0
$\begingroup$

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 } $?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.