We have a spin state

\begin{align} \ |{\Psi}\rangle=a_u|{U}\rangle+a_d|{D}\rangle \end{align}

where $|U\rangle$ and $|D\rangle$ are $up$ and $down$ basis vectors, and $a_u$,$a_d$ are their complex coefficients.

\begin{align} \ a_u=x+yi \end{align}

\begin{align} \ a_d=w+zi \end{align}

$x, y, w, z$ are the real parameters I'm asking about. Since $a_ua^*_u + a_da^*_d=1$ we can ignore one of these, let's say $z$ and still calculate the state of the system, so now we have three needed parameters.

Now, according to my book, "ignoring the overall phase factor" makes it possible to specify the spin state using only two parameters. I know what the phase factor is and that we can ignore it in the calculations, but I don't see how it relates to the number of parameters needed.

E D I T: How to do it with more basis vectors? For example in a situation like this (this should have 6 parameters, but I don't know what to do, since I can't use the sine-cosine trick now)

\begin{align} \ |{\Psi}\rangle=a_a|A\rangle+a_b|B\rangle+a_c|C\rangle+a_d|D\rangle \end{align}

  • 1
    $\begingroup$ Tip: *italics* (italics) look nicer than TeX-induced ones ($italics$). $\endgroup$ – Emilio Pisanty Sep 8 '15 at 16:23
  • $\begingroup$ Also, you can use \uparrow and \downarrow for the typical convention to get $\uparrow$ and $\downarrow$ if you prefer that notation. $\endgroup$ – danielunderwood Sep 8 '15 at 18:01

The fact that the overall phase factor does not matter means that we can choose it to be whatever we like. This gives us an extra constraint (even if is one we choose arbitrary) and so reduces the number of degrees of freedom by 1.

For example, given that $|a_u|^2 + |a_d|^2 = 1$ we can write our coefficients as \begin{align} a_u = &\cos\theta \;e^{\imath\phi_1}\\ a_d = &\sin\theta \;e^{\imath\phi_2} \end{align}

however we can rewrite $\phi_1$ and $\phi_2$ as \begin{align} \phi_1 = \Phi - \phi\\ \phi_2 = \Phi + \phi \end{align} but now $\Phi$ is simply setting the overall phase factor, so we can ignore it and we are left with 2 degrees of freedom $$ |\Psi\rangle = \cos\theta e^{-\imath\phi}|U\rangle + \sin\theta e^{\imath\phi} |D\rangle $$

  • $\begingroup$ Good answer. You can even start from the very beginning and say we have 2 complex coefficients, therefore 4 real parameters. The condition of unitarity that you gave is one constraint, reducing the number of free parameters to 3. Then you can still absorb an overall (physically insignificant) phase, taking that number down to 2. $\endgroup$ – gn0m0n Sep 8 '15 at 16:50
  • $\begingroup$ What is the boldface phi? And why can we rewrite ${\phi}_1$ and ${\phi}_2$ in this way? $\endgroup$ – Qwedfsf Sep 8 '15 at 17:24
  • $\begingroup$ $\Phi$ and $\phi$ are defined by the pair of equations relating to $\phi_1$ and $\phi_2$. you can rearrange to get $\Phi = (\phi_1 + \phi_2)/2$ and $\phi = (\phi_2 - \phi_1)/2$. We can always define coordinate transformations like this if it is convenient. $\endgroup$ – By Symmetry Sep 8 '15 at 17:31
  • $\begingroup$ Also what do you exactly mean by "${\Phi}$ is simple setting the overall phase factor, so we can ignore it"? I don't really understand this. $\endgroup$ – Qwedfsf Sep 8 '15 at 17:43
  • $\begingroup$ If you substitute for $\Phi$ and $\phi$ in the expressions for $a_u$ and $a_d$ you will find they are both proportional to $\exp(\imath\Phi)$ This means that in the expression for $|\Psi\rangle$ you can factor $\exp(\imath\Phi)$ out entirely, similar to what @EmilioPisanty has done. But that is simply an overall phase factor which does not effect the physics, so we can ignore it. (or more formally we can impose a constraint that $\Phi = 0$) $\endgroup$ – By Symmetry Sep 8 '15 at 17:56

An arbitrary normalized quantum state on two dimensions can always be written as $$ |\psi⟩=e^{i\alpha}\left(\cos\theta|\uparrow⟩+e^{i\phi}\sin\theta|\downarrow⟩\right) $$ without loss of generality.

The phase factor $e^{i\alpha}$ has no bearing on experiment, as all measurements will be proportional to $⟨\psi|\propto e^{-i\alpha}$. This means that you can set it to any arbitrary value you want without affecting your physical predictions. Thus you only have two physically relevant parameters.


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.