# How does the outcome of measurement of a qubit change when we use different basis despite the system hasn't changed? [closed]

Let's assume that the quantum state of the system is written in a standard basis {$$|0\rangle, |1\rangle$$} and when we performed a measurement we got $$|0\rangle$$ as an outcome of measurement so we ensure that the system will be in state $$|0\rangle$$ after measurement even if we make this measurement (in this basis) many times.

My question is, how we get something else than $$|0\rangle$$ {$$|+\rangle$$ or $$|-\rangle$$} when we are measuring in different basis {$$|+\rangle, |-\rangle$$} despite we didn't change the system?

• Please do a text copy-paste and ping me in a reply comment. I will fix it and give a reopen vote. – peterh Dec 31 '20 at 20:20
• @peterh-ReinstateMonica I can't clearly understand what do you mean by text copy-paste. However, I will copy my question: Based on the example above, let's assume that the quantum state of the system is written in a standard basis {|0⟩,|1⟩} and when we performed a measurement we got |0⟩ as an outcome of measurement so we ensure that the system will be in state |0⟩ after measurement even if we make this measurement (in this basis) many times. My question is, how we get something else than |0⟩ {|+⟩ or |−⟩} when we are measuring in different basis {|+⟩,|−⟩} despite we didn't change the system? – Islam Abdeen Jan 1 at 0:47
• Screenshots are highly unwelcomed network-wide. This is why you got downs and this is why your question was closed. You can read the details here (that is an MSO content, but the reasons of the QC SE are the same). I offered to make your formulas proper here, but I did not offer to manually type your image as text. – peterh Jan 1 at 1:12
• Btw, the reason that your question was not deleted on the spot is that it is anyways good (and also its answer). My reason to fix it, is the same. – peterh Jan 1 at 16:45
• @peterh-ReinstateMonica I checked the link and now I can understand why images are not the best way to explain my question. Thank you for the link. I deleted it, I think my question is clear also without it. – Islam Abdeen Jan 1 at 23:47

## 1 Answer

It is a postulate of quantum mechanics that any device that measures a two-state quantum system (a qubit) must have two preferred states $$\{|e_1\rangle, |e_2\rangle \}$$ that form an orthonormal basis for the associated vector space (here would be $$\mathbb{C}^2$$).

A measurement on the state $$|\psi\rangle$$ transforms $$|\psi\rangle$$ into one of the these basis vectors $$|e_1 \rangle$$ or $$|e_2\rangle$$. The probability that the state $$|\psi\rangle$$ is measured as $$|e_1\rangle$$ or $$|e_2\rangle$$ is the square of the magnitude of the amplitude of the component of the state in the direction of the basis vector $$|e_1\rangle$$ or $$|e_2\rangle$$ . Hence the name projection.

If you are in the state $$|+\rangle$$ and your measurement is in the computational basis, $$\{ |0\rangle, |1\rangle \}$$, then the probability of observing the state $$|0\rangle$$ is the magnitude square of the component of the projected vector. When you project the vector $$|+\rangle$$ onto the vector $$|0\rangle$$ you get the vector $$\dfrac{1}{\sqrt{2}}|0\rangle$$. Thus the probability of observing the state $$|0\rangle$$ is $$|1/\sqrt{2}|^2 = \dfrac{1}{2}$$. Similarly, $$|+\rangle$$ can be projected to $$|1\rangle$$ and the resultant vector is $$\dfrac{1}{\sqrt{2}}|1\rangle$$, hence the probability of observing $$|1\rangle$$ is also $$\dfrac{1}{2}$$.

However, if you measure in the basis $$\{|+\rangle, |-\rangle \}$$ then the projection of $$|+\rangle$$ onto itself is itself. That is $$|+\rangle$$ will be just transforms to $$|+\rangle$$. Thus the probability of observing $$|+\rangle$$ as $$|+\rangle$$ in the basis $$\{|+\rangle, |-\rangle \}$$ is 1.

From here we can see that the notion of superposition is basis-dependent. All states are superposition with respect to some bases but not with respect to others. That is, a state $$|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$$ is only a superposition with respect to the computational basis $$\{|0\rangle, |1\rangle \}$$ but not a superposition with respect to the bases $$\{ \alpha |0\rangle + \beta |1\rangle, \beta^* |0\rangle - \alpha^* |1\rangle \}$$.

Since measuring a superposition state $$|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$$ is probabilisitc, it is tempted to say that the state $$|\psi \rangle$$ is a probabilistic mixture of $$|0\rangle$$ and $$|1\rangle$$ and we just don't know which, when in fact, $$|\psi \rangle$$ is actually a definite state. That is, if we measure $$|\psi\rangle$$ in certain bases, we will get a deterministic result.

• Thanks! If quantum system is in state |+⟩ and our device measuring only basis {|0⟩,|1⟩} so it makes sense to me to not observe |0⟩ nor |1⟩, we can observe only |+⟩ with probability 1 in case of our device measuring in basis {|+⟩,|-⟩} but if our device measuring in basis {|0⟩,|1⟩} we shouldn't observe anything. So, the question here is why a measurement on the state |ψ⟩ transforms |ψ⟩ into one of the these basis vectors |e1⟩ or |e2⟩ (preferred states) and not to the basis vectors of the state itself? – Islam Abdeen Dec 24 '20 at 14:32
• @IslamAbdeen If the state is $|+\rangle$ and we measure in the computational basis $\{ |0\rangle, |1\rangle \}$ then we will see both $|0\rangle$ and $|1\rangle$ since the state $|+\rangle$ compose/made-up from some component of the state $|0\rangle$ and some component of the state $|1\rangle$. See the pic I have above. Note that this behavior of measurement is a postulate/axiom of quantum mechanics. It is not derivable or proof from other physical principles. It is derived from observation of experiments with measuring devices. – KAJ226 Dec 24 '20 at 16:32