1
$\begingroup$

Hi I have a query about the difference of two aspects of the statistical interpretation of quantum mechanics given in the popular introductory quantum mechanics books "Introduction to Quantum Mechanics" by Griffiths and "Quantum Mechanics Concepts and Applications" by Zettili.

In Griffiths we have: Consider an observable $\hat{Q}$, with eigenfunctions $f_{n}(x)$ and associated eigenvalues $q_n$. Thus since the eigenfunctions are complete, we have $\Psi(x,t) = \sum_{n}c_n f_n(x)$ which is $$\Psi(x,t) = \langle x| \Psi \rangle = \sum_n \langle x | f_n \rangle \langle f_n |\Psi \rangle$$ where $c_n = \langle f_n | \Psi \rangle$ and $|c_n|^2$ is the probability of the measurement of $\hat{Q}$ would yield eigenvalue $q_n$. After a measurement of the observable, Griffiths states that the state vector collapses to one of the eigenfunctions $f_n$.

In Zettili, he seems to deal with $c_n$ and eigenvalues $a_n$ as the same thing. He has that with eigenvalues $a_n$ and eigenfunctions $| \psi_n \rangle$ of observable $\hat{A}$ which acts on state vector $|\psi(t) \rangle$ we have $$|\psi(t) \rangle = \sum_n a_n | \psi_n \rangle.$$

As can be seen Zettili omits the coefficients $c_n$ but rather takes the eigenvalues $a_n$ as the coefficient. And thus Zettili states that the probability of getting eigenvalues $a_n$ is: $$P_n(a_n) = \frac{|a_n|^2}{\langle \psi | \psi \rangle}.$$

Which interpretation is correct (or preferable) and mostly commonly used?

Also, Zettili states that the state vector collapses to $a_n| \psi_n \rangle$ where as Griffiths just states that the state vector collapses to the eigenfunction?

Page in question:enter image description here

$\endgroup$
  • 1
    $\begingroup$ If indeed that's what Zettili has, then it's wrong, but it's so wrong that the natural candidate is that you're misquoting it. $\endgroup$ – Emilio Pisanty May 30 '16 at 12:45
  • 1
    $\begingroup$ There is no evidence of a mistake in either book. The idea that $a_n$ is an eigenvalue is purely a mistake by the OP. $\endgroup$ – Luboš Motl May 30 '16 at 12:46
  • $\begingroup$ @EmilioPisanty I updated the question to include an attachment of the page in question. Maybe I am missing something but this is what it seems to state. $\endgroup$ – user100411 May 30 '16 at 12:54
  • 1
    $\begingroup$ OK, with the picture, I agree that the book by Zettili is completely sloppy and uses the same symbol for the eigenvalues and the probability amplitudes everywhere. This is really bad enough to make the book unusable. They're completely different things, they don't even have the same unit, the eigenvalue is typically real while the amplitude is importantly complex, and so on. It seems to be more than a typo, the author really seems confused about some basic things. $\endgroup$ – Luboš Motl May 30 '16 at 12:56
  • 1
    $\begingroup$ Yes, I agree with Luboš - the misidentification of $a_n$ as two different objects on the same page is bad enough to completely disqualify the book. $\endgroup$ – Emilio Pisanty May 30 '16 at 12:58
2
$\begingroup$

There is absolutely no "freedom for interpretations" here. Both books should – and all other books that are not completely wrong do – agree with these formulae and agree that $a_n$ never indicates an eigenvalue.

In both books and all of science, $a_n$ is the complex probability amplitude such that $|a_n|^2$ represents the probability that the system has the $n$-th eigenvalue of the corresponding operator. The eigenvalue is usually called $\lambda_n$. After this answer was written, a screenshot proved that Zettili really uses the symbol $a_n$ both for the amplitude and the eigenvalue – it's a big enough confusion in the notation.

The complex probability amplitude $a_n$ or $c_n$ – both notations are widespread, and many others – are the coefficients in the expansion of a state vector $$ | \phi \rangle = \sum_n c_n |\psi_n\rangle $$ Here, the basis vectors $|\psi_n\rangle$ are eigenvectors of an operator $L$ $$ L |\psi_n \rangle = \lambda_n | \psi_n \rangle$$ where $\lambda_n$ are the eigenvalues. Also, the equation $$ P_n(a_n) = \frac{|a_n|^2}{\langle \psi | \psi \rangle} $$ says that the probability that the eigenvalue is $\lambda_n$ (which is indicated simply by the subscript $n$ of $P_n$) is the squared absolute value of the probability amplitude, $|a_n|^2$. The denominator is written there to allow the norm $\langle \psi|\psi \rangle$ to be different than one – it has the same effect as rescaling $|\psi\rangle$ for the norm to be one.

Also, in the equation, the left hand side contains $(a_n)$ which simply says that the probability that the eigenvalue $\lambda_n$ is realized is a function of the probability amplitudes $a_n$ (well, the particular one with the same $n$ is the most important one, but the others may enter through the denominator which is needed if one ignores the condition that the norm should be one).

$\endgroup$
  • $\begingroup$ You might be right but this is what it seems to be stating. See edited question for attachment. $\endgroup$ – user100411 May 30 '16 at 12:51
  • 1
    $\begingroup$ Sorry, I agree, there is a typo repeated a few times below (3,2), "eigenvalue $a_n$" should be "eigenvalue $\lambda_n$" a few times. Well, he's sloppy and clearly uses the same symbol both for the amplitudes and the eigenvalues. So the full correction of the book would need to decide about almost every equation. ;-) $\endgroup$ – Luboš Motl May 30 '16 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy