# Propagator of Hermitian operator

Is the propagator of a hermitian operator always unitary?

I am asking this because the propagator of in the Schrödinger equation is unitary and my book says this is to be expected since the Hamiltonian is hermitian. However, I constructed the propagator of 2 coupled mass system of same mass and same spring constant and found out that it is not unitary!(Hamiltonian in one dimensional case)

Edit: What I mean by unitary is: $UU^+=U^+U=I$ and the solution to Schrödinger equation: $$|\psi(t)\rangle =\sum e^{\frac{-itE}{\hbar}}|E\rangle\langle E|\psi(0)\rangle$$ is in terms of the propagator $$\psi(t)=\sum U_{H}(t)|\psi(0)\rangle$$ so this propagator $U_H(t)$ is unitary becuase the Hamiltonian is hermitian according to my book. However when I solved a coupled 2-mass system of zero initial velocity. The solution in terms of the propagator was: $$|x(t)\rangle =\sum U_{sp}(t) |x(0)\rangle$$ where $U_{sp}$ is the propagator of the spring-coupled masses.

The differential equation describing the coupled 2-mass system is(zero initial velocity for both masses):

$\begin{pmatrix} x1^{''}(t) \\ x2^{''}(t) \end{pmatrix}$=$\begin{pmatrix} \frac{-2k}{m}& \frac{k}{m}\\ \frac{k}{m}& \frac{-2k}{m} \end{pmatrix}$*$\begin{pmatrix} x1(t)\\ x2(t) \end{pmatrix}$ where $\Omega$=$\begin{pmatrix} \frac{-2k}{m}& \frac{k}{m}\\ \frac{k}{m}& \frac{-2k}{m} \end{pmatrix}$

The solution in terms of the propagator and initial displacement is $$\begin{pmatrix} x1(t)\\ x2(t) \end{pmatrix}=\begin{pmatrix} \frac{{}cos(\sqrt{}\frac{k}{m}t)+cos(\sqrt{}\frac{3k}{m}t)}{2} & \frac{{}cos(\sqrt{}\frac{k}{m}t)-cos(\sqrt{}\frac{3k}{m}t)}{2} \\ \frac{{}cos(\sqrt{}\frac{k}{m}t)-cos(\sqrt{}\frac{3k}{m}t)}{2} & \frac{{}cos(\sqrt{}\frac{k}{m}t)+cos(\sqrt{}\frac{3k}{m}t)}{2} \end{pmatrix}\begin{pmatrix} x1(0)\\ x2(0) \end{pmatrix}$$ where the propagator here is $$U_{sp}(t)=\begin{pmatrix} \frac{{}cos(\sqrt{}\frac{k}{m}t)+cos(\sqrt{}\frac{3k}{m}t)}{2} & \frac{{}cos(\sqrt{}\frac{k}{m}t)-cos(\sqrt{}\frac{3k}{m}t)}{2} \\ \frac{{}cos(\sqrt{}\frac{k}{m}t)-cos(\sqrt{}\frac{3k}{m}t)}{2} & \frac{{}cos(\sqrt{}\frac{k}{m}t)+cos(\sqrt{}\frac{3k}{m}t)}{2} \end{pmatrix}$$ which is not unitary?! But if the original matrix $\Omega$ is hermitian shouldn't the propagator be unitary?

• what exactly do you mean by propagator, and what does it mean for it to be unitary? Is it the fundamental solution of some PDE? by unitary you mean that $||f||=||G*f||$ for some norm $||\cdot||$? – AccidentalFourierTransform Dec 2 '17 at 18:33
• Just as a small comment: for the future, please, use \langle and \rangle inside the code for the < and >. OTOH, what makes you think the Hamiltonian in your problem is not self-adjoint, or your U is not unitary? – DanielC Dec 2 '17 at 18:58
• You said you've computed the propagator for the spring-couple masses and it's not unitary. Can you show us your hamiltonian and your propagator? – John Donne Dec 2 '17 at 19:31
• What you wish to examine, instead, is the unitary Mehler kernel, K, so $\psi(x,t)=\langle x|e^{-itH/\hbar}|\psi\rangle =\int dy \langle x| e^{-itH/\hbar}|y\rangle \langle y|\psi\rangle=\int dy~ \psi(y,0)~K(x,t;y,0)$. – Cosmas Zachos Dec 2 '17 at 21:55
• What you wrote classically is not the propagator. You may define something of the sort in phase-space, but I'm not sure "where you are coming from"--this is not a tutorial. For the QM propagator, the WP article is pretty self-explanatory. You might rewind your question. – Cosmas Zachos Dec 2 '17 at 22:53

Given the Schrodinger equation $$i\hbar \frac{\partial \psi}{\partial t}=H\psi$$ one can formally write the solution as: $$\psi(t)=e^{-itH/\hbar}\psi(0)$$ We can call $U(t)=e^{-itH/\hbar}$ the "propagator". Then if $H$ is hermitian, $U(t)$ is unitary, as you correctly claim. It is helpful to note that the argument of the exponential is however anti-hermitian: $(iH)^\dagger = -iH$. So we can also state the following. Consider the equation: $$\frac{\partial y(t)}{\partial t}=A y(t)$$ where $A$ doesn't depend on $t$. The solution is formally given by $y(t)=e^{At} y(0)$. So in your definition, the "propagator" is given by $U(t)=e^{At}$. So in particular, the "propagator" in this case is unitary if $A$ is anti-hermitian. We only required $H$ to be hermitian in the Schrodinger eq. because there was an $i$ which made the overall argument anti-hermitian. The argument is the same if $y$ is a vector.
The system you are considering is second order. Suppose we have a system of second order ODEs: $$\ddot{\textbf{x}}(t) = M \textbf{x}(t)$$ The solution is going to depend on the initial velocity, so it won't be in the form you seem to be interested in. Suppose that the initial velocity is zero, so that the solution is given by $\textbf{x}(t) = U(t) \textbf{x}(0)$. Can we require some condition on $M$ which would ensure that $U$ is unitary? Your example shows that it's not enough to require that $M$ is hermitian. It's easier to analyse a first order differential equation, so let's write: $$\textbf{y}(t)=\begin{pmatrix} \textbf{x} \\ \textbf{x}' \end{pmatrix}\,\,\,\,\,\,\,\,\,\, K = \begin{pmatrix} 0 & I \\ M & 0 \end{pmatrix}$$ Then the original second order system is equivalent to the first order system $\dot{\textbf{y}} = K \textbf{y}$. The solution to this problem can be written very easily as $\textbf{y}(t) = \exp{(Kt)} \textbf{y}(0)$. Can we express the exponential in a nice way? Playing with the expression for $K$ we realise its powers have the following expression: $$K^{2k} = \begin{pmatrix} M^k & 0 \\ 0 & M^k \end{pmatrix}\,\,\,\,\,\,\,\,\,\,K^{2k+1} = \begin{pmatrix} 0 & M^k \\ M^{k+1} & 0 \end{pmatrix}$$ Therefore we can write its exponential as: $$\exp{K}=\sum_{n=0}^\infty \frac{K^n}{n!}= \sum_{k=0}^\infty \begin{pmatrix} \frac{M^k}{(2k)!} & \frac{M^k}{(2k+1)!} \\ \frac{M^{k+1}}{(2k+1)!} & \frac{M^k}{(2k)!} \end{pmatrix}$$ and since we're assuming that the initial velocity is zero, we get for the "propagator": $$U(t)=\sum_{k=0}^\infty \frac{M^k t^k}{(2k)!}$$ This is the Taylor series for $\cosh{\sqrt{x}}$. From the series you can see that $M$ hermitian implies that $U$ is hermitian, like you got in your result. What if $M$ is anti-hermitian? It was the case of interest previously. It is not true in general that $M$ anti-hermitian implies $U$ unitary. In particular $$M=\begin{pmatrix} i & 0 \\ 0 & i\end{pmatrix}\implies U(1)=\begin{pmatrix} \cosh{e^{i\pi/4}} & 0 \\ 0 & \cosh{e^{i\pi/4}}\end{pmatrix}$$ but since $\cosh{e^{i\pi/4}}\cosh{e^{-i\pi/4}}\neq 1$, then $UU^\dagger \neq 1$.