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If I now see the Schrödinger equation, I just see a bunch of weird symbols, but I want to know what it actually means. So I'm taking a course of Linear Algebra and I'm planning on starting with PDE's next month. What 'other math' is needed other than Linear Algebra and (Partial) Differential Equations to gain a full understanding of this equation?

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2 Answers 2

Look at the following questions previously asked on physics.SE.

Summarizing that answers with my experience, you should know linear algebra (but, I think, it's better to know math behind Hilbert space - in infinite dimensions, which is relevant for eigenvalues), PDE, a bit of calculus (Dirac-$\delta$ function, Fourier transform etc). That's the most basic knowledge which is needed by QM. And, of course, Statistics and Probability.

If you know some common families of polynomial (Legendre, Laquerre, Airy) that would be a big plus.

For some advanced topics, group theory is required.

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But do you need to know all these mathematical formalisms to understand the essence of the equation? According to Wikipedia linked to in the question, the equation is nothing more than (Total energy) = (kinetic energy) + (potential energy). Is there more to it? –  Zeynel Oct 14 '12 at 13:25
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@Zeynel That's kind of like saying if you know force = mass * acceleration, you are a rocket scientist. At the end of the day, all physics boils down to conserving something (mass, momentum, energy, charge) but that doesn't mean if you know that something is conserved you understand the subject or equations. –  tpg2114 Oct 14 '12 at 15:57
    
More to the point, the math helps you understand the nature of the terms. Is it a source or a sink? Linear or nonlinear? What would solutions look like in each of those cases? How does the solution change when boundary conditions change? What are the relative magnitude of the terms? All of those are questions that I think one would have to be comfortable answering to say they understand the essence of the equation. –  tpg2114 Oct 14 '12 at 16:00

To have some feeling what the equation

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

as such is about, here some heuristics:

A quantum mechanical operators like $P$, which give measurable physical quantities, must contain information about real numbers (not complex ones). This is ensured by the operators being self-adjoint: $P^\dagger = P$. A consequence is that if you have a svalar product $\langle\Psi,\Phi\rangle$, then for hermitean operator, you have $\langle\Psi,P\Phi\rangle=\langle P^\dagger\Psi,\Phi\rangle=\langle P\Psi,\Phi\rangle$. I'm not going into the distinction between self-adjoint and hermitian here, but the point is that they generalize the notion of a symmetric matrix.

The Hamiltonian operator $H$ is such a "symmetric" operator

$$H^\dagger = H,$$ and then the $i$ makes the physical interpretation of $\Psi$ sensible:

The operator $$I:=-i \hbar\ \text{1}$$ is anti-self-adjoint

$$I^\dagger = (-i \hbar\ \text{1})^\dagger= -i^* \hbar\ \text{1}=-I,$$ and, as a consequence, the product $$A:=-i\hbar H=IH$$ is as well

$$A^\dagger=(IH)^\dagger = (HI)^\dagger = I^\dagger H^\dagger=-IH=-A.$$

Bringing the number $i\hbar$ in the Schrödinger equation on the other side, the equation reads

$$\frac{\partial}{\partial t}\Psi=A\ \Psi.$$

(With the formal solution $\Psi=\text e^{At}\Psi_0$.)

We are happy with $A$ being anti-self adjoint: The Schrödinger equation makes the time evolution unitary, similar to how in this example

http://www.wolframalpha.com/input/?i=matixExp%5B%7B%7B0%2C-1%7D%2C%7B1%2C0%7D%7Dt%5D

the exponential of an anti-symmetric matrix is a rotation.

For example, if $\Psi$ is a wave function, $\langle\Psi,\Psi\rangle$ is understood to be the total propability for find the particle/particles it describes. This quantity shound not change in time. And this is ensured by $A$ being anti-self-adjoint:

$$\frac{\partial}{\partial t}\langle\Psi,\Psi\rangle =\langle\frac{\partial}{\partial t}\Psi,\Psi\rangle+\langle\Psi,\frac{\partial}{\partial t}\Psi\rangle =\langle A\Psi,\Psi\rangle+\langle\Psi,A\Psi\rangle= $$ $$=\langle A\Psi,\Psi\rangle+\langle A^\dagger\Psi,\Psi\rangle =\langle A\Psi,\Psi\rangle+\langle (-A)\Psi,\Psi\rangle =0. $$

$$\Longrightarrow \langle\Psi,\Psi\rangle=\text{const.}$$

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