## 4 Observables
### Problems with Deutsch Measurement  
Because a QTM must be physically feasible, at least \"in principle\", let us analyse, in detail, whether the termination protocol proposed by [PERSON] [10] is realizable.  
As any quantum system, a QTM evolves with respect to a continuos time. This means that the time evolution operator \(U\) of a machine \(M\) is the solution of the Schrodinger equation with a given Hamiltonian \(H\) and depends on the time [21]. In particular, when \(H\) is constant, the solution of the equation is  
\[U_{t}=e^{-(i/h)Ht}\quad\text{ for }t\in\mathbb{R}^{+}\]  
Now, let \(\Delta\) be the duration of a single step of a Deutsch QTM. This means that after \(n\) steps of the QTM, we are at time \(t=n\Delta\), and that for the corresponding configurations \(|\psi_{n}\rangle\) of the QTM, it holds the following property  
\[|\psi_{n+1}\rangle =U_{(n+1)\Delta}|\psi_{0}\rangle\] \[=e^{-(i/h)H(n+1)\Delta}|\psi_{0}\rangle=e^{-(i/h)H\Delta}e^{-(i/ h)Hn\Delta}|\psi_{0}\rangle=U_{\Delta}U_{n\Delta}|\psi_{0}\rangle\] \[=U_{\Delta}|\psi_{n}\rangle\]  
In other words, at each step \(n\), the next configuration \(|\psi_{n+1}\rangle\) is obtained by applying the evolution operator \(U_{\Delta}\) to the current configuration \(|\psi_{n}\rangle\).  
Now, let us suppose that, as a consequence of the measurement of the halt qubit \(T\), after \(n\) steps, and, therefore, at time \(t_{n}=n\Delta\), the system collapses to the quantum configuration  
\[|\psi_{n}\rangle=\frac{|T=1\rangle\otimes\sum_{D\in\ell^{/in}}d_{D}|D\rangle}{ \sqrt{\sum_{D\in\ell^{/in}}|d_{D}|^{2}}}\]  
According to [PERSON]'s approach, one has to perform a second measurement to read the actual value of the output on the tape. However, since we are in a continuous time setting, what is the time of this second measurement? Since it can be done only after the measure of the halt qubit \(T\), this must take place at a time \(t_{n+\epsilon}=(n+\epsilon)\Delta\), with \(\epsilon>0\). We can distinguish two cases (analogous reasonings apply for \(\epsilon>1\)):  
1. \(0<\epsilon<1\). In this case, the tape is measured at an intermediate time between \(t_{n}\) and \(t_{n+1}\), and the machine is in an intermediate configuration \(|\psi_{n+\epsilon}\rangle\) between \(|\psi_{n}\rangle\) and \(|\psi_{n+1}\rangle\). Because the QTM is defined in terms of the unitary operator \(U_{\Delta}\), its configurations are well-defined only at time \(t_{n}\), for \(n\in\mathbb{N}\). The quantum configuration \(|\psi_{n+\epsilon}\rangle\) is therefore unknown and depends on how the unitary operator \(U_{\Delta}\) is physically realised. In any case, in general it differs from \(|\psi_{n}\rangle\).
2. \(\epsilon=1\). In this case the second measurement takes place at time \(t_{n+1}\), on the configuration \(|\psi_{n+1}\rangle=U_{\Delta}|\psi_{n}\rangle\). That implies \(|\psi_{n}\rangle\
eq|\psi_{n+1}\rangle\), since \(|\psi_{n}\rangle=|\psi_{n+1}\rangle\) may hold only in the trivial case the unitary operator \(U_{\Delta}\) is the identity and the \(|\psi_{n}\rangle=|\psi_{0}\rangle\), for every \(n\).  
As a consequence, without additional restrictions on its evolution operator, [PERSON]'s approach does not allow for extracting the result of the computation of a QTM.