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9.1: Measurement

  • Page ID
    5217
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    The result of a measurement of the observable \(A\) must yield one of the eigenvalues of \(\hat{A}\). 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 \(\hat{A}\) by \(\{a_i\}\), then each of the eigenvalues \( {a_i} \) satisfies an eigenvalue equation

    \[\hat{A}\vert a_i\rangle = a_i \vert a_i\rangle \nonumber \]

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

    \[\langle a_i\vert \hat{A} = \langle a_i\vert a_i \nonumber \]

    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

    \[P_{a_i} = \vert\langle a_i\vert\Psi(t)\rangle \vert^2 \nonumber \]

    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

    \[\langle a_i\vert a_j\rangle = \delta_{ij} \nonumber \] and thus any arbitrary vector \(\vert\phi\rangle \) can be expanded as a linear combination of these vectors:

    \[\vert\phi\rangle = \sum_i c_i \vert a_i\rangle \nonumber \] 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

    \[c_i = \langle a_i\vert\phi\rangle \nonumber \]

    The eigenvectors also satisfy a closure relation:

    \[I = \sum_i \vert a_i\rangle \langle a_i\vert \nonumber \]

    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

    \[\langle A \rangle = \langle \Psi(t)\vert A\vert\Psi(t)\rangle \nonumber \]

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

    \[\vert\Psi(t)\rangle = \sum_i \alpha_i(t) \vert a_i\rangle \nonumber \]

    where \( {a_i} \) 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

    \[ \begin{align*} \langle A \rangle (t) &= \sum_{i,j} \alpha_i^*(t) \alpha_j(t) \langle a_i\vert A\vert a_i \rangle \\[4pt] &=\sum_{i,j} \alpha_i^*(t) \alpha_j a_i(t) \delta_{ij} \\[4pt] &= \sum_i a_i \vert\alpha_i(t)\vert^2 \end{align*}\]

    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

    \[AB - BA = 0 \nonumber \]

    that is

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

    where, in the second line, the quantity \(\left [A, B \right ] \equiv AB - BA \) is know as the commutator between \(A\) and \(B\). If \(\left [ A, B \right ] = 0 \), 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

    \[ X\vert x\rangle = x\vert x\rangle \nonumber \]

    \[ P\vert p\rangle \nonumber =p\vert p\rangle \nonumber \]

    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

    \[ \left [ X, P \right ] = i \hbar I \nonumber \]

    and that the inner product between eigenvectors of \(X\) and \(P\) is

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

    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:

    \[ \begin{align*} \langle x\vert x'\rangle & = \delta(x-x') \\[4pt] \langle p\vert p'\rangle &= \delta(p-p') \end{align*}\]

    \[\begin{align*} \vert \phi \rangle &= \int dx \vert x\rangle \langle x\vert\phi\rangle \\[4pt] \vert\phi\rangle &= \int dp \vert p\rangle \langle p\vert\phi \rangle \end{align*}\]

    \[\begin{align*} I &= \int dx \vert x\rangle \langle x\vert \\[4pt] I &= \int dp \vert p\rangle \langle p\vert \end{align*}\]

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

    \[P(x,t)dx = \vert\langle x\vert\Psi(t)\rangle \vert^2 dx \nonumber \]

    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

    \[\langle x\vert X\vert\Psi(t)\rangle = x\Psi(x,t) \nonumber \]

    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

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

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

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


    This page titled 9.1: Measurement is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.