I am currently taking a course in QM and can't see how the eigenstates have been found for examples like this one:

Let $\phi _1$ and $\phi _2$ be two normalised wavefunctions orthogonal onto each other. Let the action of the operator $\hat{A}$ on these states be:

$$\hat{A} \phi _1 = \phi _2$$

$$\hat{A} \phi _2 = \phi _1$$

Then the states $\Psi _1 = \phi _1 + \phi _2$ and $\Psi _2 = \phi _1 - \phi _2$ are eigenstates of $\hat{A}$ corresponding to the eigenvalues $+1$ and $-1$ respectively, that is:

$$\hat{A} \Psi _1 = \Psi _1$$

$$\hat{A} \Psi _2 = -\Psi _2$$

So I can see the states $\Psi _1$ and $\Psi _2$ satisfy the eigenfunction equation ($\hat{A}\Psi = \lambda \Psi$) with the given eigenvalues but I can't see how the states were derived, they seem to have been plucked from thin air but just work! Lots of QM questions seem to utilise states in the form $(\alpha \phi _1 \pm \beta \phi _2)$ but I'm not sure why?!

Appreciate any help!


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  • $\begingroup$ Are $\phi_i$ a basis in this case? $\endgroup$ – gabe Mar 29 at 22:04
  • 1
    $\begingroup$ I don't think there's much to see here. They didn't come up with those vectors from somewhere . They're just saying, if these vectors satisfy these particular conditions related to this operator, then those particular linear combinations are eigenvectors of the operator. It just so happens that this particular situation comes up a lot, because we often model systems as two-level systems. $\endgroup$ – march Mar 29 at 22:17
  • $\begingroup$ @gabe it isn't explicitly said they form a basis, but I am presuming they do. The text above is exactly as given. $\endgroup$ – Matt_Ald Mar 29 at 22:24
  • $\begingroup$ @march Ah ok, so its just saying this works as an example in a generic case rather than deriving a specific situation. Interesting we usually use two-level systems, is there a reason for this or is it just convenient and usually the simplest representation? $\endgroup$ – Matt_Ald Mar 29 at 22:26
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    $\begingroup$ Write $\hat A$ as a 2x2 matrix and appreciate the point. $\endgroup$ – Cosmas Zachos Mar 29 at 22:43

Assume $\phi_{1,2}$ form a basis. Consider the eigenvalue equation for $\hat{A}$, i.e. $\hat{A}\psi=\lambda\psi$. If we apply $\hat{A}$ again we get the equation $\hat{A}^2\psi=\lambda^2\psi$. But note from the definition of $\hat{A}$, i.e. its action on the basis, that $\hat{A}^2 =\text{Id}$. Thus the previous equation gives us

$$ \lambda^2=1 \rightarrow \lambda=\pm1 $$

So we have found the eigen values pretty easily. The question remains as to how to find the eigenvectors. To do this, we begin by writing $ \phi_1 = \left( \begin{matrix} 1 \\\ 0 \end{matrix}\right)$ and $ \phi_2 = \left( \begin{matrix} 0 \\\ 1 \end{matrix}\right)$. A little thought leads us to conclude that we can represent $\hat{A}$ with the matrix

$$ \hat{A} = \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right)$$

For $\lambda = 1$ we want to solve $ \left(\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right)-\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}\right) \right) \left(\begin{matrix} a \\b\end{matrix} \right) = \left(\begin{matrix} 0 \\0\end{matrix} \right)$, or $\left( \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix}\right)\left(\begin{matrix} a \\b\end{matrix} \right) = \left(\begin{matrix} 0 \\0\end{matrix} \right)$. This yields the system tells us that $a=b$ and we are left to choose the normalization, so we say $a=b=1$. Thus the eigenvector corresponding to $\lambda=1$ is $\left(\begin{matrix} 1 \\1\end{matrix} \right) = \left(\begin{matrix} 1 \\0\end{matrix} \right) + \left(\begin{matrix} 0 \\1\end{matrix} \right) = \phi_1 + \phi_2$. The eigenvector for $\lambda=-1$ follows similarly.

This is a standard example of a two level system, which @march points out in a comment is very common in physics. As Cosmas Zachos points out in another comment, the mystery vanishes when you realize you can write operators on two level systems as a two by two matrix, and find the eigenvectors in the usual way.


gabe has already given an answer in terms of matrices and using idempotency. I shall exhibit a rather dry approach.

Assuming that the $\phi$’s form a basis, then any vector can be expressed as $\alpha\phi_1+\beta\phi_2$.

Now let $\psi=\alpha\phi_1+\beta\phi_2$ be an eigenstate of $A$. Then we have,


$\implies \alpha A\phi_1+\beta A \phi_2$

$\implies \alpha\phi_2+\beta\phi_1=\lambda\alpha\phi_1+\lambda\beta\phi_2$

$\implies \lambda\alpha=\beta , \lambda\beta=\alpha$

$\implies \lambda^2\alpha=\alpha$

$\implies \lambda = \pm 1$

Plugging these back into $\alpha-\beta$ relation gives us $\alpha=\pm\beta$

Finally $\psi = \phi_1 \pm \phi_2$


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