# Time evolution operator acting on a non-eigenket

I'm taking a course in QM at my university, and I'm trying to work out an assignment given to the class by our professor. The setup is as follows:

The problem is about a simplified description of neutrinos. The main simplification is that we only care about the neutrino type. We assume that when measured a neutrino can be found to be one, and only one, of two types: an electron neutrino, denoted $|\nu_e\rangle$, or a muon neutrino $|\nu_\mu\rangle$ (we are not including the tau neutrino).

We represent the two kets as $|\nu_e\rangle = \begin{bmatrix}1\\0\end{bmatrix}$ and $|\nu_\mu\rangle = \begin{bmatrix}0\\1\end{bmatrix}$. Furthermore we have the hamiltonian given in this representation as $$H = \begin{bmatrix} \alpha_e & g_{e\mu}\\ g_{\mu e} & \alpha_\mu \end{bmatrix}$$ where $\alpha_e,\alpha_\mu \in \mathbf{R}$ are constants related to the masses and kinetic energies of the neutrinos, and $g_{e\mu} = g_{\mu e}^*$ are complex numbers related to the interactions between different types of neutrinos.

The last point in this problem prior to where my issues arise asks us to write $H$ as a linear combination of the Pauli matrices and the (2x2) identity matrix. I believe a have this right, and it comes out to the following: $$H = \frac{g_{\mu e} + g_{e \mu}}{2}\sigma_1 + \frac{g_{\mu e} - g_{e \mu}}{2i}\sigma_2 + \frac{\alpha_e - \alpha_\mu}{2}\sigma_3 + \frac{\alpha_e+\alpha_\mu}{2}I_2.$$

Now we assume that a neutrino in the state $|\nu\rangle = a|\nu_e\rangle + b|\nu_\mu\rangle$ (with $|a|^2+|b|^2=1$) has just been measured to be an electron neutrino, so that it is in the state $|\nu\rangle=|\nu_e\rangle$ at time $t=0$. We now want to find the probability of a second measurement on the same neutrino at a later time $t$ giving the result that the neutrino is a muon neutrino.

From here I'm working on the assumption that I need to find $|\nu(t)\rangle$, and use that to find to probability. I know that to do this I can act on $|\nu(t=0)\rangle =|\nu_e\rangle$ with the time evolution operator $U$: $$U = \exp\left(-iHt/\hbar\right) = \sum_{k=0}^\infty \frac{1}{k!}\left(-\frac{iHt}{\hbar}\right)^k.$$

This is where I get stuck. Normally when we have acted with $U$ we have done so on an energy-eigenket so that $H^k|\lambda\rangle = E^k|\lambda\rangle$, and from there things are not that bad. But now, $|\nu_e\rangle$ is not an energy-eigenstate, leading to a very ugly expression.

I have been able to work out a recursive formula (that I think is correct) allowing me to compute $|\nu(t)\rangle$ numerically. Unfortunately, the assignment explicitly asks for an analytical solution for the probability. I'm also thinking that the rewrite of $H$ as a linear combination of Pauli matrices should be used, but I have not been able to figure out how this might help.

I'm also a little unsure as to what "measuring the neutrino to be a muon neutrino" means. What kind of measurement is this referring to? What eigenvalue equation has been solved to give a measurement value that we interpret to give the neutrino type? It seems to me I need to know this in order to find the probability, e.g. what to do if I actually find $|\nu(t)\rangle$?

It has been a full day working with my fellow students on this, so I'm hoping someone here has some helpful insight!

EDIT: Without a doubt the easiest way to do it was to exploit the exponential properties of Pauli matrices, as suggested in comments. Still, I managed to do it using all proposed solutions, although with different algebraic complexities.

• Find the eigenstates of $H$ and expand the initial state $|\nu_e\rangle$ in this eigenbasis. Then apply $e^{-iHt}$. – Meng Cheng Sep 27 '15 at 20:44

## 1 Answer

The trick to exponentiate the $2\times 2$ matrix is to diagonalise it. A diagonal matrix $D$ exists such that $H= C^{t}DC$, with $C$ being orthogonal due to the fact that $H$ is hermitian. Using the orthogonality of $C$ and the definition of the exponential as power series leads you to prove that $$e^{-iHt} = e^{-i(C^{t}DC)t} = C^{t}\,e^{-i(D)t}\,C$$ and now the exponential of a diagonal matrix is nothing but a matrix whose entries are the exponential of the previous ones. This is equivalent to first finding the eigenvectors of the Hamiltonian and then expressing your neutrino state in terms of those: the transformation matrix $C$ appearing is exactly the same.

I'm also a little unsure as to what "measuring the neutrino to be a muon neutrino" means.

Once you have acted upon your initial state $|\nu\rangle$ with the time evolution operator (exploiting the above formula) you will end up with a new evolved state, say $|\nu'(t)\rangle$ which can be expressed as linear combination of the muon and electron neutrino states, since you assume the two span a basis for your space; namely $$|\nu'(t)\rangle = c_1(t)|\nu_e\rangle + c_2(t)|\nu_{\mu}\rangle.$$ Taking the scalar product with either of your basis vectors, $c_1(t), c_2(t)$ represent the components of the new state onto the previous one, whose modulus square gives, in the standard interpretation of quantum mechanics, the probability that the state will collapse into either of the two once you perform a measurement (however this will be performed).

• I had thought of finding the eigenkets, but got stuck in frustration... I hadn't thought about this way with diagonalising the matrix, but it seems like a really nice way to do it. Thank you!! – Bendik Sep 28 '15 at 9:39
• Diagonalising the Hamiltonian is more or less the standard trick to calculate its exponential. In the example at hand I guess that you can even use some exponential properties of the Pauli matrices to further simplify the final expression. – gented Sep 28 '15 at 9:49
• That might be, I will look into it. Some what strange to me that this is the standard way to do it, and yet we have not been introduced to the idea. Or maybe that was the professors hope, that we would find out about it during the course of trying to solve the problem. Anyway, thanks again! – Bendik Sep 28 '15 at 9:59
• I can't seem to get this right. I think I understand the way to do it, but finding the matrices $C$ and $D$ gets really tricky because $H$ is more or less a general (hermitian) matrix. My expressions get so complicated that I have yet to end up with the same answer twice due to some kind of mistake. I feel like this shouldn't be that difficult, and that I'm missing something to simplify the calculations. I don't suppose you have any tips as to how I should start evaluating $C^t\exp(-iDt/\hbar)C$? @Gennaro – Bendik Sep 28 '15 at 20:57
• @Bendik That the expression gets down to calculations, that may certainly be, but I believe they cannot be that hard. $C$ is the matrix you diagonalise the Hamiltonian with and $D$ contains its eigenvalues (and is diagonal, of course). It's a $2\times 2$ matrix, it cannot be that impolite: where do your calculations problems lie? – gented Sep 28 '15 at 21:03