If a state $\psi $ is in the $ S_{z} $ basis represented by $\mid\psi\rangle = c_{+}\mid z\rangle + c_{-} \mid -z\rangle$ Does the associated probabilities change when I multiply $ \psi $ by $ e^{i\phi } $?


The spin state is given as a linear combination of spin up and spin down states. So

$$\psi=c_+| z +\rangle+c_-| z -\rangle$$

The square of the modulus of the complex coefficients $c_+$ and $c_-$ represent the probabilities of each associated base kets. When you multiply $\psi$ by a constant, the state will not change as a ket and a constant times that ket represents the same state. But the complex coefficients change.

$$e^{i\phi}\psi=e^{i\phi}c_+| z +\rangle+e^{i\phi}c_-| z -\rangle=c'_+| z +\rangle+c'_-| z -\rangle$$.

where $c'_+$ and$c'_-$ represents the new complex coeffecients.

Now, let's look at the probabilities.

Probability for the state $|z +\rangle$ is

$$P_+=\mid c'_+\mid^2=\mid e^{i\phi}c_+\mid^2=\mid c_+\mid^2\Rightarrow\text{probability of $| z +\rangle$ state remains unchanged on multiplication of $e^{i\phi}$}$$

and that for $| z -\rangle$ is

$$P_-=\mid c'_-\mid^2=\mid e^{i\phi}c_-\mid^2=\mid c_-\mid^2\Rightarrow\text{probability of $| z -\rangle$ state remains unchanged on multiplication of $e^{i\phi}$}$$

since, $\mid e^{i\phi}\mid^2=\mid \cos\phi+i\sin\phi\mid^2= \left(\sqrt{\cos^2\phi+\sin^2\phi} \right)^2=1.$

Hence the presence of the term $e^{i\phi}$ will not affect the probabilities of the states.

  • $\begingroup$ is it not e^{i \phi}*e^{-i \phi} since it is multiplied by a complex conjugate? $\endgroup$ – Paul777 May 11 '16 at 4:35
  • $\begingroup$ Also, is it not possible to distribute the square to cos \phi and i*sin \phi? $\endgroup$ – Paul777 May 11 '16 at 4:37
  • $\begingroup$ I don't understand what you meant $\endgroup$ – UKH May 11 '16 at 5:58
  • $\begingroup$ @Unnikrishnan Do you mean $| \cdot \rangle$ instead of $\langle \cdot \rangle$ ? $\endgroup$ – Asaf May 11 '16 at 17:30
  • $\begingroup$ @Paul777 You can do $P = |c'|^2$ or $P = c'c'^* = e^{i \phi} c c^* e^{-i \phi}$. They are equivalent. $\endgroup$ – Asaf May 11 '16 at 17:33

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