[NOTES/QM-06004] A Discussion of The Third Postulate of Quantum Mechanics

The third postulate is explained in  several parts. In this section we use the notation $A$ to denote a dynamical variable, $\hat{A}$ will denote the corresponding hermitian operator. The eigenvalues will be real and it is assumed that the eigenvectors have been chosen to be orthonormal. If we denote the eigenvalues of $\hat{A}$ by $\alpha_1,\alpha_2,\ldots,\alpha_k,\ldots$ and the corresponding orthonormal eigenvectors by $\ket{u_1},\ket{u_2},\ldots,\ket{u_k},\ldots$ then we have \begin{equation} A\ket{u_k}=\alpha_k \ket{u_k}, \qquad \innerproduct{u_k}{u_m}=\delta_{km} \label{EQ01}. \end{equation}

1. The first part of the third postulate says that the only outcome of a measurement of a dynamical variable $A$ is one of the eigenvalues of the corresponding operator $\hat{A}$. Thus, if an experiment to measure a dynamical variable $A$ is performed, the result must be only one of the eigenvalues $\alpha_k$. In particular, an answer different, from every eigenvalue, cannot be the outcome of measurement of $A$.
2. The next part of the postulate tells that if a system is represented by one of the eigenvectors $\ket{u_n}$, a measurement of the dynamical variable $A$ will give the corresponding eigenvalue $\alpha_n$.
3. The third postulate also tells us about the outcome of measurement of $A$ when the state vector $\ket{\psi}$ is not an eigenvector of $\hat{A}$. In this case, the result is some times one eigenvalue and sometimes another. We cannot predict the result of a single measurement fully. When repeated measurements are made, different eigenvalues $\alpha_n$ will be obtained with different probabilities \(p_n\) which can be predicted.
4. Applying the postulate To compute the probabilities \(p_n\), we first expand the state vector $ \ket{\psi}$ in terms of the eigenvectors of the operator $\hat{A}$ corresponding to the dynamical variable $A$ which is being measured and write \begin{equation} \ket{\psi}= \sum_k c_k \ket{u_k} \label{EQ02}. \end{equation} Then the probability \(p_k\) of getting value $\alpha_k$ is given by $|c_k|^2$. We continue to assume that the state vector $\ket{\psi}$ and the eigenvectors $\ket{u_k}$ are orthogonal {\it ,i.e,,} \begin{equation} \innerproduct{\psi}{\psi} =1, \qquad \innerproduct{u_k}{u_k} =1 \label{EQ03} \end{equation}
5. How do we compute the coefficients $c_k$ in \eqref{EQ03}? The eigenvectors of a hermitian operator are orthogonal and this helps in computing the coefficients. Taking scalar product of \eqref{EQ02} with $\ket{u_n}$ gives \begin{eqnarray} \innerproduct{u_k}{\psi} &=&\sum_n c_k \innerproduct{u_k}{u_n} =\sum_n c_n \delta_{kn} = c_k \label{EQ04} \end{eqnarray} Note that, in the right hand side of \eqref{EQ04}, only the term with $k=n$ survives, all other terms where $n\ne k$ will vanish due to orthogonality property of the eigenvectors. \begin{equation} \therefore \qquad c_k = \innerproduct{u_k}{\psi} \label{EQ07} \end{equation} and \begin{equation} \boxed{p_k = |c_k|^2 = |\innerproduct{u_k}{\psi}|^2} \end{equation}
6. The Parseval relation \begin{equation} \innerproduct{\psi}{\psi} = \sum_k |c_k|^2 \label{EQ08} \end{equation} implies that \begin{equation} \sum_k |c_k|^2 =1 \Longrightarrow \sum_k p_k=1 \label{EQ09}. \end{equation} if the state vector $\ket{\psi}$ is normalised, $\innerproduct{\psi}{\psi}=1$. This suggests that the interpretation of expressions $\displaystyle |c_k|^2 $ as probabilities \(p_k\) is consistent with the requirement that the sum of all probabilities be equal to unity. We shall call the coefficient $c_k$ as the probability amplitude for obtaining a value $\alpha_k$ for the dynamical variable $A$ when the system is in state $\ket{\psi}$.

As a consequence of the dual nature of matter and radiation, we have indeterminacy in the theoretical predictions. The origin of this indeterminacy can be traced to the superposition principle which in turn is needed to incorporate the wave nature of matter. In classical mechanics the result of measurement of position, momenta, and every other dynamical variables, can be fully predicted. This is no longer true in quantum theory. Here is a summary

 Remarks

• A {\it single} measurement of $A$ does not lead to a definite answer when the {\it state vector} is not an eigenvector $\hat{A}$.
• In general, a result of a measurement of $A$ must be one of the eigenvalues. The outcome of a {\it single} experiment is {\it indeterminate}, and the quantum theory is probabilistic by its nature in contrast to the classical theory which is deterministic. When measurement is repeated several times, we will sometimes get an eigenvalue $\alpha_j$ sometimes some other eigenvalue $\alpha_k$, and only the probabilities of each outcome can be predicted.
• A simple consequence of the above discussion is that a measurement of a dynamical variable $A$ will give a value $\alpha_m$ with probability $1$ if and only if the state is represented by corresponding eigenvector $\ket{u_m}$.
• It must be remembered that some obvious changes, described later at the end of the next section, will be needed when the eigenvalues of $\hat{A}$ are continuous.
• Finally, we leave it as an exercise for you to convince yourself that the assumption about the probabilities, as stated above, is correctly contained in the the following statement. {\it Given that the system is in a state described by the state vector $\ket{\psi}$, the probability that it will be found in the state given by the vector $\ket{\phi}$ is equal to $|\innerproduct{\phi}{\psi}|^2$.}