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I'm quite new to the field of Quantum chemistry, but I can't wrap my head around the following postulate of Quantum Chemistry, stated as Postulate 3 in [1]:

Postulate 3: In any measurement of the observable associated with the operator A, the only values that will ever be observed are the eigenvalues $a_n$ which satisfy the eigenvalue equation $$A\psi_n = a_n \psi_n.$$

My first thought was, that this was just a little bit fuzzy, because one result in QC is, that energies, as eigenvalues of the Hamiltonian, don't exists continiuosly but only discrete, so we would need restrict the set of possible $\psi_n$ to a set of basis vectors for the space of eigenfunctions of A. Otherwise we could use any linear combination of eigenvectors and thus realise any (energy) eigenvalue.

This is for example done, if we take a look at the 1 dimensional particle in the box, which is usually provided with the basis set $B_1 = \{\psi_n\}_{n \in \mathbb{N}}$ with \begin{equation} \psi_n(x) = \begin{cases}\left(\frac{2}{L}\right)^{1/2} \sin\left(\frac{n\pi x}{L}\right) & 0\leq x \leq L \\ 0 & else\end{cases}. \end{equation}

According to the postulate above, the choice of the basis is quite important as the corresponding energy values to the wave functions $\psi_n$ are the only ones that could be observed. However, and here is my problem, the choice of this basis set for the 1D particle in the box seems arbitrary to me, as we could also set \begin{equation} \psi'_1 = \frac{1}{\sqrt{2}}(\psi_1 + \psi_2) \text{ and } \psi'_2 = \frac{1}{\sqrt{2}}(\psi_1 - \psi_2), \end{equation}

so that $B_2 = \{\psi'_1, \psi'_2, \psi_3, \psi_4,\dots\}$ would also be an (orthonormal) basis set for the same space as $B_1$.

The important implication would now be, that the corresponding energy values for $B_1$ and $B_2$ are different and hence, depending on the choice of such a basis set, the system could be in different (discrete) states.

To me this seems as a contradiction, as I would expect that the energy states of the system should be independent of that mathematical framework, i.e. every mathematical formulation should yield the same physical properties. This leads me to the conclusion, that I made a mistake somewhere or I did understand something wrong. Any help would be much appreciated.

[1]: D. A. McQuarrie, “Quantum Chemistry,” Oxford University Press, Oxford, 1983.

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  • $\begingroup$ The expectation value of the energy, defined by $\langle E \rangle = \langle \psi | \hat{H} | \psi\rangle$, is not the same as the value that would be obtained if the energy were to be measured. Expectation values are continuous, and are linear combinations of eigenvalues. The eigenvalues, which are the values you would measure, are discrete. $\endgroup$ Commented Mar 5, 2023 at 12:52
  • $\begingroup$ @orthocresol What I meant was that, if $\psi_1$ and $\psi_2$ are eigenvalues of $H$ then so is any linear combination of them both. Hence the choice of the basis set of these $\psi_n$ is crucial, if that makes sense. $\endgroup$ Commented Mar 5, 2023 at 13:02
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    $\begingroup$ @student7481 Your above comment is garbled. If ψ1 and ψ2 are eigenvectors of a given matrix then a linear combination of those eigenvectors is also an eigenvector of the original matrix if and only if the corresponding eigenvalue is the same for both cases, i.e. they correspond to degenerate states. The linear combination of two non-degenerate evecs is in general not an eigenvector of the original matrix. $\endgroup$
    – Ian Bush
    Commented Mar 5, 2023 at 14:52
  • $\begingroup$ You are right, that was also the error. $\endgroup$ Commented Mar 5, 2023 at 15:53

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If you can algebraically solve the Schroedinger differential equation then generally $n$ quantised solutions are obtained and are determined by the potential energy function you use and the boundary conditions. If you cannot solve the equation algebraically then using a basis set of other functions, such as sin/cos, gaussians etc, then the solution can be found by adding these functions to form a basis set of $m$ functions and $m$ can increase until sufficient accuracy is obtained for the $n^{th}$ solution to the Schroedinger equation. Generally $m \gg n$ because the basis set functions are not solutions to your problem but will approximate to it if sufficient number of terms are used. See 'Solution of the Schrödinger Equation for One-Dimensional Anharmonic Potentials' J. Chem. Educ. 2011, 88, 7, 929–931 for examples.

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This is actually something that had me stuck for a long while as an undergraduate, so let me see if my resolution helps you. I don’t want to get into any philosophically sticky propositions about wavefunction collapse or the interpretation of quantum measurements, but suffice to say that this postulate can be presented in a rather pragmatic form using spectroscopy. In fact, McQuarrie shows that spectroscopic transitions arising from the coupling of the electric dipole of a molecule with the electromagnetic field of light cause the coefficients of a wavefunction to change. If you begin in some initial state, then the probability to transition to a final state is proportional to the transition dipole moment integral between the states. This result is called Fermi’s Golden Rule of time dependent perturbation theory. But more importantly, in the process of deriving this result you find that only when the frequency of the electromagnetic radiation is on resonance (matches energy) with the gap between the energy eigenvalues of the states will you cause any transitions to occur. This is a very practical example of why the eigenvalues of the Hamiltonian have such a prominent role in quantum measurement. For less friendly cases like the double slit experiment, we somewhat have to throw our hands up because we cannot write down an explicit Hamiltonian for the problem. In other words, we really have no mechanism to describe why we observe the results that we do. But whenever we can write out a Hamiltonian explicitly we get results like we see with Fermi’s Golden Rule. I hope this was helpful to you.

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The postulate could have been stated more clearly as

Postulate 3: In any measurement of the observable associated with the operator A, the only values that will ever be discretely observed are the individual eigenvalues $a_n$, each corresponding to an eigenstate described by wavefunction $\psi_n$, which satisfy the eigenvalue equation $$A\psi_n = a_n \psi_n.$$

One consequence being that degenerate eigenstates (those sharing the same eigenvalue) can be arbitrarily combined to generate an observable state with the same observable eigenvalue, but orthogonal states (differing in eigenvalue) cannot be combined to generate pure observable states.

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  • $\begingroup$ Hat tip to @IanBush for the enlightening comment to the OP. $\endgroup$
    – Buck Thorn
    Commented Dec 31, 2023 at 8:33
  • $\begingroup$ Another consequence is that when a system exists as a superposition the act of observation perturbs the system when not in an eigenstate and results in collapse of the wavefunction (selection of eigenstates). $\endgroup$
    – Buck Thorn
    Commented Dec 31, 2023 at 8:36

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