". $\endgroup$ – knzhou Jan 22 at 17:31
  • $\begingroup$ Thanks for your answer. $\endgroup$ – Charlie Jan 22 at 17:37
  • $\begingroup$ Ok, so, essentially, what we do is to compare the wavefunction with the complex expression of an ordinary wave, $f(x,t)=f_0e^{i(kx-\omega t)}=f_0e^{ikx}e^{-i\omega t}$, identifying the therm $C/\hbar$ we got with $\omega$, and then applying Einstein-De Broglie's relation, $\omega=E/\hbar$, so $C=E$. Then, the only assumption to get $C=E$ is that the particle behaves like a wave in the time, isn't it? $\endgroup$ – Quaerendo Jan 22 at 19:05
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    $\begingroup$ @Quaerendo Well, sure, but my point is that the real physical assumption is $\omega = E/\hbar$ in the first place. That's the new thing. You can't answer this question without implicitly assuming it somewhere. $\endgroup$ – knzhou Jan 22 at 19:08
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The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities.

In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E_K+V$ in standard systems) you know that $$\dot{A} = \{ A, H \} +\frac{ \partial A }{\partial t}$$ with $\{ a,b\}$ the Poisson brackets. Following quantization rules we replace the Poisson brackets with commutation relations (divided by $i\hbar$), and all quantities in operators, to get $$\frac{d}{dt} \langle A \rangle = -\frac{i}{\hbar}\langle \left[ A, H \right] \rangle + \frac{ \partial }{\partial t}\langle A \rangle$$ and you can see how $H$ here plays the role of the energy. This is just the Heisenberg picture, which is equivalent to the Schroedinger picture, so we can identify the term as a Hamiltonian, which measures the energy.

Another (closely related) way to see this: start from the fundamental relations of QM $[x,p]=i\hbar$, from which you can derive $[x, f(p)]=i\hbar \partial_p f$ and $[f(x),p]=i\hbar \partial_x f$. Then you take Hamilton equations from classical mechanics and quantize them

$$ \frac{d}{dt}\langle x \rangle = \langle \frac{\partial H}{\partial p} \rangle = -\frac{i}{\hbar}\langle \left[x, H\right]\rangle$$ and again you got Heisenberg picture, from which you can derive Schroedinger equation, with the $H$ the energy as it comes from classical mechanics.

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