Well, the problem is that if $x$ is continuously distributed (i.e. we have an operator $X$ whose eigenvalues form a continuous spectrum $\Sigma_X$), typically $P(x=\lambda)=0$ -- and this gives us very little information about the actual probability distribution. What we're really interested in is $P(x\in B)$ for $B$ some subset of $\Sigma_X$.

Technically, we need $B$ to be a "Borel subset", or "measurable subset". We will be omitting several such technicalities in the article, such as the need for the spectral theorem to define a "projection-valued measure" or "spectral measure" on an operator with a continuous spectrum -- this is something that will be covered in the MAO1103: Linear Algebra course.

First, let's think about $P(x\in B)$ in the countable case. One can write $B=\{\lambda_1,\ldots\lambda_n\}$, and then simply say that

$$P(x\in B)=\sum |\langle\psi|\phi_k\rangle|^2$$

But the term on the right is a Pythagorean sum -- specifically, it is the length-squared of the vector formed by summing all the projections of $|\psi\rangle$ onto the eigenstates $|\phi_1\rangle\ldots|\phi_k\rangle$. But this is the same as the length of the projection of $|\psi\rangle$ onto the span of these eigenstates.

(

**Note on notations:**From here onwards, we will use the notation $|\lambda\rangle$ to refer to the eigenvector corresponding to the eigenvalue $\lambda$ (if the eigenspace has dimension more than 1, we'll figure something out). We will use the notation $\{|B\rangle\}$ to refer to the span of the eigenvectors corresponding to the eigenvalues in $B$.)

So we could just define a Hermitian

**projection operator**$L_X(B)$ for any subset $B$ of the spectrum of $X$ -- it is an easy exercise to write down an explicit form for $L_X(B)$ in terms of the eigenvectors of $X$.

Then the probability $P(x\in B)$ is simply $|L_X(B)|\psi\rangle|^2$. Recalling that a Hermitian projection operator satisfies $L^*=L=L^2$, we can write the

**as:**

*generalised Born's rule*$$\begin{align}P(x\in B) &= |L_X(B)|\psi\rangle|^2\\ &= \langle\psi|L_X(B)|\psi\rangle\end{align}$$

Well, this is interesting! In the last article, you proved that the

**expected value**of an observable $X$ given a state $|\psi\rangle$ is given by $\langle\psi|X|\psi\rangle$. But here we have a

*given by the same expression. So we want to interpret our projection operators as some sort of "observable" -- we can omit the "Hermitian", since all observables are Hermitian.*

**probability**There's another place you might've seen something like this, and that is with

*indicator variables*in probability and statistics --

*the expected value of an indicator variable for an event is the probability that the event occurs*.

Try to interpret these projection operators as observables that are analogous to "indicators" in some sense. If you think a little about it, you might see exactly what these observables represent: the eigenvalues of $L_X(B)$ are all 1 and 0 -- if the value "1" is realised, the state has been projected into the $\{|B\rangle\}$ -- and if the value "0" is realised, it hasn't.

So projection operators are a special type of observable, measuring the answer to "

**Yes/No questions**" -- if the answer to "is the system in one of the states $\{|B\rangle\}$?" is

**yes**, the observable $L_X(B)$ takes the value 1 -- if the answer is

**no**, then it takes the value 0. So it is precisely an "

**indicator variable**" for $\{|B\rangle\}$.

We have seen such projection operators, of course, in the context of polarisation -- where the operator represented whether or not the photon has passed through. Indeed, one may formulate quantum mechanics entirely in terms of projection operators, as any question can be formulated with some number of Yes/No questions (the key reason why this can be done, as we will see -- is that these "yes/no questions" all commute, i.e. the corresponding projection operators share an eigenbasis). Let's not.

Well, this can be generalised in the straightforward way to an operator with a continuous spectrum, resulting in the same expression. We can also calculate probability

*densities*using this result. Let $X$ be an operator with continuous spectrum $\Sigma_X$ -- then we can write the state $|\psi\rangle$ in the eigenbasis of $X$:

$$ |\psi\rangle = \int_{\Sigma_X} |x\rangle\, \Psi(x)\, dx $$

Where $\Psi(x)\, dx=\langle\psi|x\rangle$ are the coefficients of the state in the eigenbasis, i.e. the probability amplitudes -- we call $\Psi(x)$ the

**wavefunction**, and it represents

*probability amplitude densities*. Then for some set $M\subseteq \Sigma_X$ of eigenvalues $L_X(B)|\psi\rangle$ is the projection:

$$ L(M)|\psi\rangle = \int_B |x\rangle\, \Psi(x)\, dx $$

And one may calculate the dot product, noting that complex dot products require taking the complex conjugate:

$$ \langle \psi | L_X(B) | \psi \rangle = \int_B \Psi^*(x)\, \Psi(x)\, dx $$

Which gives us an expression for the

**probability density function**on $\Sigma_X$ as:

$$\begin{align}\rho(x) &=\Psi^*(x)\,\Psi(x) \\

&=|\Psi(x)|^2\end{align}$$

And this applies to any operator with a continuous spectrum, like position and momentum.

Some texts define the eigenvectors $|x\rangle$ of a continuous-spectrum observable differently from us -- it is often conventional to let $|x\rangle$ be infinitely large so that $\langle x_1|x_2\rangle = \delta(x_1-x_2)$. This is so that the amplitudes $\langle\psi|x\rangle$ are not infinitesimal, but instead $\langle\psi|x\rangle=\Psi(x)$ (without multiplication by $dx$). For consistency with discrete spectra, we do not use this convention.

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