5
$\begingroup$

In various explanations I've read, from what I've gathered all particles have a wavefunction $\Psi(\mathbf{r},t)$ where $\mathbf{r}$ is the cartesian coordinates in however many dimensions you're working in. So in regular 3d space the wavefunction is $\Psi(x,y,z,t)$.

My first question is under what conditions is the wavefunction of a particle a pure quantum state? If I take an electron and completely isolate it from the rest of the universe will it be in a pure quantum state? Is this even possible and how does it relate to the eigenfunctions that appear when you act with some operator on the wavefunction? For example when the momentum operator or the position operator act on a wavefunction we get:

$$\hat{p}𝛹 = p_1𝜓_1 + p_2𝜓_2 ... + p_n𝜓_n$$

and

$$\hat{x}𝛹 = x_1𝜓_1 + x_2𝜓_2 ... + x_n𝜓_n$$

Are these functions $𝜓_n$ the same for momentum and position? Does $𝜓_1$ in the momentum equation $= 𝜓_1$ in the position equation? And what form do these functions have, I understand that a plane wave can be written in the form $e^{i(kx-\omega t)}$, but which of these functions are in this form? Are the eigenfunctions in this form or is the total wavefunction $\Psi$ in this form or are neither of them?

Sorry if this is a very convoluted or badly worded question, I'll happily clarify anything that doesn't make sense. I just have a very hard time connecting things in my head and unless I can make sense of this I don't feel I'll ever have a satisfactory understanding of this topic.

For reference I'm a 3rd year chemistry student so my math/physics understanding is nothing impressive but it's not totally non-existent.

Thanks for any responses in advance, I really do appreciate it.

$\endgroup$
3
$\begingroup$

I'll try to answer your questions in the same order they were asked.


Absolute purity is impossible in real world. However, if you limit the domain of your experiment, and look for the properties of particle(s) at specific time- and distance- scale, it is possible to create a state that behaves similarly to the pure state. (In other words, if your experiment lasts very short amount of time, and happens in limited amount of space - you can minimize the effects of the external world on your particle).


Functions $\psi_i$ are different for $\hat{x}$ and $\hat{p}$ because of the uncertainty principle. These wavefunctions are eigenstates of these operators, and since they don't commute - they can't be the same.


Momentum eigenstates have a plane wave form indeed, while position eigenstates have a form of a delta-function. This is very natural when the wavefunction is viewed as a probability distribution. The eigenstate of $\hat{x}$ is a particle that sits at fixed position. Thus there is a sharp peak at that specific position in the wavefunction and zeroes everywhere else.


As for the rest, what you're looking for is called Completeness Relation. One can decompose a wavefunction into a sum of finite (or infinite) number of eigenstates of a particular operator. The operator could be $\hat{x}$, $\hat{p}$, or even the Hamiltonian $\hat{H}$ itself.

$\endgroup$
  • $\begingroup$ 1/2 Thank you very much for taking the time to answer. If you don't mind I have a few follow up questions, I'll number them as you've separated your answers. 1. So it is impossible to 100% isolate a particle in space because its wavefunction always has some overlap with every other particle's wavefunction in the universe, as wavefunctions (in principle at least) extend infinitely. 2. I think I understand, so do commuting operators have identical eigenfunctions and eigenvalues? Is that why a measurement of one doesn't affect the other because they are the same? (mind blown if true). $\endgroup$ – Terry Nov 22 '17 at 17:36
  • $\begingroup$ 2/2 - 3. This actually rationalises something I've been confused about for a long time, the form that the eigenfunctions of momentum and position take wrt. each other, thank you. In this case, what form does the total wavefunction 𝛹(r,t) have? Is the wavefunction just the function you get when you take all eigenfunctions for all operators and add them all together with ther eigenvalues? Is this why solving the Schrodinger equation exactly is so monstrously complicated for anything other than the most basic, single particle systems? 4. I will read up on completeness, thank you for the link. $\endgroup$ – Terry Nov 22 '17 at 17:44
  • $\begingroup$ Commuting operators/observables can be measured without affecting each other indeed. / Wavefunction is a normalized probability distribution that is also a solution to the Schrodinger equation, it can be decomposed into a sum of eigenstates, and it can have any form that satisfies S.eq. It is hard to solve it exactly except for few cases indeed. $\endgroup$ – Darkseid Nov 22 '17 at 18:00
  • $\begingroup$ Wait what?! Why do you claim that momentum eigenstates have a plane wave form and the position eigenstates have a form of a delta-function? We're dealing with a free particle. Why do you assume we know exactly its position? This is utterly strange. I'd tend to claim that both the momentum and position eigenstates are a sum of plane waves, to keep the problem in the most general case possible. $\endgroup$ – thermomagnetic condensed boson Nov 22 '17 at 21:15
  • 1
    $\begingroup$ @no_choice99 that's the whole point of QM; the wavefunction initially represents a probabilistic superposition of possible states. If we were to measure the position or momentum of the particle, the wavefunction would collapse into an eigenstate of that operator. The eigenstates of the position operator are delta functions. The Completeness Relation then tells us that these form a basis for our relevant Hilbert space, free particle or not. $\endgroup$ – tusky_mcmammoth Nov 22 '17 at 23:02
3
$\begingroup$

A basis is a basis so if you have a complete set of functions you can use it to expand anything you want, including free particle wavefunctions.

Why you'd want to do this is another matter: this presumably would depend on the physics of your problem. After all, choosing one basis set rather than another - say spherical rather than Cartesian - is usually done because some features are more naturally obvious in one basis set rather than in another.

$\endgroup$
  • $\begingroup$ That actually raises another issue I've had while trying to learn, the way we approach "basis sets" in chemistry is very de-abstracted from the actual mathematical meaning of a basis set. My biggest problem I have with learning QM is that if I can't completely unite and connect everything I've tried to learn in my head and understand how they all relate the entire image just falls apart for me and I feel like I've learned nothing. My real goal is to understand how basic (and eventually more advanced) QM can be derived mathematically, I feel that is the only way I'll truly feel I understand it. $\endgroup$ – Terry Nov 22 '17 at 18:10
  • $\begingroup$ I've looked online for an explanation but a lot of them dip into, or use notations or ideas from areas of mathematics I have never in my life touched which makes understanding it hard. For example to my understanding the basis vectors i and j can be used to represent any vector in R^3 as a linear combination of themselves. But what does it mean to "change basis sets", does this exclusively involve changing coordinates for example to spherical polar coordinates? Does each coordinate system have only one "basis set"? If not what would be another example of a basis set in R^3? Are there infinite? $\endgroup$ – Terry Nov 22 '17 at 18:12
  • $\begingroup$ @Terry yes there are infinitely many. Think for instance of simply two Cartesian sets with the second rotated w/r to the first. The key mathematical element is the connection with Sturm-Liouville systems: the solutions to some types of differential equations "naturally" admit a complete set of functions. It turns out that the Schrodinger equation is a Sturm-Liouville problem in most circumstances (there are always mathematical caveats). $\endgroup$ – ZeroTheHero Nov 22 '17 at 18:39
2
$\begingroup$

First of all, wavefunctions don't describe particles, they describe systems. For example if you have a 2 particles system (say two electrons in Helium) the wavefunction describes the system.

Now if your system consists of a single particle and a Hamiltonian $\hat{H}$, then $\Psi _n$ is a pure state if: $$ \hat{H} \Psi _n = E \Psi _n$$ and $\Psi$ can not be expressed as a linear combination of other pure states $\Psi _l$; in other words $\Psi$ is a state of definite energy.

For the second part of your question lets consider a general operator $\hat{A}$:

If $\hat{A}$ is the operator associated with the observable $A$, then: $$\hat{A} \psi _n = A_n \psi_n $$ means that $A_n$ (the eigenvalues of $\hat{A}$) are the only possible results of a measurement of $A$. $\psi$ (the eigenfunctions of $\hat{A}$) are the states of definite $A$ (in your examples $A$ is position and momentum).

The set of eigenfunctions $\psi _n$ have the property that $\Psi$ can be expressed as a linear combination of the states of definite $A$.

$$\Psi = \sum _n a_n \psi_n $$

$$ \hat{A}\Psi = \sum _n a_n A_n\psi_n $$

If you have different operators you would have different eigenfunctions, but you still can express $\Psi$ as a linear combination of these other eigenfunctions.

$\endgroup$
  • $\begingroup$ Thank you for your answer. I hope you don't mind me asking some follow up questions. Where you've said that a wavefunction Ψ(r,t) describes the system as a whole, is it possible to represent this total system WF as a sum of the individual WFs that would describe the particles within it? If so what form does the WF of a single particle system take if you were to write it out mathematically? (If it's too complex to write generically would it be the sum of a load of imaginary exponential functions e^i(kx-wt) for example? $\endgroup$ – Terry Nov 22 '17 at 17:58
  • $\begingroup$ A multi-particle wavefunction cannot generically be decomposed into single-particle wavefunctions. For example, the Bell state cannot be expressed as a product of single-particle states. When it can, these are called product states, and are really useful as it means the quantum numbers characterising each state can be studied separately (eg when solving the radial and angular components of the TISE for the hydrogen atom). $\endgroup$ – tusky_mcmammoth Nov 22 '17 at 22:57

protected by Qmechanic Nov 22 '17 at 18:32

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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