There are other similar questions. But I don't understand the answers.

Suppose I express $a|0⟩+b|1⟩$ in the form $\frac{c}{\sqrt2}(|0⟩+|1⟩)+\frac{d}{\sqrt 2}(|0⟩−|1⟩)$ where $a,b,c,d∈\mathbb C$. Then in my new basis the vector $|0⟩$ is represented like $(1,1)$ . So what does that mean?

The vector $(1,0)$ in computational basis means, I 100% get $|0⟩$ in measurement. But what does it mean on other measurements? Does it mean I get $\frac{1}{\sqrt 2}(|0⟩+|1⟩), \frac{1}{\sqrt 2}(|0⟩−|1⟩)$ with equal chances? If so, then how is it possible that in one measurement you get $|0⟩$ for sure, but in new basis you could get, say $\frac{1}{\sqrt 2}(|0⟩+|1⟩)$ with 50% chance which doesn't mean you are guaranteed to get $|0\rangle$ on measurement but only half of the time.

Basically, I am asking, if in one basis I have a vector, say, $(1,0,0,0)$ and in a different basis the same vector appears to be $(1,0,1,0)$ does that mean I can collapse to $(0,0,1,0)$? Which in turn and in the end will collapse me to something different than the original $(1,0,0,0)$?


1 Answer 1


The state $\mid \psi \rangle$ is fixed. You can write it as $a |0 \rangle + b |1 \rangle$. If you write that in the other basis and get

$ (a,b) = (1,0) \implies (c,d) = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) $

They are the same state so it is still true that if you do a measurement in the $|0 \rangle$, $|1 \rangle$ basis, you will surely get $\mid 0 \rangle$ if you start with $(a,b)=(1,0)$.

For shorthand let us call the states $\frac{1}{\sqrt{2}} (|0 \rangle \pm |1 \rangle$ as $\mid \pm \rangle$ respectively. That way don't have to keep rewriting.

So we are in state $\frac{\sqrt{2}}{2} \mid + \rangle + \frac{\sqrt{2}}{2} \mid - \rangle$. If we measure in the $\mid + \rangle$ and $\mid - \rangle$ basis, we have a a $(\frac{\sqrt{2}}{2})^2=\frac{1}{2}$ probability for measuring each of $\mid + \rangle$ and $\mid - \rangle$.

Your confusion is about measuring in $\mid 0 \rangle$, $\mid 1 \rangle$ basis vs measuring in the $\mid \pm \rangle$. You don't do them both at the same time. Either you pick the first, and surely get $\mid 0 \rangle$ or you do the second and get $\mid + \rangle$ or $\mid - \rangle$ with equal probabilities. They are different operations.

Your original $(1,0,0,0)$ is in basis 1, then your $(1,0,1,0)$ is in basis 2. The measurement in basis 2 collapses this to $(0,0,1,0)$. Conversely if you measure in basis 1 you get $(1,0,0,0)$ in basis 1. These are different states, but that is expected because you did different measurement operations on the starting state.

You might still be thinking too classically as being able to do lots of different passive measurements in whatever order.

Edit (clarifying the confusion in the comments):

The measurement is not simply reading the state as a vector. Rewriting the vector in different basis doesn't do anything. As in linear algebra, it doesn't matter at all.

The measurement in a basis is an operation you do. It is computed by doing the rewrite first, but that is the perspective of you as an omnipetent being when doing the math. You rewrite in the desired basis first because that way the projection operator in that basis is easy to write down.

The measurement changes the vector and different measurements change the vector in different ways. Measurement 0 in basis 1 will apply some projection operator $P_1$, measurement + in basis 2 will do some other projection operator $P_2$. They are different operators, so of course, you can't expect to get the same thing even if the input state was the same. If you write $P_1$ in basis 1 you get a matrix $M_1$, and if you write $P_2$ in basis 2 as $M_2$ you get those same matrix entries, but that doesn't mean that the operators were the same. If you write $P_2$ in basis 1 you will get something totally different from $M_1$. Doing the rewrite of the state into basis 2 was so you wouldn't have to write down that matrix and could just work with matrix $M_2$.

  • $\begingroup$ So basically doing measurements in different basis yeald different results? This is counterintuitive to me because in linear algebra you learn that representing a vector in different basis doesn't matter at all and if you perform in basis1 operation1 on Vector1 you will get the same as performing operation2 in basis2 on vector2. All of them are instrinctly the same but translated in syntax from one basis to another $\endgroup$
    – bilanush
    Mar 2, 2019 at 18:23
  • $\begingroup$ Also, suppose I measure one observable in computational basis and see it's |0> then how is it possible that in different basis you can get in a copy -,+ with 50% which means you then can collapse to |1> even though you got |0> in one observable? $\endgroup$
    – bilanush
    Mar 2, 2019 at 18:35
  • $\begingroup$ Tried explaining more in the edit $\endgroup$
    – AHusain
    Mar 2, 2019 at 22:08
  • $\begingroup$ Accepted . I believe I what causes me the confusion is the bell basis measurement. So I asked another question can you try answering it ? quantumcomputing.stackexchange.com/questions/5605/… $\endgroup$
    – bilanush
    Mar 3, 2019 at 10:53

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