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Nielsen and Chuang, Chapter 2 (Box 2.6):

Suppose $M$ is any observable on a system $A$, and we have some measuring device which is capable of realizing measurements of $M$. Let $\tilde M$ denote the corresponding observable for the same measurement, performed on the composite system $AB$. Our immediate goal is to argue that $\tilde M$ is necessarily equal to $M \otimes I_B$. Note that if the system $AB$ is prepared in the state $|m\rangle |\psi\rangle$, where $|m\rangle$ is an eigenstate with an eigenvalue $m$ and $|\psi\rangle$ is any state of $B$, then the measuring device must yield the result $m$ for the measurement, with probability one. Thus, if $P_m$ is the projector onto the $m$ eigenspace of the observable $M$, then the corresponding projector for $\tilde M$ is $P_m \otimes I_B$. We therefore have

$\tilde M = \sum_{m} m P_m\otimes I_B = M \otimes I_B$.

Our professor mentioned that for a system of two qubits $A$ and $B$, if we consider the wavefunction to be $|\phi\rangle = a |00\rangle + b |11\rangle$, then the expectation value of the observable $\tilde M$ will be:

$$\langle \phi| M \otimes I_B |\phi \rangle$$

$$= (a^*\langle00|+b^*\langle 11|) M \otimes I_B (a |00\rangle + b |11\rangle)$$

The expression for the observable $\tilde M = M\otimes I_B$ is only valid if the system $AB$ was initially prepared in the state $|m\rangle |\psi\rangle$ where $|m\rangle$ is an eigenstate with eigenvalue $m$. That means it is necessary that both the eigenstates $|0\rangle$ and $|1\rangle$ of $A$ had the same eigenvalue $m$. Otherwise the professors' expression for the expectation value is simply not true! I don't understand why the professor mention it explicitly. Am I missing something?

Or, are eigenvalues for the states $|0\rangle$ and $|1\rangle$ (for measurement operator $M$) always same, by default?

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  • $\begingroup$ I don't understand what you are asking. Different measurement operator will have different eigenvalues and eigenvectors, and generally speaking the eigenvalues corresponding to $|0\rangle$ and $|1\rangle$ (provided that these states are eigenvectors) will be different. Also you wrote the equation for the expectation value of $\tilde M$ over $|\phi\rangle$, but then you mention $|m\rangle|\psi\rangle$. What is the relation between these states? $\endgroup$ – glS Jan 26 '18 at 17:19
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The text says, if you start in an arbitrary eigenvector $|m \rangle$, then you get the eigenvalue $m$. However, $m$ has no meaning, it's just a dummy variable. In your specific case, this means that if you start in $|0 \rangle$ you get $0$ and if you start in $|1 \rangle$ you get $1$. The equivalence $\tilde{M} = M \otimes I_B$ is true for all vectors since it's true for all eigenvectors .

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  • $\begingroup$ Okay, but I don't think you addressed my question! I was asking about the eigenvalues for the two states...are they equal? If not, is the professor's expression for expectation value legitimate? $\endgroup$ – user182786 Jan 26 '18 at 16:40
  • $\begingroup$ @user182786 Whoops, I read too fast. New answer! $\endgroup$ – knzhou Jan 26 '18 at 16:48
  • $\begingroup$ So, I understand that the eigenvalue is not same for $|0\rangle$ and $|1\rangle$. They are $0$ and $1$ respectively. OK. Then, I don't think the professor's expression for the expectation value of $\tilde M$ is correct if $\phi = a|00\rangle + b|11\rangle$ as the eigenvalue for $|0\rangle_A$ and $|1\rangle_A$ are not same. It would be correct if it were something like $\phi = a|00\rangle + b |01\rangle$ or $\phi = a|10\rangle + b|11\rangle$ where the eigenvalues for the first member's (state of $A$) are same. Am I right? $\endgroup$ – user182786 Jan 26 '18 at 16:56
  • $\begingroup$ I mean for this $\tilde M = \sum_{m} m P_M\otimes I_B = M \otimes I_B$ to be valid, it is necessary that the system of two qubits was prepared such that $A's$ state was an eigenstate in itself with some eigenvalue $m$. $a|0\rangle + b|1\rangle$ is clearly not an eigenstate with a certain eigenvalue $m$. So the expression cannot be valid for the original $\phi = a|00\rangle + b|11\rangle$ $\endgroup$ – user182786 Jan 26 '18 at 16:59
  • $\begingroup$ @user182786 The derivation is more general than that. It says, $\tilde{M}$ and $M \otimes I_B$ are equivalent for any eigenvector with an arbitrary eigenvalue, which we might call $m$. So they're equivalent, period. $\endgroup$ – knzhou Jan 26 '18 at 17:02

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