Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The reasoning leading to the Schrödinger equation goes as follows:

A plane wave in empty space has the following form:

$$\psi = e^{i(kx-\omega t)}$$

Einstein had previously explained the photoelectric effect, i.e. the emission of electrons from a metal surface through light, by suggesting that light is made from photons containing the momentum:

$$p_\text{Photon}=\hbar k$$

and energy:

$$E_\text{Photon}=\hbar\omega$$

This was proven to be correct by experiment.

De Broglie then suggested that the same relations hold for electrons, so that, given the momentum and energy of an electron, one could find the wavelength and frequency of an electron's plane wave:

$$\begin{gather} k_\text{Electron}=\frac{p_\text{Electron}}{\hbar}\\ \omega_\text{Electron}=\frac{E_\text{Electron}}{\hbar} \end{gather}$$

Given the plane-wave function of an electron, one can then use certain operators to extract momentum and energy from it:

$$\begin{gather} \hat{p}\psi = -i\hbar\partial_x\psi = \hbar k \psi\\ \hat{E}\psi = i\hbar\partial_t\psi = \hbar\omega\psi \end{gather}$$

But for a free electron outside of a force field, the relationship between energy and momentum is given by:

$$E = \frac{p^2}{2m}$$

The Schrödinger Equation for a free electron can be derived from this equation by replacing energy and momentum by the extraction operators:

$$\hat{E}\psi = i\hbar\partial_t\psi = \frac{\hat{p}^2}{2m}\psi=-\frac{ \hbar^2\partial^2_x}{2m}\psi$$

Schrödinger now argued that placing the free electron in a Potential would modify the equation simply through the addition of the Potential Energy:

$$i\hbar\partial_t\psi =\biggl(-\frac{\hbar^2\partial^2_x}{2m}+V(x)\biggr)\psi$$

Now my question is: The introduction of a potential is going to mess up the plane waves very seriously, transforming them into something quite different, so the extraction operators for Energy and Momentum - which only work for plane waves, for which they were designed - are no longer going to work! How can the Schrödinger equation still hold up?

share|improve this question
3  
The whole argument using plane waves etc. leading to the Schrodinger equation is a plausibility argument, not a derivation. The Schrodinger equation and the canonical commutation relations $\left[\hat{x},\hat{p}\right]=i\hbar$ are really the more fundamental principles. –  Michael Brown Aug 7 '13 at 0:45
3  
Also, an arbitrarily complicated function can be written as a superposition of plane waves. :) –  Michael Brown Aug 7 '13 at 0:47
    
... "extraction operator"? Never heard about this ... –  Dilaton Aug 7 '13 at 10:50

3 Answers 3

While plane waves aren't eigenstates of Hamiltonians with nonconstant potentials, they're still perfectly valid states. They also form a complete set, so any state can be expressed as a sum of plane waves. The Schrödinger equation is linear, so if it applies to every plane wave, it applies to every sum of plane waves, and thus every state.

share|improve this answer
    
Plane waves do form a complete set if all frequencies can be used - but in case of the hydrogen atom in its lowest state, only ONE frequency can be used. I don't at all see that the ground state of the hydrogen atom is a superposition of plane waves of that one frequency. One could choose different directions of the plane wave, but I don't know if this is enough to recreate the wave equation of hydrogen. –  Sebastian Henckel Apr 29 at 18:55
    
@Sebastian Henckel: Don't think frequency, think wavelength. Wavelength determines momentum, and the dispersion relation between frequency and wavelength becomes complicated in the presence of nonconstant potentials. While there's only one frequency, the ground state of the hydrogen atom has nonzero wavelength components almost everywhere. –  Dan Apr 29 at 19:10
    
@Sebastian Henckel: Also, recall that the expansion in terms of plane waves is just the Fourier transform. Are you implying that the ground state of hydrogen can't be fourier transformed? –  Dan Apr 29 at 19:17
    
That dispersion relations frequency vs. wavelength become complicated for non-homogeneous potentials is a different way of saying that the plane-wave derived operators for momentum and energy should no longer be valid for non-homogeneous potentials. –  Sebastian Henckel Apr 29 at 22:21
    
As to the Fourier transform of the ground state of hydrogen: Everything of course can be written as a superposition of Fourier functions - aka plane waves - if all wave-vectors can be used. But I thought your answer implied that the hydrogen ground state can be built by superimposing only those wave-vectors with $\lvert \vec{k} \rvert = \sqrt{\frac{2m\omega}{\hbar}}$ See my follow-up question: physics.stackexchange.com/q/110529/27232 –  Sebastian Henckel Apr 29 at 22:29

Think of classical mechanics. For a free particle we have a Lagrangian:

$$L=\frac{m}2 v^2.$$

Trajectories of such a particle would be just straight lines, and momentum would conserve. The momentum is defined as $$p=\partial_{v}L=mv.$$ Energy of this particle is:

$$E=v\;\partial_vL-L=\frac{mv^2}2$$

Now, similarly, we can subtract potential energy to get Lagrangian for particle in non-uniform potential:

$$L'=\frac{m}2v^2-U.$$

Now as you say, introduction of the potential is going to mess up the trajectories very seriously, transforming them into something other. Does this mean that equation for momentum should change? In no way: $\partial_vU=0$, thus $$p'=mv.$$ Should expression for energy change? Surely it does:

$$E'=v\;\partial_vL-L=\frac{mv^2}2+U$$

Now what is different is that $p$ is no longer a conserved quantity, so defining a single value of momentum doesn't describe motion of our particle in potential.

Quantum mechanics is quite similar. Momentum operator by definition is an operator, whose eigenstates are states with definite momentum, and corresponding eigenvalues are those definite values of momentum. Hamiltonian is by definition an operator, whose eigenstates are states with definite energy, and corresponding eigenvalues are those definite values of energy.

Now, as inhomogeneous potential makes momentum not conserve, we no longer can say that eigenstates of momentum operator remain eigenstates of energy operator. Moreover, as in classical mechanics, we can't specify single value of momentum to describe eigenstate of Hamiltonian: in state with definite total energy momentum can be different.

Another way to think of this: position operator is defined as an operator, whose eigenvalues are positions and corresponding eigenstates are states with definite position, i.e. in position representation the states are $\delta(x-x_0)$. Now, this operator would commute with energy operator only for some very weird unphysical potentials (I'm not even sure how to construct them if it's possible at all). Do we change this operator for normal potentials? Of course not, position as a physical quantity doesn't depend on potential. Yet its operator in momentum representation has the same form as momentum operator in position representation:

$$\left\langle p\right|\hat r\left| p'\right\rangle=i\hbar\nabla\delta(p-p').$$

Compare with momentum operator in position representation:

$$\left\langle r\right|\hat p\left| r'\right\rangle=-i\hbar\nabla\delta(r-r').$$

share|improve this answer
up vote 0 down vote accepted

I found a proof that the momentum operator does not change when the wave it operates upon is not a plane wave:

Let $\phi(k) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-ikx}\psi(x)dx$

be the Fourier transform of an (arbitrary) wave function in position space.

The expectation value of k - using the wave function $\psi(x)$ in position space - is:

$\overline{k}=\int\limits_{-\infty}^{\infty}\psi^*(x)k\psi(x)dx$

Rewriting the above expression with the Fourier Transform of $\psi(x)$ we get:

$\overline{k}=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)ke^{ikx}\phi(k) dxdk = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)ke^{ikx}e^{-ikx'}\psi(x')dxdkdx'$

We can now write $ke^{-ikx'}$ in the above expression as $i\frac{\partial}{\partial x'} e^{ikx'}$:

$\overline{k}=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)e^{ikx}[i\frac{\partial}{\partial x'} e^{ikx'}]\psi(x')dxdkdx'$

Let us integrate the two last factors containing x' by parts, using the fact that the wave equation $\psi(x')$ will vanish at infinity:

$\overline{k}= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)\frac{1}{i}\frac{\partial\psi(x')}{\partial x'}e^{ik(x-x')}dx dkdx'$

The Integral over k of the last exponential together with the factor $\frac{1}{2\pi}$is the Dirac Delta Function, so when we integrate over x, we end up with:

$\overline{k}=\int\limits_{-\infty}^{\infty}\psi^*(x')\frac{1}{i}\frac{\partial\psi(x')}{\partial x'}dx'$

Using the de Broglie equation p = $\hbar k$ and using x instead of x' for clarity we get:

$\overline{p}=\int\limits_{-\infty}^{\infty}\psi^*(x)\frac{\hbar}{i}\frac{\partial\psi(x)}{\partial x}dx$

where we see that the momentum operator $\hat{p}=-i\hbar\partial_x$

Exactly the same demonstration can be made with $\omega$ and $t$ instead of $k$ and $x$, and the result is that the Energy operator is $\hat{E}= i\hbar\partial_t$.

However, I'm still not completely convinced by this argument, because I don't really know what physical interpretation to give to the waves $e^{ikx}$ and $e^{-i\omega t}$ in which we have Fourier-analysed the wave function $\psi (x,t)$ under consideration. We have presumed that they are plane waves with Momentum $\hbar k $ and Energy $\hbar \omega$, respectively, but $e^{ikx}$ is time independent and $e^{-i\omega t}$ is space independent - Something like this is completely unphysical, much more so than a plane wave $e^{i(kx-\omega t)}$ which might not exist for practical purposes, but could at least theoretically exist and has definite momentum AND Energy.

But I believe this is is as good as it gets.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.