For a report I'm writing on Quantum Computing, I'm interested in understanding a little about this famous equation. I'm an undergraduate student of math, so I can bear some formalism in the explanation. However I'm not so stupid to think I can understand this landmark without some years of physics. I'll just be happy to be able to read the equation and recognize it in its various forms.

To be more precise, here are my questions.

Hyperphysics tell me that Shrodinger's equation "is a wave equation in terms of the wavefunction".

  1. Where is the wave equation in the most general form of the equation?

    $$\mathrm{i}\hbar\frac{\partial}{\partial t}\Psi=H\Psi$$

    I thought wave equation should be of the type


    It's the difference in order of of derivation that is bugging me.

    From Wikipedia

    "The equation is derived by partially differentiating the standard wave equation and substituting the relation between the momentum of the particle and the wavelength of the wave associated with the particle in De Broglie's hypothesis."

  2. Can somebody show me the passages in a simple (or better general) case?

  3. I think this questions is the most difficult to answer to a newbie. What is the Hamiltonian of a state? How much, generally speaking, does the Hamiltonian have to do do with the energy of a state?

  4. What assumptions did Schrödinger make about the wave function of a state, to be able to write the equation? Or what are the important things I should note in a wave function that are basilar to proof the equation? With both questions I mean, what are the passages between de Broglie (yes there are these waves) and Schrödinger (the wave function is characterized by)?

  5. It's often said "The equation helps finds the form of the wave function" as often as "The equation helps us predict the evolution of a wave function" Which of the two? When one, when the other?

  • 3
    $\begingroup$ Philisophically I always find requests to explain an equation for the laymen to be a little strange. The point of writing it in math is to have a precise, and complete representation of the theory... $\endgroup$ Commented Dec 15, 2012 at 16:13
  • $\begingroup$ You're right. That's why I tried to make it clear I'm not asking an explanation of the "equation" as you mean it, rather the meaning of the "symbols in it". In particulart question number 1 is the most important for me now. $\endgroup$
    – Temitope.A
    Commented Dec 15, 2012 at 17:04
  • $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$
    – Qmechanic
    Commented Dec 15, 2012 at 18:21

3 Answers 3


What you write is the time-dependent Schrödinger equation. This is not the equation of a true wave. He postulated the equation using a heuristic approach and some ideas/analogies from optics, and he believed on the existence of a true wave. However, the correct interpretation of $\Psi$ was given by Born: $\Psi$ is an unobservable function, whose complex square $|\Psi|^2$ gives probabilities. In older literature $\Psi$ is still named the wavefunction, In modern literature the term state function is preferred. The terms "wave equation" and "wave formulation" are legacy terms.

In fact, part of the confusion that had Schrödinger, when he believed that his equation described a physical wave, is due to the fact he worked with single particles. In that case $\Psi$ is defined in an abstract space which is isomorphic to the tri-dimensional space. However, when you consider a second particle and write $\Psi$ for a two-body system, the isomorphism is broken and the superficial analogy with a physical wave is completely lost. A good discussion of this is given in Ballentine textbook on quantum mechanics (section 4.2).

The Schrödinger equation cannot be derived from wave theory. This is why the equation is postulated in quantum mechanics.

There is no Hamiltonian for one state; the Hamiltonian is characteristic of a given system with independence of its state. Energy is a possible physical property of a system, one of the possible observables of a system; it is more correct to say that the Hamiltonian gives the energy of a system in the cases when the system is in a certain state. A quantum system always has a Hamiltonian, but not always has a defined energy. Only certain states $\Psi_E$ that satisfy the time-independent Schrödinger equation $H\Psi_E = E \Psi_E$ are associated to a value $E$ of energy. The quantum system can be in a superposition of the $\Psi_E$ states or can be in more general states for which energy is not defined.

Wavefunctions $\Psi$ have to satisfy a number of basic requirements such as continuity, differentiability, finiteness, normalization... Some texts emphasize that the wavefunctions would be single-valued, but I already take this in the definition of function.

The Schrödinger equation gives both "the form of the wave function" and "the evolution of a wave function". If you know $\Psi$ at some initial time and integrate the time-dependent Schrödinger equation you obtain the form of the wavefunction to some other instant: e.g. the integration is direct and gives $\Psi(t) = \mathrm{Texp}(-\mathrm{i}/\hbar \int_0^t H(t') dt') \Psi(0)$, where $\mathrm{Texp}$ denotes a time-ordered exponential. This equation also gives the evolution of the initial wavefunction $\Psi(0)$. When the Hamiltonian is time-independent, the solution simplifies to $\Psi(t) = \exp(-\mathrm{i}Ht/\hbar) \Psi(0)$.

For stationary states, the time-dependent Schrödinger equation that you write reduces to the time-independent Schrödinger equation $H\Psi_E = E \Psi_E$; the demonstration is given in any textbook. For stationary states there is no evolution of the wavefunction, $\Psi_E$ does not depend on time, and solving the equation only gives the form of the wavefunction.

  • 1
    $\begingroup$ Good answer. I would only add that regarding the last point, I think the confusion comes from references to the "time-independent" Schrodinger eigenvalue equation $H\psi_E = E\psi_E$ being conflated with the "time-dependent" evolution equation $\mathrm{i}\hbar \dot{\psi} = H\psi$, when of course the two are entirely different beasts. $\endgroup$
    – user10851
    Commented Dec 15, 2012 at 21:07
  • $\begingroup$ @ChrisWhite Good point. Made. $\endgroup$
    – juanrga
    Commented Dec 16, 2012 at 2:33
  • $\begingroup$ 6 paragraph: maybe you should add that the equation only holds if H is time-independent. $\endgroup$
    – ungerade
    Commented Dec 16, 2012 at 12:19
  • $\begingroup$ @ungerade Another good point! Added evolution when H is time-dependent. $\endgroup$
    – juanrga
    Commented Dec 16, 2012 at 12:49

You should not think of the Schrödinger equation as a true wave equation. In electricity and magnetism, the wave equation is typically written as

$$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$$

with two temporal and two spatial derivatives. In particular, it puts time and space on 'equal footing', in other words, the equation is invariant under the Lorentz transformations of special relativity. The one-dimensional time-dependent Schrödinger equation for a free particle is

$$ \mathrm{i} \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$

which has one temporal derivative but two spatial derivatives, and so it is not Lorentz invariant (but it is Galilean invariant). For a conservative potential, we usually add $V(x) \psi$ to the right hand side.

Now, you can solve the Schrödinger equation is various situations, with potentials and boundary conditions, just like any other differential equation. You in general will solve for a complex (analytic) solution $\psi(\vec r)$: quantum mechanics demands complex functions, whereas in the (classical, E&M) wave equation complex solutions are simply shorthand for real ones. Moreover, due to the probabilistic interpretation of $\psi(\vec r)$, we make the demand that all solutions must be normalized such that $\int |\psi(\vec r)|^2 dr = 1$. We're allowed to do that because it's linear (think 'linear' as in linear algebra), it just restricts the number of solutions you can have. This requirements, plus linearity, gives you the following properties:

  1. You can put any $\psi(\vec r)$ into Schrödinger's equation (as long as it is normalized and 'nice'), and the time-dependence in the equation will predict how that state evolves.

  2. If $\psi$ is a solution to a linear equation, $a \psi$ is also a solution for some (complex) $a$. However, we say all such states are 'the same', and anyway we only accept normalized solutions ($\int |a\psi(\vec r)|^2 dr = 1$). We say that solutions like $-\psi$, and more generally $e^{i\theta}\psi$, represent the same physical state.

  3. Some special solutions $\psi_E$ are eigenstates of the right-hand-side of the time-dependent Schrödinger equation, and therefore they can be written as $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_E}{\partial x^2} = E \psi_E$$ and it can be shown that these solutions have the particular time dependence $\psi_E(\vec r, t) = \psi_E(\vec r) e^{-i E t/\hbar}$. As you may know from linear algebra, the eigenstates decomposition is very useful. Physically, these solutions are 'energy eigenstates' and represent states of constant energy.

  4. If $\psi$ and $\phi$ are solutions, so is $a \psi + b \phi$, as long as $|a|^2 + |b|^2 = 1$ to keep the solution normalized. This is what we call a 'superposition'. A very important component here is that there are many ways to 'add' two solutions with equal weights: $\frac{1}{\sqrt 2}(\psi + e^{i \theta} \phi)$ are solutions for all angles $\theta$, hence we can combine states with plus or minus signs. This turns out to be critical in many quantum phenomena, especially interference phenomena such as Rabi and Ramsey oscillations that you'll surely learn about in a quantum computing class.

Now, the connection to physics.

  1. If $\psi(\vec r, t)$ is a solution to the Schrödinger's equation at position $\vec r$ and time $t$, then the probability of finding the particle in a specific region can be found by integrating $|\psi^2|$ around that region. For that reason, we identify $|\psi|^2$ as the probability solution for the particle.

    • We expect the probability of finding a particle somewhere at any particular time $t$. The Schrödinger equation has the (essential) property that if $\int |\psi(\vec r, t)|^2 dr = 1$ at a given time, then the property holds at all times. In other words, the Schrödinger equation conserves probability. This implies that there exists a continuity equation.
  2. If you want to know the mean value of an observable $A$ at a given time just integrate $$ <A> = \int \psi(\vec r, t)^* \hat A \psi(\vec r, t) d\vec r$$ where $\hat A$ is the linear operator associated to the observable. In the position representation, the position operator is $\hat A = x$, and the momentum operator, $\hat p = - i\hbar \partial / \partial x$, which is a differential operator.

The connection to de Broglie is best thought of as historical. It's related to how Schrödinger figured out the equation, but don't look for a rigorous connection. As for the Hamiltonian, that's a very useful concept from classical mechanics. In this case, the Hamiltonian is a measure of the total energy of the system and is defined classically as $H = \frac{p^2}{2m} + V(\vec r)$. In many classical systems it's a conserved quantity. $H$ also lets you calculate classical equations of motion in terms of position and momentum. One big jump to quantum mechanics is that position and momentum are linked, so knowing 'everything' about the position (the wavefunction $\psi(\vec r))$ at one point in time tells you 'everything' about momentum and evolution. In classical mechanics, that's not enough information, you must know both a particle's position and momentum to predict its future motion.

  • $\begingroup$ Thank you! One last question. How do somebody relate the measurment principle to the equations, that an act of measurment will cause the state to collapse to an eigenstate? Or is time a concept indipendent of the equation? $\endgroup$
    – Temitope.A
    Commented Dec 16, 2012 at 11:37
  • $\begingroup$ Can states of entanglement be seen in the equation to? $\endgroup$
    – Temitope.A
    Commented Dec 16, 2012 at 11:47
  • $\begingroup$ Note that user10347 talks of a potential added to the differential equation. To get real world solutions that predict the result of a measurement one has to apply the boundary conditions of the problem. The "collapse" vocabulary is misleading. A measurement has a specific probability of existing in the space coordinates or with the fourvectors measured. The measurement itself disturbs the potential and the boundary conditions change, so that after the measurement different solutions/psi functions will apply. $\endgroup$
    – anna v
    Commented Dec 16, 2012 at 13:23
  • $\begingroup$ One type of measurement is strong measurement, where we the experimentalists, measure some differential operator $A$, and find some particular (real) number $a_i$, which is one of the eigenvalues of $A$. (Important detail: for $A$ to be measureable, it must have all real eigenvalues.) Then, we know the wavefunction "suddenly" turns into $\psi_i$, which is the eigenfunction of $A$ whose eigenvalue was that number $a_i$ we measured. The system has lost of knowledge of the original wavefunction $\psi$. The probability of measuring $a_i$ is $|<\psi_i | \psi>|^2$. $\endgroup$
    – emarti
    Commented Dec 18, 2012 at 7:12
  • $\begingroup$ @Temitope.A: Entanglement isn't obvious in anything here because I've only written single-particle wavefunctions. A two-particle wavefunction $\Psi(\vec r_1, \vec r_2)$ gives a probability $\int_{V_1}\int_{V_2}|\Psi|^2 d \vec r_1 d \vec r_2$ of detecting one particle in a region $V_1$ and a second particle in a region $V_2$. A simple solution for distinguishable particles is $\Psi(\vec r_1, \vec r_2) = \psi_1(\vec r_1) \psi_2(\vec r_2)$, and it can be shown that this satisfies all our conditions. An entangled state cannot be written so simply. (Indistinguishable particles take more care.) $\endgroup$
    – emarti
    Commented Dec 18, 2012 at 9:32

If you take the wave equation $$\nabla^2\phi = \frac{1}{u^2}\frac{d^2\phi}{dt^2}\text{,}$$ and consider a single frequency component of a wave while taking out its time dependence, $\phi = \psi e^{-i\omega t}$, then: $$\nabla^2 \phi = -\frac{4\pi^2}{\lambda^2}\phi\text{,}$$ but that means the wave amplitude should satisfy an equation of the same form: $$\nabla^2 \psi = -\frac{4\pi^2}{\lambda^2}\psi\text{,}$$ and if you know the de Broglie relation $\lambda = h/p$, where for a particle of energy $E$ in a potential $V$ has momentum $p = \sqrt{2m(E-V)}$, so that: $$\underbrace{-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi}_{\hat{H}\psi} = E\psi\text{,}$$ Therefore, the time-independent Schrödinger equation has a connection to the wave equation. The full Schrödinger equation can be recovered by putting time-dependence back in, $\Psi = \psi e^{-i\omega t}$ while respecting the de Broglie $E = \hbar\omega$: $$\hat{H}\Psi = (\hat{H}\psi)e^{-i\omega t} = \hbar\omega \psi e^{-i\omega t} = i\hbar\frac{\partial\Psi}{\partial t}\text{,}$$ and then applying the principle of superposition for the general case.

However, in this process the repeated application of the de Broglie relations takes us away from either classical waves or classical particles; to what extent the resulting "wave function" should be considered a wave is mostly a semantic issue, but it's definitely not at all a classical wave. As other answers have delved into, the proper interpretation for this new "wave function" $\Psi$ is inherently probabilistic, with its modulus-squared representing a probability density and the gradient of the complex phase being the probability current (scaled by some constants and the probability density).

As for the de Broglie relations themselves, it's possible to "guess" them from making an analogy from waves to particles. Writing $u = c/n$ and looking for solutions close to plane wave in form, $\phi = e^{A+ik_0(S-ct)}$, the wave equation gives: $$\begin{eqnarray*} \nabla^2A + (\nabla A)^2 &=& k_0^2[(\nabla S)^2 - n^2]\text{,}\\ \nabla^2 S +2\nabla A\cdot\nabla S &=& 0\text{.} \end{eqnarray*}$$ Under the assumption that the index of refraction $n$ changes slowly over distances on the order of the wavelength, then $A$ does not vary extremely, the wavelength is small, and so $k_0^2 \propto \lambda^{-2}$ is large. Therefore the term in the square brackets should be small, and we can make the approximation: $$(\nabla S)^2 = n^2\text{,}$$ which is the eikonal equation that links the wave equation with geometrical optics, in which motion of light of small wavelengths in a medium of well-behaved refractive index can be treated as rays, i.e., as if described by paths of particles/corpuscles.

For the particle analogy to work, the eikonal function $S$ must take the role of Hamilton's characteristic function $W$ formed by separation of variables from the classical Hamilton-Jacobi equation into $W - Et$, which forces the latter to be proportional to the total phase of the wave, giving $E = h\nu$ for some unknown constant of proportionality $h$ (physically Planck's constant). The index of refraction $n$ corresponds to $\sqrt{2m(E-V)}$.

This is discussed in, e.g., Goldstein's Classical Mechanics, if you're interested in details.

  • $\begingroup$ Your first equation is a wave equation, only if you substitute the total time derivatives by partial ones. Moreover, you introduce a $\Psi = \psi e^{-i\omega t} = \phi$, but the wavefunction $\Psi$ does not satisfy the first equation for a wave. $\endgroup$
    – juanrga
    Commented Dec 18, 2012 at 11:21

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