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I have studied an introductory course in quantum mechanics, and yet I still do not understand the significance of a phase difference between quantum states that a system is in a superposition of. In my lecture notes, it is stated that

Form a linear combination of two quantum states:

$|\psi \rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle $

$|\psi \rangle = e^{i\theta_1}|\phi_1\rangle + e^{i\theta_2}|\phi_2\rangle $

$|\psi \rangle = e^{i\theta_1}(|\phi_1\rangle + e^{i(\theta_2-\theta_!)}|\phi_2\rangle) $

The resultant vector, and therefore the outcome of any experiemt, depends on the relative quantum phase difference between the two states

This has me confused on many levels. Aside from the fact that the state is simple not normalised here (we would require a factor of $\frac{1}{\sqrt2}$ at the front), the magnitudes of each of the coefficients are equal, unless we allow $\theta$ to be complex which I do not think is intended here.

Also, I initially thought that the outcome of some measurement is only dependent on the magnitudes of the coefficients, and not their phase. This is immediately obvious if the eigenstates of the observable that you are measuring are the $|\phi_i\rangle$, although I considered whether it might be dependent on the relative phase of the coefficients if $|\phi_i\rangle$ are not the eigenstates. I think I then get a phase-dependent factor in the expectation value, although I would like verification on this. An additional problem that then brings up is that the result is sensitive to multiplying a basis vector by an arbitrary phase factor $e^{i\theta}, which it shouldn't be, although I think this might be resolved in considering the coefficients of some state in terms of the eigenstates of the observable.

Emilio Pisanty's answer to this question gos some way towards answering my questions, however it only mentions the significance of the phases when two states are superposed. However experiemntally, I don't understand what thi means. Does it mean having two non-entangled photons, say, and performing an experiment on them? I thought that this becomes a new system and the basis vectors form this are found from the cartesian product of basis vectors of the two states, and not their sum. I feel like this is such a basic concept, however I do not understand it! I get how a system can be in a superposition of states of some chosen basis and you can choose a different basis so that the state is in one of the basis states. However what does it mean to have two existing states, and then superpose them? Surely this means you have two systems and are interacting them in some way, but I don't see how...

Summary of question:

In what way and in what contexts does the phase between coefficients in a quantum superposition matter, and does the term 'relative quantum phase different' refer only to phase of coefficients, or also to their magnitudes which would seem to make more sense.

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    $\begingroup$ I initially thought that the outcome of some measurement is only dependent on the magnitudes of the coefficients. This is only true if the basis states used expanding your state are eigenstates of what you measure. If you have the combination of angular momentum 1 states $a\psi_{11}+b\psi_{10}+c\psi_{1,-1}$ with $\psi_{1m}$ eigenstate of $L_z$ with eigenvalue $m$, then the outcomes and probabilities for measuring $L_x$ will depend on the relative phases of the $a$, $b$ and $c$ coefficients. $\endgroup$ – ZeroTheHero Apr 21 '18 at 12:55
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There are a few distinct questions here, so I'll answer quote by quote. $\newcommand{\ket}[1]{|#1 \rangle}$

[...] the state is simple not normalised here (we would require a factor of $\frac{1}{\sqrt2}$ at the front)

People often omit explicit normalization to save space; you should just imagine the $1/\sqrt{2}$ is really there in the background. After you write $1/\sqrt{2}$ a few hundred times you'll start doing the same!

I initially thought that the outcome of some measurement is only dependent on the magnitudes of the coefficients, and not their phase. This is immediately obvious if the eigenstates of the observable that you are measuring are the $|\phi_i\rangle$, although I considered whether it might be dependent on the relative phase of the coefficients if $|\phi_i\rangle$ are not the eigenstates. I think I then get a phase-dependent factor in the expectation value, although I would like verification on this.

Absolutely. For example, suppose that the states $\ket{\phi_1}$ and $\ket{\phi_2}$ represent 'spin up' and 'spin down' respectively. That is, we have chosen the states to be eigenstates of $L_z$. Then the 'spin left' and 'spin right' states are $$\ket{\phi_1} - \ket{\phi_2}, \quad \ket{\phi_1} + \ket{\phi_2}$$ and only differ by a phase factor. The expectation value of $L_x$ depends on the phase.

An additional problem that then brings up is that the result is sensitive to multiplying a basis vector by an arbitrary phase factor $e^{i\theta}$, which it shouldn't be...

Physical results never depend on these choices, but our intermediate expressions could be. For example, suppose we defined a new basis $$\ket{\psi_1} = \ket{\phi_1}, \quad \ket{\psi_2} = - \ket{\phi_2}.$$ Then the spin left and spin right states are now $$\ket{\psi_1} + \ket{\psi_2}, \quad \ket{\psi_1} - \ket{\psi_2}.$$ That is, the expressions for these states have changed because we changed the phase convention, but the states themselves haven't. The phase difference between two coefficients is of course dependent on the convention, as is its physical meaning, but once the convention is fixed it is meaningful. For example, the phases for the spin states I've been talking about are usually fixed by demanding the $L_i$ are proportional to the Pauli matrices $\sigma_i$.

However what does it mean to have two existing states, and then superpose them? Surely this means you have two systems and are interacting them in some way, but I don't see how...

Usually 'superposition' shouldn't be used in verb form in quantum mechanics. This usage is referring to the classical intuition where waves superpose: if two sound waves are sent at each other then they superpose when they meet. But you're right that this is usually not how we prepare quantum superpositions: you don't get a $1s$ and $2s$ superposition in the hydrogen atom by slamming two hydrogen atoms in these states into each other. I recommend mentally substituting "prepare a superposition" whenever you hear "superpose".

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  • $\begingroup$ Thank you for your reply. I was wondering, when we 'prepare a superposition', how do we know that we are preparing the state as a linear combination of two other states with some definite coefficients (which may be time-depedent) versus just adding two pure ensembles with different weightings to create a new density matrix? $\endgroup$ – 21joanna12 Apr 26 '18 at 13:32
  • $\begingroup$ @21joanna12 Ah, you're asking if the superposition will be coherent or not. That depends on all the details of how you do it, but the very general principle is, if it's possible to tell "which state" it is by measuring anything besides the system itself, then you don't have a coherent superposition. $\endgroup$ – knzhou Apr 26 '18 at 13:34
  • $\begingroup$ @21joanna12 I asked a similar question about this here. For example, if you pass something through a half-silvered mirror (50% chance of going through) you do get a coherent superposition because you can't tell from the mirror's state which way the particle went. But if the particle has a 50% chance of hitting an air molecule, the superposition is not coherent because you can measure how the air molecule recoils. $\endgroup$ – knzhou Apr 26 '18 at 13:35
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The relative phase matters in all kinds of circumstances. Imagine a spin-1/2 system with $\vert +\rangle$ and $\vert -\rangle$ the spin-up and spin-down states in the $\hat z$ direction. Consider the state $$ \vert \psi\rangle =\frac{1}{\sqrt{2}}\vert +\rangle + \frac{i}{\sqrt{2}}\vert -\rangle. \tag{1} $$
The average value $\sigma_z$ is $$ \langle \sigma_z\rangle =\frac{1}{2}(+1)+\frac{1}{2}(-1)=0 $$ and in this case only depends on the magnitude of the coefficients. However, if you ask what is the probability of obtaining $s_y=+\hbar/2$, the answer to this is $1$, whereas the probability of obtaining $s_y=-\hbar/2$ is $0$. This is because (1), with the correct phase, is an eigenstate of $\sigma_y$ with eigenvalue $+\hbar/2$. Surely you can see here that the relative phase matters.

As another example, consider the superposition \begin{align} \Psi(x,t)&= \alpha \Psi_0(x,t)+\beta\Psi_1(x,t)\\ &=\alpha e^{i\omega t/2}\psi_0(x)+\beta e^{3i\omega t/2}\psi_1(x) \end{align} of harmonic oscillator states. The probability density for such a state is $$ \Psi(x,t)^*\Psi(x,t)= \vert\alpha\vert^2 \psi_0(x)^2+ \vert\beta\vert^2 \psi_1(x)^2 + (e^{i\omega t}\alpha^*\beta + e^{-i\omega t}\alpha\,\beta^*)\psi_0(x)\psi_1(x) $$ where I’ve taken $\psi_0(x)$ and $\psi_1(x)$ to be real, as is usual convention. Clearly this probability density depends the relative phase of $\alpha$ and $\beta$ (as well as their magnitude).

In the simplest example of above, imagine $a=\frac{\sqrt{3}}{2}$ and $b=\frac{i}{2}$. At $t=0$ the probability density would be $$ \vert\Psi(x,0)\vert^2= \frac{3}{4}\psi_0^2(x)+\frac{1}{4}\psi_1(x)^2 $$ but on the other hand, if I take $a=\frac{\sqrt{3}}{2}$ but $b=\frac{1}{2}$, then the probability density becomes $$ \vert\Psi(x,0)\vert^2 = \left(\frac{\sqrt{3}}{2}\psi_0+\frac{1}{2}\psi_1(x)\right)^2\ne \frac{3}{4}\psi_0(x)^2+\frac{1}{4}\psi_1(x)^2 $$ because of the cross terms that appear in expanding the square in the middle part. Both of these have the same average energy, the same probabilities of getting either $E=\frac{1}{2}\hbar\omega$ or $\frac{3}{2}\hbar\omega$, but will produce different $\langle x\rangle$.

In general, the relative phase will control the way states superpose, either completely constructively, completely destructively, or partially. It is the exception rather than the rule when this relative phase is unimportant.

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  • $\begingroup$ Hi- thank you for your reply! I was re-reding these answers when something struck me about your post. When considering the superposition of states using the Schrodinger formlism of wavefunctions as functions rather than kets, we can expand $\Psi(x,t)^*\Psi(x,t)= \vert\alpha\vert^2 \psi_0(x)^2+ \vert\beta\vert^2 \psi_1(x)^2 + (e^{i\omega t}\alpha^*\beta + e^{-i\omega t}\alpha\,\beta^*)\psi_0(x)\psi_1(x)$ as you did, which makes it obvious that the interaction of the states even though they are orthogonal, does make a contribution to experimental results. In particular, you can integrate over $\endgroup$ – 21joanna12 Apr 26 '18 at 13:17
  • $\begingroup$ a small region to find the probability of finding the particle there, and since you are not integrating over the whole of space the orthogonality of the basis functions does not make the last term cancel. On the other hand, when using the ket formalism the probability disribution is $\langle \psi | \psi \rangle$, and if the wavefucntion is in a superposition of orthogonal bassis states then taking this inner product makes all of the 'interference' terms cancel out. Of course, this is because the inner porduct of kets is equivalent to integrating over space as well, when comapring to the $\endgroup$ – 21joanna12 Apr 26 '18 at 13:21
  • $\begingroup$ Schrodinger formalism. It seems I have found a big gap in my understanding. What is the equivalent of finding the probability of the state being in some finite subset of the possible states using the ket formalism? I was taught that $|\psi\rangle$ is the equivalent to $psi$, but it cannot be since $\langle\psi|\psi\rangle$ is already 1... $\endgroup$ – 21joanna12 Apr 26 '18 at 13:25
  • $\begingroup$ @21joanna12 Since $\langle x\vert\psi\rangle=\psi(x)$ (see this answer: physics.stackexchange.com/a/364219/36194) you would do $\Psi(x,t)^*\Psi(x,t) = \vert \langle x\vert\psi(t)\rangle\vert^2$ and you can take it from there. $\endgroup$ – ZeroTheHero Apr 26 '18 at 14:18
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Suppose $\phi_1$ and $\phi_2$ are 1s states of two hydrogen atoms in an $H_2$ molecule. Then the bonding orbital has $c_1=c_2$ and the antibonding orbital has $c_1=-c_2$. So the phase matters. Note that I skipped over the fact that the $\phi_i$ overlap, but this does not invalidate the argument.

For a crystal orbital the phase vector looks like $e^{i\vec{k}\cdot\vec{r}}$ where $\hbar\vec{k}$ is the crystal momentum.

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