Consider the complex plane wave \[\Psi (x,t) = A{e}^{i(kx-\omega t)}.\] Show that \[i\hbar \frac{\partial \Psi}{\partial t} = \frac{-{\hbar}^{2}}{2m} \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t)\] for some potential function \(V(x)\).

**Solution**

The Hamiltonian of a system is \[H = T + V\] where \(T\) is the kinetic energy and \(V\) is the potential energy. Since the quantity \(H\) is the total energy, let us rewrite the Hamiltonian as \[E = \frac{{p}^{2}}{2m} + V(x).\]

Now, we take the derivatives:

\[\frac{\partial \Psi}{\partial t} = -i \omega A{e}^{i(kx-\omega t)} = -i \omega \Psi (x,t)\] \[\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -{k}^{2}A{e}^{i(kx-\omega t)} = -{k}^{2}\Psi (x,t)\]

Since \(p = \frac{2\pi \hbar}{\lambda}\) and \(k = \frac{2\pi}{\lambda}\), where \(k\) is the wavenumber and \(\lambda\) is the wavelength, we have \[k = \frac{p}{\hbar}.\]

Therefore, \[\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -\frac{{p}^{2}}{{\hbar}^{2}}\Psi (x,t).\]

Next, we multiply \(\Psi (x,t)\) to the Hamiltonian:

\[E\Psi (x,t) = \frac{{p}^{2}}{2m}\Psi (x,t) + V(x)\Psi (x,t).\]

Notice that the above equation can be expressed as

\[E\Psi (x,t) = \frac{{-{\hbar}^{2}}}{2m}\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t).\]

Since the energy of matter waves is given by \(E = \hbar \omega\), we can show that

\[E\Psi (x,t) = \frac{\hbar \omega}{-i \omega} \Psi (x,t) .\]

Combining all the right parts, we assemble the Schrödinger equation: \[i\hbar \frac{\partial \Psi}{\partial t} = \frac{-{\hbar}^{2}}{2m} \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t).\]

Check out my other notes at Proof, Disproof, and Derivation

## Comments

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TopNewestHow do we know that from the start the wave function "must" be complex? – Snehal Shekatkar · 2 years, 5 months ago

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– Ethan Robinett · 2 years, 5 months ago

We don't. The assumption that the wave function is a complex plane wave is reasonable, however, because any wave can be represented as a linear combination of such waves (Fourier proved this in a more rudimentary way). Also, all of the operators you see above are derived from the complex plane wave solution.Log in to reply

– Steven Zheng · 2 years, 5 months ago

Yup. But my set Proof, Disproof, and Derivation is still a set of problems. I could have asked to show some other wavefunction that satisfies the Schrödinger equation. It's just the complex wave is most general.Log in to reply

– Snehal Shekatkar · 2 years, 5 months ago

So why don't we start with complex function when we derive sound wave equation?Log in to reply

– Ethan Robinett · 2 years, 5 months ago

Well if I'm not mistaken I believe a sound wave would actually just be a case of the wave equation, which also describes oscillations on a string and suchLog in to reply

EM Waves vs Sound Wave – Steven Zheng · 2 years, 5 months ago

The complex plane waves describes EM waves. We could in principle use complex plane waves for everything, but that is simply overkill because on of the phases is has amplitude 0.Log in to reply

What if I convert this equation in three dimensions and take partial derivatives of all of them, would it be correct. – Racchit Jain · 1 year ago

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Schrodinger's equation cannot be derived from anything. It is as fundamental and axiomatic in Quantum Mechanics as Newton's Laws is in classical mechanics (we can prove the Newton's Laws as an approximation of the Schrodinger's equation in the classical level). If you scrutinize the definition given above, one will find that the relation H=T+V that is being used is nothing but the energy conservation principle. So Schrodinger's equation is actually the energy conservation principle from a quantum perspective. Just like one has no proof for the energy conservation other than experiments which always seem to satisfy it, Schrodinger's equation has no pen-and-paper proof. The only evidences of its validity are experiments that have never violated the equation till date. – Kuldeep Guha Mazumder · 1 year, 1 month ago

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– Jack Lindon · 7 months ago

You can derive the Schrodinger equation from Feynman's path integral formalism. In 1st year classes typically the Schrodinger equation is took as axiomatic as it is too difficult to derive quantum mechanics from its more fundamental axioms. In addition it is not true that there is no proof of energy conservation, energy conservation is easily proven from Noether's theorem and temporal symmetry.Log in to reply

– Kuldeep Guha Mazumder · 6 months, 2 weeks ago

Yes that might be true and I might have overlooked that since I am not very conversant with higher physics. But what I want to say is that whenever you want to prove something in physics you use some theory. That theory rests on some axioms. In this course you will get a chain of theories and axioms which has to end somewhere. That "somewhere" has no proof. If you want to prove that "somewhere" you need a more fundamental theory with more fundamental axioms. But ultimately are you really proving anything?Log in to reply