Measurement

The result of a measurement of the observable $A$ must yield one of the eigenvalues of $A$. Thus, we see why $A$ is required to be a hermitian operator: Hermitian operators have real eigenvalues. If we denote the set of eigenvalues of $A$ by $\{a_i\}$, then each of the eigenvalues $a_i$ satisfies an eigenvalue equation

$$A\vert a_i\rangle = a_i \vert a_i\rangle$$

where $\vert a_i\rangle$ is the corresponding eigenvector. Since the operator $A$ is hermitian and $a_i$ is therefore real, we have also the left eigenvalue equation

$$\langle a_i\vert A = \langle a_i\vert a_i$$

The probability amplitude that a measurement of $A$ will yield the eigenvalue $a_i$ is obtained by taking the inner product of the corresponding eigenvector $\vert a_i\rangle$ with the state vector $\vert\Psi(t)\rangle$, $\langle a_i\vert\Psi(t)\rangle$. Thus, the probability that the value $a_i$ is obtained is given by

$$P_{a_i} = \vert\langle a_i\vert\Psi(t)\rangle \vert^2$$

Another useful and important property of hermitian operators is that their eigenvectors form a complete orthonormal basis of the Hilbert space, when the eigenvalue spectrum is non-degenerate. That is, they are linearly independent, span the space, satisfy the orthonormality condition

$$\langle a_i\vert a_j\rangle = \delta_{ij}$$

and thus any arbitrary vector $\vert\phi\rangle$ can be expanded as a linear combination of these vectors:

$$\vert\phi\rangle = \sum_i c_i \vert a_i\rangle$$

By multiplying both sides of this equation by $\langle a_j\vert$ and using the orthonormality condition, it can be seen that the expansion coefficients are

$$c_i = \langle a_i\vert\phi\rangle$$

The eigenvectors also satisfy a closure relation:

$$I = \sum_i \vert a_i\rangle \langle a_i\vert$$

where $I$ is the identity operator.

Averaging over many individual measurements of $A$ gives rise to an average value or expectation value for the observable $A$, which we denote $\langle A \rangle$ and is given by

$$\langle A \rangle = \langle \Psi(t)\vert A\vert\Psi(t)\rangle$$

That this is true can be seen by expanding the state vector $\vert\Psi(t)\rangle$ in the eigenvectors of $A$:

$$\vert\Psi(t)\rangle = \sum_i \alpha_i(t) \vert a_i\rangle$$

where $\alpha_i$ are the amplitudes for obtaining the eigenvalue $a_i$ upon measuring $A$, i.e., $\alpha_i = \langle a_i\vert\Psi(t)\rangle$. Introducing this expansion into the expectation value expression gives

$$\langle A \rangle (t) = \sum_{i,j} \alpha_i^*(t) \alpha_j(t) \langle a_i\vert A\vert a_i \rangle$$

$$= \sum_{i,j} \alpha_i^*(t) \alpha_j a_i(t) \delta_{ij}$$

$$= \sum_i a_i \vert\alpha_i(t)\vert^2$$

The interpretation of the above result is that the expectation value of $A$ is the sum over possible outcomes of a measurement of $A$ weighted by the probability that each result is obtained. Since $\vert\alpha_i\vert^2 = \vert\langle a_i\vert\Psi(t)\rangle \vert^2$ is this probability, the equivalence of the expressions can be seen.

Two observables are said to be compatible if $AB=BA$. If this is true, then the observables can be diagonalized simultaneously to yield the same set of eigenvectors. To see this, consider the action of $BA$ on an eigenvector $\vert a_i\rangle$ of $A$. $BA\vert a_i\rangle = a_i B\vert a_i\rangle$. But if this must equal $AB\vert a_i\rangle$, then the only way this can be true is if $B\vert a_i\rangle$ yields a vector proportional to $\vert a_i\rangle$ which means it must also be an eigenvector of $B$. The condition $AB=BA$ can be expressed as

$$AB-BA = 0$$

$$\left[A,B\right] = 0$$

where, in the second line, the quantity $[A,B] \equiv AB-BA$ is know as the commutator between $A$ and $B$. If $[A,B]=0$, then $A$ and $B$ are said to commute with each other. That they can be simultaneously diagonalized implies that one can simultaneously predict the observables $A$ and $B$ with the same measurement.

As we have seen, classical observables are functions of position $x$ and momentum $p$ (for a one-particle system). Quantum analogs of classical observables are, therefore, functions of the operators $X$ and $P$ corresponding to position and momentum. Like other observables $X$ and $P$ are linear hermitian operators. The corresponding eigenvalues $x$ and $p$ and eigenvectors $\vert x\rangle$ and $\vert p\rangle$ satisfy the equations

$$X\vert x\rangle = x\vert x\rangle$$

$$P\vert p\rangle = p\vert p\rangle$$

which, in general, could constitute a continuous spectrum of eigenvalues and eigenvectors. The operators $X$ and $P$ are not compatible. In accordance with the Heisenberg uncertainty principle (to be discussed below), the commutator between $X$ and $P$ is given by

$$[X,P]= i\hbar I$$

and that the inner product between eigenvectors of $X$ and $P$ is

$$\langle x\vert p\rangle = {1 \over \sqrt{2\pi\hbar}}e^{ipx/\hbar}$$

Since, in general, the eigenvalues and eigenvectors of $X$ and $P$ form a continuous spectrum, we write the orthonormality and closure relations for the eigenvectors as:

$$\langle x\vert x'\rangle = \delta(x-x')\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\langle p\vert p'\rangle = \delta(p-p')$$

$$\vert\phi \rangle = \int dx \vert x\rangle \langle x\vert\phi\rangle \;\;\;\;\;\;\;\;... ...\;\;\;\;\; \vert\phi\rangle = \int dp \vert p\rangle \langle p\vert\phi \rangle$$

$$I = \int dx \vert x\rangle \langle x\vert\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; I = \int dp \vert p\rangle \langle p\vert$$

The probability that a measurement of the operator $X$ will yield an eigenvalue $x$ in a region $dx$ about some point is

$$P(x,t)dx = \vert\langle x\vert\Psi(t)\rangle \vert^2 dx$$

The object $\langle x\vert\Psi(t)\rangle$ is best represented by a continuous function $\Psi(x,t)$ often referred to as the wave function. It is a representation of the inner product between eigenvectors of $X$ with the state vector. To determine the action of the operator $X$ on the state vector in the basis set of the operator $X$, we compute

$$\langle x\vert X\vert\Psi(t)\rangle = x\Psi(x,t)$$

The action of $P$ on the state vector in the basis of the $X$ operator is consequential of the incompatibility of $X$ and $P$ and is given by

$$\langle x\vert P\vert\Psi(t)\rangle = {\hbar \over i}{\partial \over \partial x}\Psi(x,t)$$

Thus, in general, for any observable $A(X,P)$, its action on the state vector represented in the basis of $X$ is

$$\langle x\vert A(X,P)\vert\Psi(t)\rangle = A\left(x,{\hbar\over i}{\partial \over \partial x}\right)\Psi(x,t)$$