On pages 9 & 10 in Chapter 1 of Landau & Lifshitz Quantum Mechanics, the authors argue that eigenfunctions $\psi_n$ & $\psi_m$ are orthogonal (where $\psi_n$ and $\psi_m$ correspond to results $f_n$ & $f_m$ of some observable $f$).

The assumptions used to argue the conclusion appear to be that, when in state $$ \psi = \sum_n a_n \psi_n, $$ the probability of getting a result $f_n$ for some fixed $n$ is determined by:

  • An expression "bilinear in $\psi$ & $\psi^*$"
  • $\implies$ bilinear in "$a_n$ and $a_n^*$"

The authors then come to conclude that "the only essentially positive quantity satisfying these conditions is the square of the modulus $|a_n|^2$." The relevant section goes as follows:

The expansion (3.2) makes it possible to determine the probability of finding (i.e. the probability of getting the corresponding result on measurement), in a system in a state with wavefunction $\Psi$, any given value $f_n$ of the quantity $f$. For, according to what was said in the previous section,these probabilities must be determined by some expressions bilinear in $\Psi$ and $\Psi^*$, and therefore must be bilinear in $a_n$ and $a_n^*$. Furthermore, these expressions must, of course, be positive. Finally, the probability of the value $f_n$ must become unity if the system is in a state with wave function $\Psi=\Psi_n$, and must become zero if there is no term containing $\Psi_n$ in the expansion (3.2) of the wave function $\Psi$. This means that the required probability must be unity if all the coefficients $a_n$ except one (with the given $n$ are zero, that one being unity; the probability must be zero, if the $a_n$ concerned is zero. The only essentially positive quantity satisfying these conditions is the square of the modulus of the coefficient $a_n$. Thus we reach the result that the squared modulus $|a_n|^2$ of each coefficient in the expansion (3.2) determines the probability of the corresponding value $f_n$ of the quantity $f$ in the state with wave function $\Psi$. The sum of the probabilities of all possible values $f_n$ must be equal to unity; in other words, the relation $$\sum_n |a_n|^2 = 1. \tag{3.3}$$

(Note expansion (3.2) is simply the expression for $\psi$ given above.)

I'm having trouble understanding the details of their argument - why does it follow from only those conditions that the probability must be given by $|a_n|^2$ ?

  • $\begingroup$ Possibly a math question. I think it belongs here but I tend to lose these arguments. $\endgroup$
    – daniel
    May 4, 2017 at 8:46
  • $\begingroup$ I think this belongs here, too. $\endgroup$ May 4, 2017 at 11:15
  • $\begingroup$ It definetely belongs here because it's not really mathematical problem rather understanding the quirks of L&L=))) $\endgroup$
    – OON
    May 4, 2017 at 11:40

1 Answer 1


The requirement that the probability for $f_n$ be a bilinear in $a_n$ and $a_n^*$ means that it must have the explicit component form $$ p_n = \sum_{i,j} c^{(n)}_{ij} a_i^*a_j. \tag 1 $$ You then require that if $a_i = \delta_{im}$, i.e. if you have $\psi$ along $\psi_m$, the probability for $f_n$ be zero unless $n=m$, i.e. you require that $p_n = \delta_{nm}$, and you know that in these conditions $(1)$ reduces to $$ p_n = \sum_{i,j} c^{(n)}_{ij} \delta_{im}\delta_{jm} = c^{(n)}_{mm}, $$ so this fixes the diagonal elements to $c^{(n)}_{mm} = \delta_{mn}$.

The argument thus far, of course, does not say anything about the off-diagonal elements. To deal with these, we need to add a very important requirement to your list:

  • $p_n$ needs to be a positive-definite bilinear form (in L&L's language, "essentially positive").

Given the strong restrictions on the diagonal elements we already have, one can then show that the positivity requirement kills all the off-diagonal elements. To see why, it's useful to split this into two cases:

  1. First consider the case of $c^{(n)}_{kl}$ where neither of $k$ and $l$ are equal to $n$. Here, if you restrict the coefficient matrix to those two rows and columns, you get a two-by-two matrix that has the essential structure $$c^{(n)} = \begin{pmatrix}0&c\\c^*&0\end{pmatrix},$$ and this is obviously not positive-definite. To be a bit more concrete, if $c^{(n)}_{kl}$ is not zero, then $p_n$ for the wavefunction $\psi = \frac{1}{\sqrt{2}}(\psi_k - \sqrt{c^*/c}\,\psi_l)$ is negative.

  2. For the case where one of the two indices is equal to $n$, we can repeat the same argument, except that now the reduced two-by-two matrix has a one on the diagonal, $$c^{(n)} = \begin{pmatrix}0&c\\c^*&1\end{pmatrix},$$ so at least naively there is some hope to make it positive definite, but that is not the case. This can be seen immediately from the eigenvalues of the reduced matrix, which are given by $\frac12\left(1\pm \sqrt{1+4|c|^2}\right)$, one of which is always negative unless $c=0$. That means, then, that by taking $\psi$ along the corresponding eigenvector, you would obtain a negative probability $p_n$.

This completes the proof - you require that the only nonzero coefficient is $c^{(n)}_{nn}$.

  • $\begingroup$ Thanks a lot for this detailed answer. Forgot to include it in the question but dealing with the off-diagonal elements was exactly where I was stuck. Using positive-definite matrices to analyse it is pretty neat! (As you can probably tell I'm quite new to this...) $\endgroup$
    – Platehead
    May 4, 2017 at 12:27

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