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I'm quite new to the field of Quantum mechanics, but I can't wrap my head around the following postulate of Quantum mechanics, stated as Postulate 3 in [1]. Although I approached quantum mechanics from the view of a chemist, I feel that this is a rather mathematical problem, that's why I search for help here.

The above mentioned Postulate 3 says:

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 QM is, that energies, as eigenvalues of the Hamiltonian, don't exists continuously but only discrete, so we would need to restrict the set of possible $\psi_n$ to a set of basis vectors for the space of eigenvectors (or eigenfunctions in this case) of A. Otherwise we could use any linear combination of eigenvectors (which are themselves eigenvectors) and thus realise any 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 $B_1 = \{\psi_n\}_{n \in \mathbb{N}}$ for the space of eigenvectors, where \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 & \rm 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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2 Answers 2

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Linear combinations of eigenvectors with different eigenvalues are not eigenvectors. So there is no contradiction here.

If you compute matrix elements of the Hamiltonian in the basis $B_1$, it will be diagonal allowing you to read off the eigenvalues. \begin{align} H = \frac{\hbar^2 \pi^2}{2m L^2} \mathrm{diag}(1, 4, 9, \dots) \end{align} If you use the basis $B_2$ then $H$ will not be diagonal. You'll have to diagonalize it to see the eigenvalues and this is equivalent to just going back to $B_1$.

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    $\begingroup$ "Linear combinations of eigenvectors with different eigenvalues are not eigenvectors. So there is no contradiction here." That was it! Thanks. Stupid mistake, this sounded so right to me, that I didn't challenge it properly. $\endgroup$ Mar 5, 2023 at 15:30
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The eigenfunctions of the Hamilton operator in your example are indeed the functions $\psi_n(x)$ ($n=1,2, \ldots$) given above with associated energy eigenvalues $E_n=\hbar^2 n^2 \pi^2/2m L^2$. The postulate, you mentioned, says that whenever you measure the energy of this system, the outcome can only be one the eigenvalues $E_n$.

Suppose, you prepare a large number $N$ of identical copies of your system in the energy eigenstate $\psi_n$, the outcome of an energy measurement will always (for all $N$ copies) be $E_n$. On the other hand, if you prepare $N$ copies of your system in the state $\psi_1^\prime =(\psi_1+\psi_2)/\sqrt{2}$ (being not an energy eigenstate), you will measure the energy eigenvalue $E_1$ in $50 \% =(1/\sqrt{2})^2$ of the cases and $E_2$ also in $50 \% =(1/\sqrt{2})^2$ of your copies (disregarding the usual statistical fluctuations disappearing in the limit $N \to \infty$).

Simply regarded as a basis of your Hilbert space, $B_1 = \{\psi_1, \psi_2, \psi_3, \ldots\}$ (consisting of energy eigenvectors) is just as good as the other one, $B_2= \{\psi_1^\prime, \psi_2^\prime, \psi_3, \ldots \}$ ($\psi_1^\prime$ and $\psi_2^\prime$ being not energy eigenstates). An arbitrary state $\psi$ can equally well be expanded with respect to $B_1$ or $B_2$. The advantage of the basis $B_1$ is, however, that the matrix elements of the Hamilton operator, $\langle \psi_m | H | \psi_n \rangle = E_n \delta_{m n}$, are diagonal, whereas $\langle \psi_1^\prime | H \psi_2^\prime \rangle \ne 0$ with respect to $B_2$.

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