# 3.9: Measurement


Suppose that $$A$$ is an Hermitian operator corresponding to some dynamical variable. By analogy with the discussion in Section [scoll], we expect that if a measurement of $$A$$ yields the result $$a$$ then the act of measurement will cause the wavefunction to collapse to a state in which a measurement of $$A$$ is bound to give the result $$a$$. What sort of wavefunction, $$\psi$$, is such that a measurement of $$A$$ is bound to yield a certain result, $$a$$? Well, expressing $$\psi$$ as a linear combination of the eigenstates of $$A$$, we have $\label{e4.128} \psi = \sum_i c_i\,\psi_i,$ where $$\psi_i$$ is an eigenstate of $$A$$ corresponding to the eigenvalue $$a_i$$. If a measurement of $$A$$ is bound to yield the result $$a$$ then $\langle A\rangle= a,$ and $\label{e4.130} \sigma_A^{\,2} = \langle A^2\rangle - \langle A\rangle = 0.$ Now, it is easily seen that \begin{aligned} \label{e4.131} \langle A\rangle &= \sum_i |c_i|^{\,2}\,a_i,\\[0.5ex] \langle A^2\rangle &= \sum_i |c_i|^{\,2}\,a_i^{\,2}.\end{aligned} Thus, Equation ([e4.130]) gives $\sum_i a_i^{\,2}\,|c_i|^{\,2} - \left(\sum_i a_i\,|c_i|^{\,2}\right)^2=0.$ Furthermore, the normalization condition yields $\label{e4.134} \sum_i |c_i|^{\,2} = 1.$

For instance, suppose that there are only two eigenstates. The previous two equations then reduce to $$|c_1|^{\,2}=x$$, and $$|c_2|^{\,2}=1-x$$, where $$0\leq x\leq 1$$, and $\label{e4.126} (a_1-a_2)^2\,x\,(1-x) = 0.$ The only solutions are $$x=0$$ and $$x=1$$. This result can easily be generalized to the case where there are more than two eigenstates. It follows that a state associated with a definite value of $$A$$ is one in which one of the $$|c_i|^{\,2}$$ is unity, and all of the others are zero. In other words, the only states associated with definite values of $$A$$ are the eigenstates of $$A$$. It immediately follows that the result of a measurement of $$A$$ must be one of the eigenvalues of $$A$$. Moreover, if a general wavefunction is expanded as a linear combination of the eigenstates of $$A$$, like in Equation ([e4.128]), then it is clear from Equation ([e4.131]), and the general definition of a mean, that the probability of a measurement of $$A$$ yielding the eigenvalue $$a_i$$ is simply $$|c_i|^{\,2}$$, where $$c_i$$ is the coefficient in front of the $$i$$th eigenstate in the expansion. Note, from Equation ([e4.134]), that these probabilities are properly normalized: that is, the probability of a measurement of $$A$$ resulting in any possible answer is unity. Finally, if a measurement of $$A$$ results in the eigenvalue $$a_i$$ then immediately after the measurement the system will be left in the eigenstate corresponding to $$a_i$$.

Consider two physical dynamical variables represented by the two Hermitian operators $$A$$ and $$B$$. Under what circumstances is it possible to simultaneously measure these two variables (exactly)? Well, the possible results of measurements of $$A$$ and $$B$$ are the eigenvalues of $$A$$ and $$B$$, respectively. Thus, to simultaneously measure $$A$$ and $$B$$ (exactly) there must exist states which are simultaneous eigenstates of $$A$$ and $$B$$. In fact, in order for $$A$$ and $$B$$ to be simultaneously measurable under all circumstances, we need all of the eigenstates of $$A$$ to also be eigenstates of $$B$$, and vice versa, so that all states associated with unique values of $$A$$ are also associated with unique values of $$B$$, and vice versa.

Now, we have already seen, in Section 1.8, that if $$A$$ and $$B$$ do not commute (i.e., if $$A\,B\neq B\,A$$) then they cannot be simultaneously measured. This suggests that the condition for simultaneous measurement is that $$A$$ and $$B$$ should commute. Suppose that this is the case, and that the $$\psi_i$$ and $$a_i$$ are the normalized eigenstates and eigenvalues of $$A$$, respectively. It follows that $(A\,B-B\,A)\,\psi_i = (A\,B-B\,a_i)\,\psi_i = (A-a_i)\,B\,\psi_i = 0,$ or $A\,(B\,\psi_i) = a_i\,(B\,\psi_i).$ Thus, $$B\,\psi_i$$ is an eigenstate of $$A$$ corresponding to the eigenvalue $$a_i$$ (though not necessarily a normalized one). In other words, $$B\,\psi_i\propto \psi_i$$, or $B\,\psi_i = b_i\,\psi_i,$ where $$b_i$$ is a constant of proportionality. Hence, $$\psi_i$$ is an eigenstate of $$B$$, and, thus, a simultaneous eigenstate of $$A$$ and $$B$$. We conclude that if $$A$$ and $$B$$ commute then they possess simultaneous eigenstates, and are thus simultaneously measurable (exactly).

## Continuous Eigenvalues

In the previous two sections, it was tacitly assumed that we were dealing with operators possessing discrete eigenvalues and square-integrable eigenstates. Unfortunately, some operators—most notably, $$x$$ and $$p$$—possess eigenvalues that lie in a continuous range and non-square-integrable eigenstates (in fact, these two properties go hand in hand). Let us, therefore, investigate the eigenstates and eigenvalues of the displacement and momentum operators.

Let $$\psi_x(x,x')$$ be the eigenstate of $$x$$ corresponding to the eigenvalue $$x'$$. It follows that $x\,\psi_x(x,x') = x'\,\psi_x(x,x')$ for all $$x$$. Consider the Dirac delta-function $$\delta(x-x')$$. We can write $x\,\delta(x-x') = x'\,\delta(x-x'),$ because $$\delta(x-x')$$ is only non-zero infinitesimally close to $$x=x'$$. Evidently, $$\psi_x(x,x')$$ is proportional to $$\delta(x-x')$$. Let us make the constant of proportionality unity, so that $\psi_x(x,x') = \delta(x-x').$ It is easily demonstrated that $\int_{-\infty}^{\infty} \delta(x-x')\,\delta(x-x'')\,dx = \delta(x'-x'').$ Hence, $$\psi_x(x,x')$$ satisfies the orthonormality condition $\label{e4.143} \int_{-\infty}^\infty \psi_x^\ast(x,x')\,\psi_x(x,x'')\,dx = \delta(x'-x'').$ This condition is analogous to the orthonormality condition ([e3.125]) satisfied by square-integrable eigenstates. Now, by definition, $$\delta(x-x')$$ satisfies $\int_{-\infty}^\infty f(x)\,\delta(x-x')\,dx = f(x'),$ where $$f(x)$$ is a general function. We can thus write $\label{e4.144} \psi(x) = \int_{-\infty}^\infty c(x')\,\psi_x(x,x')\,dx',$ where $$c(x')=\psi(x')$$, or $\label{e4.145} c(x') = \int_{-\infty}^\infty \psi_x^\ast(x,x')\,\psi(x)\,dx.$ In other words, we can expand a general wavefunction $$\psi(x)$$ as a linear combination of the eigenstates, $$\psi_x(x,x')$$, of the displacement operator. Equations ([e4.144]) and ([e4.145]) are analogous to Equations ([e3.123]) and ([e3.126]), respectively, for square-integrable eigenstates. Finally, by analogy with the results in Section 1.9, the probability density of a measurement of $$x$$ yielding the value $$x'$$ is $$|c(x')|^{\,2}$$, which is equivalent to the standard result $$|\psi(x')|^{\,2}$$. Moreover, these probabilities are properly normalized provided $$\psi(x)$$ is properly normalized [cf., Equation ([e3.127])]: that is, $\int_{-\infty}^\infty |c(x')|^{\,2}\,dx'= \int_{-\infty}^\infty |\psi(x')|^{\,2}\,dx' =1.$ Finally, if a measurement of $$x$$ yields the value $$x'$$ then the system is left in the corresponding displacement eigenstate, $$\psi_x(x,x')$$, immediately after the measurement. That is, the wavefunction collapses to a “spike-function”, $$\delta(x-x')$$, as discussed in Section [scoll].

Now, an eigenstate of the momentum operator $$p\equiv -{\rm i}\,\hbar\,\partial/\partial x$$ corresponding to the eigenvalue $$p'$$ satisfies $-{\rm i}\,\hbar\,\frac{\partial \psi_p(x,p')}{\partial x} = p'\,\psi_p(x,p').$ It is evident that $\label{e4.148} \psi_p(x,p') \propto {\rm e}^{+{\rm i}\,p'\,x/\hbar}.$ We require $$\psi_p(x,p')$$ to satisfy an analogous orthonormality condition to Equation ([e4.143]): that is, $\int_{-\infty}^\infty \psi_p^\ast(x,p')\,\psi_p(x,p'')\,dx = \delta(p'-p'').$ Thus, it follows from Equation ([e3.72]) that the constant of proportionality in Equation ([e4.148]) should be $$(2\pi\,\hbar)^{-1/2}$$: that is, $\label{e4.148a} \psi_p(x,p') =\frac{ {\rm e}^{+{\rm i}\,p'\,x/\hbar}}{(2\pi\,\hbar)^{1/2}}.$ Furthermore, according to Equations ([e3.64]) and ([e3.65]), $\label{e4.152} \psi(x) = \int_{-\infty}^\infty c(p')\,\psi_p(x,p')\,dp',$ where $$c(p') = \phi(p')$$ [see Equation ([e3.65])], or $\label{e4.153} c(p') = \int_{-\infty}^\infty \psi_p^\ast(x,p')\,\psi(x)\,dx.$ In other words, we can expand a general wavefunction $$\psi(x)$$ as a linear combination of the eigenstates, $$\psi_p(x,p')$$, of the momentum operator. Equations ([e4.152]) and ([e4.153]) are again analogous to Equations ([e3.123]) and ([e3.126]), respectively, for square-integrable eigenstates. Likewise, the probability density of a measurement of $$p$$ yielding the result $$p'$$ is $$|c(p')|^{\,2}$$, which is equivalent to the standard result $$|\phi(p')|^{\,2}$$. The probabilities are also properly normalized provided $$\psi(x)$$ is properly normalized [cf., Equation ([e3.83])]: that is, $\int_{-\infty}^\infty |c(p')|^{\,2}\,dp'= \int_{-\infty}^{\infty} |\phi(p')|^{\,2}\,dp' = \int_{-\infty}^\infty |\psi(x')|^{\,2}\,dx' =1.$ Finally, if a mesurement of $$p$$ yields the value $$p'$$ then the system is left in the corresponding momentum eigenstate, $$\psi_p(x,p')$$, immediately after the measurement.

## Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 3.9: Measurement is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.