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For a quantum state on the form $$|\psi \rangle = \alpha |0 \rangle + \beta |1 \rangle$$ which possible qubit states can you construct from this? I know that $\alpha$ and $\beta$ must satisfy $$|\alpha|^2 + |\beta|^2 = 1$$ but are there any other requirements besides this?

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  • $\begingroup$ You need to clarify. What operations are you allowed to do? Are $\alpha$ and $\beta$ fixed at the beginning? $\endgroup$ – AHusain Jun 3 at 3:03
  • $\begingroup$ I don't thiik the possible operations matter. The question is just about what the allowed Qubit states can be $\endgroup$ – Annonymus Jun 3 at 6:04
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Any unitary transformation preserves angles and length of vectors it is applied upon.

Therefore, an unitary transformation applied on your states $|\psi\rangle$ will lead to state $|\psi_1\rangle = \alpha_1|0\rangle + \beta_1|1\rangle$ satisfying $|\alpha_1|^2+|\beta_1|^2=1$.

Since any gate in a quantum computer must be unitary (if we do not count measurement which is special case), you can do any operation you want and requirements on qubit will be preserved.

Taking appropriate unitary trasformation $U$, you can convert any state $|\psi\rangle$ to any other state $|\psi_1\rangle$.

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  • $\begingroup$ I guess then in theory, it should be possible to construct a qubit in the state $|\psi \rangle=\frac{\sqrt{2}}{2}|0\rangle - \frac{\sqrt{2}}{2}|1\rangle$. Am I right? $\endgroup$ – Annonymus Jun 3 at 8:35
  • $\begingroup$ Yes, to do it, you can start with $|0\rangle$, apply the Pauli gate $X$ to obtain $|1\rangle$, then apply the Hadamard gate $H$. $\endgroup$ – Michele Amoretti Jun 3 at 8:41
  • $\begingroup$ I am sorry. I meant to write $\alpha = \frac{\sqrt{3}}{2}$ and $\beta = \frac{1}{2}$. I cannot edit the previous question $\endgroup$ – Annonymus Jun 3 at 8:49
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There are no additional constraints on $\alpha$ or $\beta$. This means that we cannot directly answer the question of "which possible qubit states can you construct" because that suggests an enumeration, but this corresponds to a continuum of states!

Note that a particularly convenient way of writing these is $$ e^{i\gamma}(\cos\theta|0\rangle+\sin\theta e^{i\phi}|1\rangle), $$ with $\theta\in[0,\pi)$ and $\phi\in[0,2\pi)$, as this automatically takes care of the normalisation in the parametrisation. Also, the $e^{i\gamma}$ term is strictly unnecessary because it has no observable consequences, but I left it in there so that it gives a mathematical identity with the $\alpha,\beta$ formulation.

If you are asking how to go from the state $|0\rangle$ to any state $|\psi\rangle$ that you want, then start from the parameterisation above, and consider $$ R_y(2\theta)|0\rangle=(I\cos\theta-i\sin\theta Y)|0\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle, $$ so that you can then just follow it up with $$ R_z(\phi)R_y(2\theta)|0\rangle=(I\cos\frac{\phi}{2}-i\sin\frac{\phi}{2} Z)(\cos\theta|0\rangle+\sin\theta|1\rangle)=e^{-i\phi/2}|\psi\rangle $$ So you can prep any state you want with two single-qubit rotations (up to an irrelevant global phase).

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