What math is needed to understand the Schrödinger equation?

Question

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?

Answer 1

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

• What is the math knowledge necessary for starting Quantum Mechanics?
• study quantum mechanics without physics background
• Mathematics for Quantum Mechanics

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.

Answer 2

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:

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

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

Answer 3

The simple answer: everything. I think it's the part where you need to understand more than basic level in each and every sector of maths. Mastering on Calculus II, Linear Algebra II, Statistics and Probability II is a must. Not to mention multi-variable calculus, differential equations up to differential geometry (which of course includes Tensor calculus). And lots and lots of patience. I'm afraid it's not something prodigious and requires much more than perseverance.