Bra-ket formalism of quantum mechanical measurement

The state of a quantum mechanical system is described by an element $|\psi\rangle$ , called a ket, from a Hilbert space, i.e., a vector space that is complete and equipped with an inner product as well as a norm related to this inner product. Using bra-ket notation, the inner product of vectors $|\phi\rangle$ and $|\psi\rangle$ is represented as $\langle\phi|\psi\rangle$ . In quantum mechanics, this inner product is not necessarily a real number, and we have the rule $\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^{\ast}$ .

An observable is represented by a hermitian operator, say $\hat{Q}$ . A general operator $\hat{A}$ acting on a ket $|\psi\rangle$ gives another ket, $\hat{A} |\psi\rangle$ . The inner product rule is then $\langle\phi|\hat{A}|\psi\rangle = \langle\psi|\hat{A}^{\dag}|\phi\rangle^{\ast}$ , where $\hat{A}^{\dag}$ is the hermitian conjugate of $\hat{A}$ . However, a hermitian operator $\hat{Q}$ has the inner product rule $\langle\phi|\hat{Q}|\psi\rangle = \langle\psi|\hat{Q}|\phi\rangle^{\ast}$ , and therefore $\hat{Q} = \hat{Q}^{\dag}$ . This is the defining property of a hermitian operator, and it is straightforward to show that satisfaction of this inner product rule implies the eigenvalues of $\hat{Q}$ must be real.

For an observable represented by the hermitian operator $\hat{Q}$ , we can always find an orthonormal basis of the state space that consists of the eigenvectors of $\hat{Q}$ . In the discrete spectrum case, for example, the eigenvectors of $\hat{Q}$ might be $\{|\phi_i\rangle\}$ and the corresponding eigenvalues $\{k_i\}$ . (The discussion can easily be adapted to the continuous spectrum case by replacing summation by integration, and sequences of discrete coefficients by continuous functions, etc.) In order to perform the measurement of the observable on the quantum state $|\psi\rangle$ , we would first need to expand the state using as basis vectors the eigenvectors of $\hat{Q}$ . We can do this using a projection operator

$\hat{P} = \sum_i |\phi_i\rangle\langle\phi_i| \qquad \qquad \qquad (1)$

Thus the expansion of $|\psi\rangle$ in terms of the eigenvectors of $\hat{Q}$ would be given by

$|\psi\rangle = \hat{P}|\psi\rangle = \sum_i |\phi_i\rangle \langle\phi_i|\psi\rangle = \sum_i a_i |\phi_i\rangle \qquad \qquad \qquad (2)$

where the expansion coefficient $a_i \equiv \langle \phi_i|\psi\rangle$ is the projection of $|\psi\rangle$ on the axis represented by the i-th eigenvector of $\hat{Q}$ . (Notice that the quantum state $|\psi\rangle$ itself remains unchanged by the action of projection operators like $\hat{P}$ . Only the representation of $|\psi\rangle$ changes, with respect to the eigenvectors of different hermitian operators).

Given the expansion of the ket

$|\psi\rangle = \sum_i a_i |\phi_i\rangle \qquad \qquad \qquad (3)$

we then have a corresponding expansion of the bra, given by

$\langle\psi| = \sum_i \langle\phi_i| a_i^{\ast} \qquad \qquad \qquad (4)$

The quantum state $|\psi\rangle$ is normalised so that

$\langle\psi|\psi\rangle = \sum_i \langle \phi_i| a_i^{\ast} a_i |\phi_i \rangle$

$= \sum_i \langle \phi_i| |a_i|^2 |\phi_i \rangle$

$= \sum_i |a_i|^2 \langle\phi_i|\phi_i\rangle$

$= \sum_i |a_i|^2$

$= 1 \qquad \qquad \qquad (5)$

where the absence of cross-products of the eigenvectors of $\hat{Q}$ , and the penultimate equality in (5), follow from the orthonormality of the eigenvectors of $\hat{Q}$ . Thus, the sequence of squares of the absolute values of the expansion coefficients, $\{|a_i|^2\}$ , can be regarded as representing a probability distribution.

We can use this probability distribution to calculate the expected value of the measurement of the observable $\hat{Q}$ :

$\langle \hat{Q} \rangle \equiv \langle \psi|\hat{Q}|\psi \rangle$

$= \big(\sum_i \langle \phi_i | a_i^{\ast}\big)\big(\sum_i a_i \hat{Q} |\phi_i \rangle \big)$

$= \big(\sum_i \langle \phi_i | a_i^{\ast}\big)\big(\sum_i a_i k_i |\phi_i \rangle \big)$

$= \sum_i k_i \langle \phi_i|a_i^{\ast} a_i |\phi_i\rangle$

$= k_i |a_i|^2 \langle \phi_i|\phi_i\rangle$

$= |a_i|^2 k_i \qquad \qquad \qquad (6)$

In the case of two non-commuting hermitian operators, $\hat{Q}$ and $\hat{R}$ , we can easily derive Heisenberg’s uncertainty principle using this mathematical structure, as follows. Let

$\langle \hat{Q} \rangle \equiv \langle \psi | \hat{Q} |\psi\rangle \qquad \qquad \qquad (7)$

and

$\langle \hat{R} \rangle \equiv \langle \psi | \hat{R} |\psi\rangle \qquad \qquad \qquad (8)$

be the expectations of $\hat{Q}$ and $\hat{R}$ , computed as per (6). Let

$\hat{Q}^{\prime} \equiv \hat{Q} - \langle \hat{Q} \rangle \qquad \qquad \qquad (9)$

$\hat{R}^{\prime} \equiv \hat{R} - \langle \hat{R} \rangle \qquad \qquad \qquad (10)$

be corresponding deviations from the mean. (Note that $\hat{Q}^{\prime}$ and $\hat{R}^{\prime}$ must be hermitian if $\langle \hat{Q} \rangle$ and $\langle \hat{R} \rangle$ are, since they are obtained simply by subtracting a real number). Then the mean squared deviations are

$\Delta q^2 \equiv \langle \psi |\hat{Q}^{\prime} \hat{Q}^{\prime} |\psi \rangle \equiv \langle \hat{Q}^{\prime \ 2} \rangle \qquad \qquad \qquad (11)$

$\Delta q^2 \equiv \langle \psi |\hat{R}^{\prime} \hat{R}^{\prime} |\psi \rangle \equiv \langle \hat{R}^{\prime \ 2} \rangle \qquad \qquad \qquad (12)$

And by the Cauchy-Schwarz inequality, we can write

$\Delta q^2 \Delta r^2$

$= \langle \hat{Q}^{\prime \ 2} \rangle \cdot \langle \hat{R}^{\prime \ 2} \rangle$

$\ge \langle \hat{Q}^{\prime}\hat{R}^{\prime} \rangle^2$

$= \bigg(\frac{1}{2} \langle \hat{Q}^{\prime}\hat{R}^{\prime} - \hat{R}^{\prime} \hat{Q}^{\prime}\rangle + \frac{1}{2} \langle \hat{Q}^{\prime}\hat{R}^{\prime} + \hat{R}^{\prime} \hat{Q}^{\prime}\rangle\bigg)^2$

$\ge \frac{1}{4} |\langle \hat{Q}^{\prime}\hat{R}^{\prime} - \hat{R}^{\prime} \hat{Q}^{\prime}\rangle|^2$

$= \frac{1}{4} |\langle [\hat{Q}^{\prime}, \hat{R}^{\prime}] \rangle|^2 \qquad \qquad \qquad (13)$

Therefore we have

$\Delta q \Delta r \ge \frac{1}{2} |\langle [\hat{Q}^{\prime}, \hat{R}^{\prime}] \rangle| \qquad \qquad \qquad (14)$

(Notice that we introduced the norm brackets in (13) to allow for the fact that $\langle \hat{Q}^{\prime}\hat{R}^{\prime} - \hat{R}^{\prime} \hat{Q}^{\prime}\rangle$ will in general be a complex number. The expression $\hat{Q}^{\prime}\hat{R}^{\prime} - \hat{R}^{\prime} \hat{Q}^{\prime}$ is not hermitian, even though $\hat{Q}^{\prime}$ and $\hat{R}^{\prime}$ are).

Finally, observe that

$[\hat{Q}^{\prime}, \hat{R}^{\prime}] = \hat{Q}^{\prime}\hat{R}^{\prime} - \hat{R}^{\prime} \hat{Q}^{\prime} = [\hat{Q}, \hat{R}] \qquad \qquad \qquad (15)$

Using this in (14) gives us the key inequality of the Heisenberg uncertainty principle:

$\Delta q \Delta r \ge \frac{1}{2} |\langle [\hat{Q}, \hat{R}] \rangle| \qquad \qquad \qquad (16)$

For example, in the case of the momentum and position operators in one dimension, we have

$[\hat{P}, \hat{X}] = -i \hbar \qquad \qquad \qquad (17)$

Putting (17) into (16) then gives the canonical inequality of the Heisenberg uncertainty principle:

$\Delta p \Delta x \ge \frac{1}{2} \hbar$

Bra-ket formalism of quantum mechanical measurement

Published by Dr Christian P. H. Salas

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