Lecture (xx minutes)

Having established the connection between the wave function and the column vector as representations of the quantum state, for the continuous and discrete observables, respectively, it now remains to make some of the other connections that will be familiar to students who have studied the spin-1/2 system.

The complex conjugate of the wave function follows easily: $\left \langle x| \psi \right\rangle = \psi ^*\left( x \right)$ $\left\langle \psi \right| \buildrel\textstyle.\over= {\psi ^*}\left( x \right)$

The notion of a probability density is new. It is a continuous function as opposed to the discrete probability functions the students encountered in Spins. (Relate to other continuous functions like height, for example.) Although the probability density was introduced in an earlier Modern Physics course, it now has much more impact.

$\wp \left( x \right) \equiv \psi ^*\left( x \right)\psi \left( x \right) = {\left| {\psi \left( x \right)} \right|^2}$

Discuss the dimensions of probability density and introduce the integral between two values of $x$ to calculate probability, which is dimensionless.

${\wp _{a < x < b}} = \int\limits_a^b {\psi ^*\left( x \right)\psi \left( x \right)dx}$

Discuss normalized probability and relate to the same concept in the Spins course.

Note from Winter 2012 (Mary Bridget Kustusch):

Students were really bothered by the probability of finding a particle somewhere in space being $\langle{\psi}\vert{\psi}\rangle$ and not $\vert\langle{\psi}\vert{\psi}\rangle\vert^2$. It helped to go back to the idea of completeness and projectors to show how they actually were already doing the norm squared when looking at normalization. I think it also illuminated where the integral comes from $$\langle{\psi}\vert{\psi}\rangle=\bra{\psi}\left(\sum_n \ket{x_n}\bra{x_n}\right)\ket{\psi}=\sum_n \langle{\psi}\vert{x_n}\rangle\langle{x_n}\vert{\psi}\rangle= \sum_n \vert\langle{x_n}\vert{\psi}\rangle\vert^2 =\int_{-\infty}^{\infty} dx\,\, \psi^*(x)\psi(x)=\int_{-\infty}^{\infty} \vert\psi(x)\vert^2 dx=1$$

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