Constant phase factors in wavefunctions

[Ibix]

TL;DR Summary

I can't get rid of constant phase factors when superposing wave functions. What am I not getting?

I'm trying to repair the dismal state of my knowledge of QM, so I downloaded Tong's notes and have read through them a couple of times and I have a question.

Tong says (section 1.1.1, p7 in the pdf) that an overall constant phase factor $e^{i\alpha}$ infront of a wavefunction describes an equivalent state - $\psi(x,t)\equiv e^{i\alpha}\psi(x,t)$. In particular, he notes that the observable probability density function is $P(x,t)=\left(e^{i\alpha}\psi(x,t)\right)\left(e^{i\alpha}\psi(x,t)\right)^*$, and the complex conjugation causes the exponentials to cancel. OK, fine.

My question is about superposition. If I superpose two states $e^{i\alpha}\psi_\alpha$ and $e^{i\beta}\psi_\beta$ and compute the probability density of the superposed state I get$$\begin{eqnarray*}
P&=&\left(e^{i\alpha}\psi_\alpha+e^{i\beta}\psi_\beta\right)\left(e^{i\alpha}\psi_\alpha+e^{i\beta}\psi_\beta\right)^*\\
&=&|\psi_\alpha|^2+e^{i(\alpha-\beta)}\psi_\alpha\psi^*_\beta+e^{i(\beta-\alpha)}\psi_\beta\psi^*_\alpha+|\psi_\beta|^2\\
&=&|\psi_\alpha|^2+2\mathrm{Re}\left(e^{i(\alpha-\beta)}\psi_\alpha\psi^*_\beta\right)+|\psi_\beta|^2
\end{eqnarray*}$$which does depend on the phase factors in front of the contributing states. To me, that implies that those states aren't equivalent. Am I misunderstanding something? Reading too much into "equivalent"? Making a stupid arithmetic error?

 

[PeroK]

TL;DR Summary: I can't get rid of constant phase factors when superposing wave functions. What am I not getting?

I'm trying to repair the dismal state of my knowledge of QM, so I downloaded Tong's notes and have read through them a couple of times and I have a question.

Tong says (section 1.1.1, p7 in the pdf) that an overall constant phase factor $e^{i\alpha}$ infront of a wavefunction describes an equivalent state - $\psi(x,t)\equiv e^{i\alpha}\psi(x,t)$. In particular, he notes that the observable probability density function is $P(x,t)=\left(e^{i\alpha}\psi(x,t)\right)\left(e^{i\alpha}\psi(x,t)\right)^*$, and the complex conjugation causes the exponentials to cancel. OK, fine.

My question is about superposition. If I superpose two states $e^{i\alpha}\psi_\alpha$ and $e^{i\beta}\psi_\beta$ and compute the probability density of the superposed state I get$$\begin{eqnarray*}
P&=&\left(e^{i\alpha}\psi_\alpha+e^{i\beta}\psi_\beta\right)\left(e^{i\alpha}\psi_\alpha+e^{i\beta}\psi_\beta\right)^*\\
&=&|\psi_\alpha|^2+e^{i(\alpha-\beta)}\psi_\alpha\psi^*_\beta+e^{i(\beta-\alpha)}\psi_\beta\psi^*_\alpha+|\psi_\beta|^2\\
&=&|\psi_\alpha|^2+2\mathrm{Re}\left(e^{i(\alpha-\beta)}\psi_\alpha\psi^*_\beta\right)+|\psi_\beta|^2
\end{eqnarray*}$$which does depend on the phase factors in front of the contributing states. To me, that implies that those states aren't equivalent. Am I misunderstanding something? Reading too much into "equivalent"? Making a stupid arithmetic error?

The states are only equivalent in that the possible values of all observables are the same. They are not identical states if we combine them with other states. In particular:
$$\psi_1 + \psi_2 \ne \psi_1 + e^{i\beta}\psi_2$$PS but, $\psi_2$ and $e^{i\beta}\psi_2$ have the same possible values for all observables.

The famous example is that a $2\pi$ rotation of an electron spin state induces a minus sign. The state looks the same, except if you manage to superimpose the original spin state with the rotated one and the resultant state (at that point in space) is zero.

 

[Ibix]

Right. So I take that to mean that I'm reading "equivalent" a little more generally than Tong intends. Thanks.

 

[anuttarasammyak]

TL;DR Summary: I can't get rid of constant phase factors when superposing wave functions. What am I not getting?

Tong says (section 1.1.1, p7 in the pdf) that an overall constant phase factor eiα infront of a wavefunction describes an equivalent state - ψ(x,t)≡eiαψ(x,t). In particular, he notes that the observable probability density function is P(x,t)=(eiαψ(x,t))(eiαψ(x,t))∗, and the complex conjugation causes the exponentials to cancel. OK, fine.

Wave function is written as superposition of two or more wavefunctions, e.g.
$$\psi(x,t)=\psi_1(x,t)+ \psi_2(x,t)$$
"Tongs' rule" seems to say
$$\psi(x,t) \equiv e^{i\alpha}\psi(x,t)=e^{i\alpha}\psi_1(x,t)+ e^{i\alpha}\psi_2(x,t) \neq e^{i\alpha}\psi_1(x,t)+ e^{i\beta}\psi_2(x,t)$$

 

[DrClaude]

Right. So I take that to mean that I'm reading "equivalent" a little more generally than Tong intends. Thanks.

Not exactly. You have to distinguish between absolute phase (which is irrelevant) and relative phase (which leads to all interesting physics ).

In you example, you have
$$
\begin{align*}
\Psi &= e^{i\alpha}\psi_\alpha+e^{i\beta}\psi_\beta \\
&= e^{i\alpha} \left( \psi_\alpha + e^{i(\beta-\alpha)} \psi_\beta \right)
\end{align*}
$$
The absolute phase $\alpha$ does not change the state of the system, $\Psi$, but the relative phase $\beta-\alpha$ can change many an observation.

 

[Orodruin]

The absolute phase $\alpha$ does not change the state of the system, $\Psi$, but the relative phase $\beta-\alpha$ can change many an observation.

If it didn’t I would probably be unemployed …