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Apologies if my question is worded poorly or unclear. I am still new to quantum mechanics and am having trouble understanding this concept.

In my textbook, it says:

Instead of measuring |ψ⟩ in a rotated basis ( |v⟩, |⊥v⟩ ), we achieve the same effect by rotating the entire space, so that |v⟩ is mapped to |0⟩ and |⊥v⟩ is mapped to |1⟩, and then measuring in the computational basis ( |0⟩ , |1⟩ ).

Such rigid body transformations of the vector space are called unitary transformations. For example, rotations and reflections are unitary. A postulate of quantum physics is that quantum evolution is unitary.

Does this mean that a measurement in any orthonormal basis can be transformed to the computational basis ( |0⟩ , |1⟩ ) while keeping the measurement consistent?

If I had a measurement for |ψ⟩ in the basis { |u1⟩, |u2⟩, |u3⟩, ... , |ud⟩ } , how can I prove that the measurement can be "mimicked" in some sense by a unitary transformation followed by a measurement in the computational basis?

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Since you have an orthonormal basis $\{|u_i\rangle\}$, you can define a unitary $$ U=\sum_i|i\rangle\langle u_i|, $$ which transforms it into the computational basis $\{|i\rangle=U|\psi_i\rangle\}$. It's straightforward to check that this is unitary by evaluating $UU^\dagger=I$.

So, imagine that you have a state $|\psi\rangle$ and you're going to measure it in the basis $\{|u_i\rangle\}$. You will get answer $i$ with probability $$ p_i=|\langle u_i|\psi\rangle|^2=|\langle i|U| \psi\rangle|^2 $$ This is exactly the same as applying $U$ to $|\psi\rangle$ and measuring the state $U|\psi\rangle$ in the standard basis.

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  • $\begingroup$ I see how $|\langle u_{i}|\psi\rangle|^{2}=|\langle i|U \psi\rangle|^{2}$, as $|\langle i|U \psi\rangle|=|\langle U^{\dagger} i| \psi\rangle|=|\langle u_{i}| \psi\rangle|$. But how does $|\langle u_{i}|\psi\rangle|^{2}=|\langle i|U^{\dagger} \psi\rangle|^{2}$? $\endgroup$ Commented Feb 20, 2022 at 12:00
  • $\begingroup$ I probably got some hermitian conjugates jumbled... serves me right for not writing it down first! $\endgroup$ Commented Feb 21, 2022 at 9:23
  • $\begingroup$ XD I spent like an hour trying to reason out how $|\langle i |U^{\dagger} \psi \rangle|=|\langle u_{i} |\psi \rangle|$. Thought I had completely misunderstood something fundamental. $\endgroup$ Commented Feb 21, 2022 at 13:58
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Given $|\psi\rangle = \sum_{i}a_{i}|u_{i}\rangle$, the outcome $p_{i}=\langle \psi|u_{i}\rangle \langle u_{i}|\psi\rangle=|\langle u_{i}|\psi\rangle|^{2}$

If ${|i\rangle}$ is the computational basis, then a unitary $U=\sum_{i}|i\rangle \langle u_{i}|$ can be created that maps the basis you are currently measuring in to the computational one.

Then, you need simply map your state vector to the new basis, the computational in this case, and then perform the measurement that is the image of the measurement you would have desired before:

$$U|\psi\rangle = \sum_{i}a_{i}U|u_{i}\rangle=\sum_{i}a_{i}|i\rangle=|\phi\rangle$$

Now $p_{i}=\langle \phi|i\rangle \langle i|\phi\rangle=|\langle i|\phi\rangle|^{2}$, and it remains unchanges as in both cases $p_{i}=|a_{i}|^{2}$, as the amplitudes are left unaffected by the unitary transformation.

Also note $p_{i}=\langle \phi|i\rangle \langle i|\phi\rangle = \langle \psi|U^{\dagger}|i\rangle \langle i|U|\psi\rangle=\langle \psi|u_{i}\rangle \langle u_{i}|\psi\rangle$

It is worth noting however that while you do receieve the same probability, the output state, and $U|\psi\rangle$ are not the same states that you started with. So in a sense, no, the measurement is not preserved.

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