particle in a box problem

  • the schrodinger’s equation (SE) is used for discrete states at very specific energy levels

linearity and normalization

  • schrodinger’s equation is linear
  • the wavefunction \( \psi \) appears only in first order
    • there is no second or higher order terms
    • such as \(\psi^2\) or \(\psi^3\)
  • so, if \( \psi \) is a solution, so is \(a\cdot\psi \)
    • regardless of the constant \(a\) value
  • quantum mechanics linearity is very general
    • QM equations are linear in the QM amplitude for which the equation is being solved
  • unlike CM, this is an absolute property in QM
  • allows use of linear algebra for QM math

normalization

  • since SE is linear, the wave function can be normalized

  • born’s postulate says that the probability of finding a particle near some point vector \( \textbf{r} \)
    • proportional to the modulus squared of the wave function
    • mathematically, \( P(\textbf{r}) \propto \lvert \psi(\textbf{r}) \rvert^2 \)
  • assume \(P(\textbf{r})\) to be a probability density
    • for some infinitesimal volume \(d^3 \textbf{r}\) around \(\textbf{r}\)
  • the probability of finding the particle in that volume is \( P(\textbf{r})d^3\textbf{r} \)
  • the sum of all such probabilities should be 1
    • \( \int P(\textbf{r})d^3\textbf{r} = 1 \)
  • this integral usually gives some other real positive number
    • this necessitates normalization
  • consider the integral output to be
    • \( \frac{1}{\lvert a \rvert^2} \)
    • \(a\) is possibly complex

  • so \( \int \lvert \psi(\textbf{r}) \rvert^2 d^3\textbf{r} = \frac{1}{\lvert a \rvert^2}\)

  • we know that \( \psi(\textbf{r}) \) is a solution to SE
    • so even \(a\psi(\textbf{r})\) is a solution
    • consequence of linearity of SE
  • so, assume a solution \( \psi_N = a\psi \) instead of \( \psi \)
    • then \( \int \lvert \psi_N (\textbf{r}) \rvert^2 d^3\textbf{r} = 1 \)
  • we can use \( \int \lvert \psi_N (\textbf{r}) \rvert^2 \) as the probability density
    • \(P(\textbf{r}) = \int \lvert \psi_N (\textbf{r}) \rvert^2 \)
    • this is the “normalized wavefunction”

steps to normalize

  • take the solution \( \psi \) obtained from SE
  • integrate \( \lvert \psi(\textbf{r}) \rvert^2 \) to get \( \frac{1}{\lvert a \rvert^2} \)
  • find \(a\) to obtain \( \psi_N = a\psi \)
    • for which \( \int \lvert \psi_N (\textbf{r}) \rvert^2 d^3 \textbf{r} = 1 \)
  • the new probability density is \( \lvert \psi_n(\textbf{r}) \rvert^2 \)

note

  • normalization only sets the magnitude of \(a\)
    • not the phase
    • free to choose any phase for \(a\)
  • if space is infinite, \( \sin(kx), \cos(kz) and \exp{(i\textbf{k}\cdot\textbf{r})} \) cannot be normalized this way
    • their squared modulus is not “Lebesgue integrable” i.e. not L2 functions
    • result of over-idealizing the math to get functions that are simple to use
      • this is a mathematical difficulty, not physical
  • workarounds:
    1. work in finite volumes in actual problems
    2. use normalization to a delta function introduces another infinity to compensate for the first one