The Schrödinger equation is just another way of writing the conservation of energy, right? So how can you use it to find the quantum wavefunction? I mean in every example I've seen the wavefunction is calculated using sines and cosines so I don't see how you can use the Schrödinger equation to find a wavefunction. All I can see it being useful in is finding energy levels which doesn't help calculate the wavefunction. I'm pretty sure I'm wrong about everything I said because the Schrödinger equation is a big deal in QM, so can you please explain me what this is actually used for? I've had this question for ages now. Any help would be helpful.

  • 1
    $\begingroup$ Are you talking about the time-dependent or the time-independent Schrödinger equation? $\endgroup$
    – ACuriousMind
    Commented Sep 27, 2020 at 15:56
  • $\begingroup$ well...both if you could... first time independent then time dependent . thanx $\endgroup$ Commented Sep 27, 2020 at 16:18
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/581926/2451 $\endgroup$
    – Qmechanic
    Commented Sep 27, 2020 at 16:45
  • $\begingroup$ One use of the wavefunction of a particle is to compute its expectation value of position, momentum, etc. That is a big subject but you can find numerous examples of them on the web. $\endgroup$
    – Bill Watts
    Commented Sep 27, 2020 at 17:58
  • 1
    $\begingroup$ The Schrödinger equation is just another way of writing the conservation of energy, right? No, this is a mischaracterization of it. $\endgroup$
    – G. Smith
    Commented Sep 27, 2020 at 19:02

3 Answers 3


The Schrödinger equation is just another way of writing the conservation of energy, right?

Wrong. The original motivation for Erwin Schrödinger can be read in his 1926 paper An Undulatory Theory of the Mechanics of Atoms and Molecules (behind paywall, but freely readable copies can easily be found on the web).

Namely, the ansatz is the Hamilton-Jacobi equation, whose solution $W(x,y,z,t)$ is seen as a set of constant-value surfaces in $(x,y,z)$ propagating in space as time $t$ changes, one (possibly disjoint) surface for a chosen constant value. Reinterpreting these surfaces as the constant-phase surfaces, or wavefronts, of some waves, gives us the idea of the quantum wavefunction (whose actual interpretation was not known until Max Born formulated the rule named after him).

Now, when we try to solve the time-dependent Schrödinger's equation, we find that it can be in general variable-separated for the time dimension $t$ to yield an expansion of the wavefunction in eigenfunctions of the Hamiltonian operator. This then gives us the time-independent Schrödinger's equation for the individual eigenfunctions.

This is not much different from what we do when we solve e.g. the wave equation for oscillation of a drum-like membrane. The nodes of the eigenmodes of vibration of such a membrane form the famous Chladni figures.

I mean in every example I've seen the wavefunction is calculated using sines and cosines so I don't see how you can use the Schrödinger equation to find a wavefunction.

The cases when you can form a wavefunction with merely sines and cosines are very rare. They are the most trivial types of problems that you can solve with Schrödinger's equation. Even the harmonic oscillator in quantum mechanical description requires the use of a set of functions that are definitely not sines nor cosines.

To understand how to use the Schrödinger's equation to find a wavefunction (I'm assuming you mean the eigenfunction of Hamiltonian, rather than time evolution of an arbitrary wavefunction), try reading the derivations of the solutions of various problems like quantum harmonic oscillator and hydrogen atom.


There are two equations commonly called the Schrödinger equation. Most physicists I know would prefer to call*:

$$i\hbar \frac{d}{dt}|\psi\rangle = \hat{H}|\psi\rangle$$

'the' Schrödinger equation. This is also called the 'time dependent' Schrödinger equation. It is the quantum mechanical replacement for Newton's second law:

$$ m \frac{d^2}{dt^2}\vec{r} = F(\vec{r}).$$

The TDSE tells you what $|\psi\rangle$ will be at time $t+dt$ if you know $|\psi\rangle$ at time $t$ and the operator $\hat{H}$. By incrementing by small time steps in this manner you can predict the state of a quantum system arbitrarily far into the future if you can solve the TDSE.

The other equation people call the Schrödinger equation is the 'time independent Schrodinger equation'. This is a mathematical tool used in the process of solving the full TDSE. This equation reads something like:

$$\hat{H}|E\rangle = E|E\rangle$$

where now we interpret the state $|E\rangle$ not an arbitrary state, but a state with a definite value of energy equal to $E$. The TISE then is read as an equation that, given the form of $\hat{H}$ tells lets you solve for the shape of states with each definite energy $E$. For many models it will turn out that this is only solvable for certain discrete values of $E$. That means that there are not states with intermediate definite energies, which in turn gives rise to the discreteness of atomic spectra etc (and the name 'quantum', ultimately).

*P.S. the notation $|\text{stuff}\rangle$ just means 'quantum state of system'. It's convenient because it means you can label a state $|E\rangle$ because it has energy $E$ and not get the symbol for $|E\rangle$ the state and $E$ the number confused.


Essential knowledge about the Schrödinger-equation SE (I limit my post to the time-independent equation) is that it is an eigen-value equation.

I assume that you know the basics of linear algebra so if you take a, for instance a 2x2 matrix $A$, an eigen-value problem can be set up: Let's assume

$$A= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$

there is the following eigenvalue problem:

$$ A v = \lambda v$$

that has a simple solution: $\lambda_{1,2} = \pm 1$ and $v_1 =\left( \begin{array}{c} 1\\ 1 \end{array} \right)$ and $v_2 =\left( \begin{array}{c} -1\\ 1 \end{array} \right)$. In particular the space of solutions

$$ v= \langle v,v_1\rangle v_1 + \langle v,v_2\rangle v_2$$

spanned by $v_1$ and $v_2$ is 2-dimensional.

The Schroedinger-equation is basically the same with a couple of differences: the matrix $A$ is substituted by the Hamilton-operator $H$ and $\lambda$ are the energy (eigen)-values and $v$ corresponds to the wave-function $\psi$.

$$ H\psi = E\psi$$

However, the subtle difference of the SE to the linear algebra problem is that the space the eigen-vectors of this equation belong to is infinite (and called Hilbert-space). One strategy of finding the wave-function is searching for a basis in this infinite space. Evidently this is not so trivial. But the tools of mathematical physics provide us with a whole pool of special functions.

For instance let's look at the 1-dim. quantum harmonic oscillator. In this case we are lucky since eigen-vectors can be constructed from the Hermite-polynomials $H_n$ that have the following form (and are of infinite number):

$$\psi_n(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}\frac{1}{\sqrt{2^n n!}} \exp(-\frac{m\omega}{2 \hbar} x^2) H_n(x\sqrt{\frac{m\omega}{\hbar}})\quad \text{for} \quad n=0,1,2,3,\ldots $$

and the solution of the SE can be constructed as an infinite linear combination of these eigen-vectors:

$$\psi = \sum_{n=0}^{\infty} \langle \psi, \psi_n\rangle \psi_n $$

with $\langle \cdot, \cdot\rangle$ is the scalar product on the Hilbert space. We are even more lucky: Here (that's not always the case!) each element of the infinite basis even fulfills the eigen-value equation:

$$H\psi_n =E_n \psi_n \quad \text{with} \quad E_n =\hbar\omega (n+\frac{1}{2})$$

Of course, there might be problems which do not require to compute the eigen-vectors, where the computation of the eigenvalues is sufficient. This also happens in linear algebra. But the eigenvectors, i.e. for the SE the wave-function, fulfills an essential role in the SE-eigenvalue problem.

A final word: the domain where the problem is posed also plays an essential role. In case of the harmonic oscillator it is $(-\infty,\infty)$, but for infinitely deep potential well it is $[-a, a]$. In the latter case the basis of the Hilbert space (here you might indeed get $\sin$'s and $\cos$'s) is a completely different one to the basis of the Hilbert space in case of the 1-dim. harmonic oscillator.

Moreover, symmetry and dimension plays a crucial role: Again, an appropiate basis of the Hilbert space for the 3-dim. hydrogen atom is completely different to the one's already mentioned.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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