Derivation of the Time-Dependent Schrödinger Equation

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+VH = T + V where TT is the kinetic energy and VV is the potential energy. Since the quantity HH is the total energy, let us rewrite the Hamiltonian as E=p22m+V(x).E = \frac{{p}^{2}}{2m} + V(x).

Now, we take the derivatives:

Ψt=iωAei(kxωt)=iωΨ(x,t)\frac{\partial \Psi}{\partial t} = -i \omega A{e}^{i(kx-\omega t)} = -i \omega \Psi (x,t) 2Ψx2=k2Aei(kxωt)=k2Ψ(x,t)\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -{k}^{2}A{e}^{i(kx-\omega t)} = -{k}^{2}\Psi (x,t)

Since p=2πλp = \frac{2\pi \hbar}{\lambda} and k=2πλk = \frac{2\pi}{\lambda}, where kk is the wavenumber and λ\lambda is the wavelength, we have k=p.k = \frac{p}{\hbar}.

Therefore, 2Ψx2=p22Ψ(x,t).\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -\frac{{p}^{2}}{{\hbar}^{2}}\Psi (x,t).

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

EΨ(x,t)=p22mΨ(x,t)+V(x)Ψ(x,t).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

EΨ(x,t)=22m2Ψx2+V(x)Ψ(x,t).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=ωE = \hbar \omega, we can show that

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

Combining all the right parts, we assemble the Schrödinger equation: iΨt=22m2Ψx2+V(x)Ψ(x,t).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

#Mechanics #QuantumMechanics #SchrodingerEquation

Note by Steven Zheng
6 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

How do we know that from the start the wave function "must" be complex?

Snehal Shekatkar - 6 years, 9 months ago

Log in to reply

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.

A Former Brilliant Member - 6 years, 9 months ago

Log in to reply

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.

Steven Zheng - 6 years, 9 months ago

Log in to reply

@Steven Zheng So why don't we start with complex function when we derive sound wave equation?

Snehal Shekatkar - 6 years, 9 months ago

Log in to reply

@Snehal Shekatkar 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 such

A Former Brilliant Member - 6 years, 9 months ago

Log in to reply

@A Former Brilliant Member 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. EM Waves vs Sound Wave

Steven Zheng - 6 years, 9 months ago

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 - 5 years, 6 months ago

Log in to reply

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.

Jack Lindon - 4 years, 11 months ago

Log in to reply

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?

Kuldeep Guha Mazumder - 4 years, 11 months ago

What if I convert this equation in three dimensions and take partial derivatives of all of them, would it be correct.

Racchit Jain - 5 years, 5 months ago

What are the major operators in quantum mechanics?

Samuel Barasa - 2 years, 8 months ago

Derive e=mc²

Kotte Sivaprasad - 9 months, 3 weeks ago

Write a comment or ask a question...what are the major operators in quantum mechanics?

Samuel Barasa - 2 years, 8 months ago
×

Problem Loading...

Note Loading...

Set Loading...