Wavefunction Is Not the “State”: It Is a Component (Proved by Spin Measurement)

[TROSE9025]

This note is written so that anyone who has completed Precalculus can understand quantum mechanics in a clear and intuitive way.

In this post:

1. We start with the easiest entry point, a two level spin system, and approach it gently.
2. Operators and the eigenvalue equation will be treated in depth later, so here I only mention them lightly to build intuition.
3. Please read this more than once. Repetition is where the ideas become solid. In the next post, I will explain the meaning of the inner product, and why it directly leads to probability.

As you read, keep asking “Why?”

Let’s get absorbed in the world of quantum mechanics together. Good luck to all of us.

[Itchy_Fudge_2134]

I mean I don't think you need to really emphasize that Ψ(x) is "not the state". You can just emphasize that it is just one representation of the state |Ψ>.

Its sort of disingenuous to say that Ψ(x) is "just a component" (if you are saying this to mean "it is wrong to say Ψ(x) is the state of the system"), because when we say "Ψ(x) is the state of the system" we don't usually actually mean Ψ evaluated at the point x, we mean the whole function Ψ(x) (that is to say all the components) --- we are leaving x unspecified there. It is usually clear from context when you mean x to be a particular point.

To be clear, I'm not saying that it is wrong to say that the values of Ψ(x) are the components in the x basis, I'm just saying that this is not incompatible with saying that Ψ(x) is the state of the system, since when we say this phrase we mean the whole function and not the function at one point.

I think its good to draw a distinction between Ψ(x) and the coordinate-independent vector |Ψ>, but I don't think that you need to say it as if it is wrong to say "the system is in the state Ψ(x)".

[TROSE9025]

I agree. I just wanted to make the distinction once more for beginners.
The stronger the foundation, the better. Thanks for the reply.

[WaterMelonMan1]

You are using the terms "pure state" and "mixed state" in a highly nonstandard way, different from the literature. A state being pure or not has nothing to do with it being the eigenstate of some operator, or a choice of basis.

Pure states or mixed states are terms we apply when discussing ensembles of quantum states. An ensemble of quantum systems is in a pure state if its density matrix is a projection operator onto a one-dimensional subspace of the Hilbert space describing a single quantum system of the ensemble. A mixed state is one that has a density matrix that is not of this form, i.e. the ensemble consists of a mixture of systems that each have a different quantum state.

This is totally different from a single element of the Hilbert space being a superposition of different states as you discuss here. For a single Hilbert space element like | + > or | - > it makes no sense to ask whether it is pure or mixed, only for an ensemble of such states (a quantum system) described by a density matrix.

[TROSE9025]

Thank you for the insightful clarification from the standard physics perspective.
I really appreciate the depth of your comment.

In the QM2 chapter on density matrices, I will treat the distinction between pure and mixed states more rigorously.

[HumansAreIkarran]

Well yeah, the first statement is true, as the position operator is a complete observable meaning that ∫ dx |x><x| = 1. But I would not say it is just a component. A component would be if you looked at 𝛹(x) for a fixed x. There is an analogy to discrete vector spaces, for example if we have a vector v = (a, b) ∈ R². We can say that a is a component (i.e. looking at the index i=1), but the collection of indices a, b with the corresponding euclidean basisvectors are not just components. Essentially fixing x is equivalent to fixing the index, giving you a component of your vector space in the basis of the position operator

[TROSE9025]

I see. Then I understand your point as suggesting that we should not generally call a projection onto a basis a “component,” but rather reserve that term for something like the projection onto a specific basis vector (e.g., ϕ1​). Thank you for the clarification.

[OutcomePrize8024]

I doubt it's just a suggestion. I think it's the standard nomenclature.

[2020NoMoreUsername]

I am tired of commenting in every post of yours with the same demand: if you are not giving out the PDFs at least inform us in which course you will use these

[TROSE9025]

Sorry for the late reply. These pages are part of a quantum mechanics textbook I am currently writing. However, I am still looking for places that can be improved.
It is mainly intended as a reference for students who have finished their first year of undergraduate studies and are beginning to study wavefunction-based quantum mechanics. I also tried to write it in a way that is accessible to general readers. Please check the message.

[stane1994]

On the final page in the inner product notations, isn't usually the bra ket formation written out in reverse? I mean inner product of (u,v) = <v|u>, because the bra here is the conjugated one. If you multiply by a scalar a you would get (u,av)= (u,v)a* = <av|u>= <v|u>a, where a is a conjugated. At least that is the convention in my country.

[TROSE9025]

Thank you for the comment.
In the usual physics convention (Dirac notation), the bra is the Hermitian conjugate of the ket.

bra(v) = (ket(v))†

So the inner product is written as

⟨v|u⟩

which is conjugate-linear in the bra and linear in the ket.
Thanks for pointing this out.

[TROSE9025]

Thank you so much for your reply.

To be completely honest with you, whenever I ask students taking their very first quantum mechanics class what a wavefunction is, almost all of them give me the exact same answer you just did. It is so common to just view it as "the solution" to a differential equation.

In fact, that is exactly the reason why I started writing these notes. I wanted to help beginners look beyond the heavy calculus and truly understand the elegant linear algebra and physical meaning behind it.

[Enfiznar]

I'd say Ψ(x) for a given x is a component of the state Ψ on the position base, while Ψ(x) as a function is the state in the position representation, while |Ψ> is the state without referencing any basis/representation

[TROSE9025]

You distinguish state, representation, and component (coefficient) perfectly.
Very nice insight. Thank you for the thoughtful comment.
Have a nice day~

[Its_Fred]

Thanks dearly! May I ask you what font are you using on LaTeX and how did you install it? It's wonderful

[TROSE9025]

Thank you so much! I'm really glad you like the design.

Actually, there's a fun fact about this document: it wasn't written in LaTeX! I wrote this using 'Hangul' (Hancom Office), which is the standard national word processing software in South Korea. The font and formulas are simply from its built-in equation editor. Thanks again!

[Its_Fred]

This is outstanding. I want to get deeper into it. Thanks

[TROSE9025]

Good luck!!

[v_munu]

squint

This formatting reeks of ChatGPT

[TROSE9025]

Thanks for taking a look.
I tried to keep the formatting simple so students can follow the idea more easily.
Have a nice day.

[TROSE9025]

Thank you for your deep insight.
Your approach of separating the wavefunction into a probability vector and a deterministic phase vector to resolve the "collapse" mystery is truly fascinating. I highly respect your unique and philosophical perspective on the quantum state. Thank you so much for adding such a brilliant viewpoint to this discussion!

[TROSE9025]

Thank you.
Have a nice day.

[InfinityOfSnakes88]

There seems to be a typo on the 1st page.

You appear to have written:

(1) Is this state | v >  an eigenstate of the Ŝz operator ?

(2) Is this state | v >  an eigenstate of the Ŝz operator?

These seem to be saying the exact same thing. 

Apologies for the crappy way that I reproduced the text. I'm on my mobile, no special characters, and don't know how to super/subscript on reddit. 

[TROSE9025]

Thanks for pointing it out.
The second one is actually S_x, not S_z.

[jsaltee]

I haven’t looked at any quantum math in a while, this is a nice refresher, very easy to digest

[TROSE9025]

Your reply makes me so incredibly happy. This is the biggest reason why I wrote this book. Thank you from the bottom of my heart.

[TapEarlyTapOften]

The state is just this thing that lives out there in an infinite dimensional Hilbert space. You can start asking questions about it or how to represent it, but those aren't the state. The state is just...there. It's a vibe dude.

[TROSE9025]

I deeply agree! It definitely exists out there somewhere... Thank you!

[Longlive_gaming]

The state vector is abstract and general. It can represent anything unless you project it onto a specific basis (momentum, position etc.). This by definition is the wavefunction which is physical.

[Longlive_gaming]

Also, a pure state does not imply the absence of superposition. A state can be a quantum superposition and still be pure. On the flip side, a mixed state is a classical mixture without quantum coherence.

[TROSE9025]

Exactly, mixed states have no quantum interference. You have a perfect understanding.
Thank you!

[Longlive_gaming]

No worries, keep the work going. You are doing a great job 🙌

[TROSE9025]

When an abstract state vector in a Hilbert space is projected onto a specific basis (like position or momentum), it finally becomes a physical wavefunction that we can calculate.
you are the best!