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The identity operation means that we are not disturbing the system at all. However, if we consider the following operation

$$ \begin{pmatrix} 1 + \epsilon & 0\\ 0 & 1 + \epsilon \end{pmatrix} $$ where $0\le \epsilon\le1$. What would this kind of operation (measurement) correspond to?

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    $\begingroup$ If we call your operator $\hat{O}$, then $\hat{O}\psi = (1+\epsilon)\psi$ for all $\psi$. So this is a measurement that always gives a result of $1+\epsilon$. $\endgroup$ – Ricky Tensor Dec 28 '18 at 18:23
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    $\begingroup$ @RickyTensor That's an answer, not a comment $\endgroup$ – Aaron Stevens Dec 28 '18 at 18:24
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I went through a few quantum mechanics books and articles, including Dirac's. The following is my interpretation of the situation you presented. Feel free to raise objections.

It is quite unclear, even today, as to what exactly constitutes a 'measurement' in Quantum Mechanics. However, there are some properties associated with operators, which can be 'measured' through experiment, thereby obtaining some information on the collapsed state of the wave-function.

Let $|\psi\rangle$ denote any arbitrary state of a system, and $\hat{I}$ denote the identity operator. The expectation of the 'measurement' $\hat{I}$ on the system, is $$\langle \psi|\hat{I}|\psi\rangle = 1$$, assuming $|\psi\rangle$ to be normalised. This is true for all possible states of the system. In other words, $\hat{I}$ is an operator with only one eigenvalue, 1, corresponding to an infinite,continuous degeneracy, spanning the Hilbert space we are working in.

Moving to a density matrix representation of the system,(see here) let the density matrix of the initial system be $\rho$. Then, you can show that for the operator $\hat{I}$, the value $$Tr(\hat{I}\rho) = \langle \hat{I} \rangle = 1$$ since '1' is the only possible eigenvalue. Also, the degenerate subspace of the eigenvalue '1' of $\hat{I}$, is the original Hilbert space. Thus, the projection operator $P_1$(as defined in the article linked above), is just the identity operator $\hat{I}$ in the original Hilbert space. Therefore, the new density matrix of the ensemble is given by, $$\rho'' = \hat{I}\rho\hat{I} = \rho$$

In other words, the same density matrix remains.

Changing operator $\hat{I}$ to $\hat{I'}=\lambda\hat{I}$ ($\lambda \ne 0$), we will still get the eigenvalue $\lambda$ for any arbitrary system. Thus, we still have, $$Tr(\rho\hat{I'})=\lambda Tr(\rho) = \lambda = \langle \hat{I'}\rangle$$ This is perfectly consistent. However, the projection operator $P'_\lambda$(as defined in the linked wiki article) is still the identity operator $\hat{I}$ in the original Hilbert space. In other words, the new density matrix after application of $\hat{I'}$ is, $$\eta = \hat{I}\rho\hat{I} = \rho$$ In this case as well, the density matrix remains unchanged.

Interpretation: One may say that the operator $\hat{I}$ constitutes a 'measurement', which gives the value '1' irrespective of the system state. This in no way, improves our knowledge of the system, and one might say the information is conserved, i.e: the change in Von-Neumann Entropy is zero for the measurement $\hat{I}$(pretty useless measurement!).

As we derived above, $\hat{I'}$ is also a 'measurement' which 'does' nothing to the system, and gives us the value $\lambda$ irrespective of the system state. Again, no information is 'leaked' about the system to us(the change in Von-Neumann Entropy is zero), through this measurement(i.e: it is equally lame!).

One might also say, that such an operation is 'not' a 'measurement' at all, as it gives us no information about the system.

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    $\begingroup$ Kindly mention your objections instead of just down-voting the answer. The latter won't help anyone. $\endgroup$ – Lelouch Dec 28 '18 at 18:45
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    $\begingroup$ Thanks, @Lelouch. That looks nice. Just came to know from a friend that such measurement is called Biased measurement in the literature. $\endgroup$ – Zilch Dec 29 '18 at 5:13

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