I'm confused as to whether the sine function can be technically considered an eigenfunction for momentum operator.

Once the sine function is decomposed, it can be decomposed as a linear sum of two eigenfuntions for the momentum operator since $$\sin(kx)=\frac{1}{2i}\left[e^{ikx}+e^{-ikx}\right] \, .$$ Applying the momentum operator on each of these functions gives $hi$ and $-hi$ for the momentum values. However, applying the momentum operator on $\sin(kx)$ itself clearly shows that it is not an eigenfunction one cannot recover the original sine function after applying the operator. Why is sine not an eigenfunction despite the fact that it is a linear sum of two eigenfunctions?

Also, for a particle in a box, one can use its wavefunction (a sine function $\sqrt{2/L} \sin(n\pi x/L)$) to calculate the average momentum as 0. If sine is not an eigenfunction for momentum, how can we calculate the average for the momentum observable?

  • 1
    $\begingroup$ The wavefunction for a particle in a finite box is NOT $\sqrt(2/L) * sin(n \pi x/L)$. It is $\sqrt(2/L) * sin(n \pi x/L) \times Rect(x/L)$, where Rect() is the standard rect function. Alternatively, it's defined piece-wise. Its a very important difference, especially when you discuss things like momentum eigenfunctions! The Fourier transform of a sine and the above function look completely different. $\endgroup$ – aquirdturtle Dec 27 '16 at 6:19


$\hat p\sin(kx)$ is not a multiple of itself. It is true that $\hat p\,e^{\pm i k x}= \pm\hbar k e^{\pm i k x}$ but a (complex) linear combination of two eigenfunctions is only an eigenfunction if both eigenfunctions in the sum have the same eigenvalue, which is NOT the case here as the eigenvalues $\pm \hbar k$ differ by a sign.

$\sin(kx)$ is an eigenfunction of $\hat p^2$ (although be careful as it is not a normalizable eigenfunction in the sense that $\int_{-\infty}^\infty \sin^2(kx) dx$ is not finite).

In response to comments: Let $\hat T$ be your favorite operator, and let $\phi_A(x)$ and $\phi_B(x)$ be such that $$ \hat T\phi_A(x)=A\phi_A(x)\, ,\qquad \hat T\phi_B(x)=B\phi_B(x)\, . $$ Take the complex linear combination $$ \psi(x)=\alpha \phi_A(x)+ \beta \phi_B(x) $$ and look at \begin{align} \hat T\psi(x)&= A\alpha \psi_A(x)+B\beta\phi_B(x)\\ &=A(\alpha \psi_A(x)+\beta\phi_B(x))+(B-A)\beta \phi_B(x)\\ &=A\psi(x)+(B-A)\beta\phi_B(x) \end{align} For this to be a multiple of the original $\psi(x)$ one must "cancel" the extra $\phi_B(x)$ term and so have $A=B$, i.e. the eigenvalues are the same, or $\beta=0$, meaning you did not have a linear combination to start with.

The calculation of average values is done using \begin{equation} \int dx\, \psi^*(x) \left[ \hat T\psi(x)\right]\, \end{equation} You can use the explicit expression of $\psi(x)$ above and find, if your functions $\phi_A(x)$ and $\phi_B(x)$ are orthonormal in the sense that $$ \int dx \phi_i^*(x) \phi_j(x)=\delta_{ij} $$ that the average value of $\hat T$ comes out as $\alpha^*\alpha A+\beta^*\beta B$. Of course if $A=B$ and $\alpha^*\alpha +\beta^*\beta=1$ then the average value is just the eigenvalue, but this is really a special case of the general expression for the average value. The average value makes sense as the integral above for any $\psi(x)$, irrespective of whether or not one know the expansion of $\psi(x)$ in terms of eigenfunctions of $\hat T$.

In the case of $\sin(n\pi x/L)$ and $\hat p$, the mathematical machinery can bypassed by a physical argument. The quantity $$ \psi^*(x)\left[\hat p \psi(x)\right]= i\hbar (n\pi/L) \sin(n\pi x/L)\cos(n \pi x/L) $$ is purely imaginary, so the integral will also be purely imaginary. However, $\hat p$ is an observable so its average values must be real. This is a contradiction unless the average value is $0$. This "trick" will apply whenever the wavefunction is real and the observable is represented by an operator of the form $i\times$(something real). Another situation where this trick would be useful would be the average value of $\hat p$ for a wave function that is a real linear combination of harmonic oscillator states.

  • $\begingroup$ Thanks for the reply. Could you explain why the eigenfunctions need to have the same eigenvalue in order for the superposition to be valid? Also, could you address the second question as to why we can still find the average of the observable if it's not an eigenfunction? $\endgroup$ – John Smith Dec 27 '16 at 16:26

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