B Understanding Quantum Theory: Wave vs Vector in Hilbert Space

[fanieh]

I read from a textbook. It's about single outcome. When the vector is in the coordinate axis. It has single outcome, correct?

In QM, when it is in the eigenvalue of an observable. It has single outcome too correct? ( I use the word single outcome instead of collapse).

Hence aren't they related since they have both single outcome?

To elaborate. when a quantum state is measured, the quantum state "jumps" to one of the axes of Hilbert space determined by that measurement, correct? hence I said there was a value.

 

[PeterDonis]

when a quantum state is measured, the quantum state "jumps" to one of the axes of Hilbert space determined by that measurement, correct?

Sort of. First, you are assuming a collapse interpretation, but the basic math of QM does not require a collapse interpretation. But let's put that aside for now.

Second, a Hilbert space does not have a single unique basis. There are an infinite number of possible bases for every Hilbert space. For a given observable, the Hermitian operator associated with that observable picks out one basis from the infinite number of possible ones. The elements of this basis are called "eigenstates" (or eigenvectors) in state vector language, or "eigenfunctions" in wave function language (though the latter is used much less often in this connection, even if you are using the wave function representation). So, if we assume a collapse interpretation, then the measurement makes the quantum state "jump" to one of the eigenstates, i.e., to one of the elements of the basis picked out by the observable being measured.

hence I said there was a value

The fact that a particular measurement results in a value does not mean a particular basis, or a particular element of that basis, "has a value" in a general sense. At most, it means that state is associated with that value (the "eigenvalue") if we are making that particular measurement. But that is a very restricted association.

 

[fanieh]

,

Sort of. First, you are assuming a collapse interpretation, but the basic math of QM does not require a collapse interpretation. But let's put that aside for now.

Second, a Hilbert space does not have a single unique basis. There are an infinite number of possible bases for every Hilbert space. For a given observable, the Hermitian operator associated with that observable picks out one basis from the infinite number of possible ones. The elements of this basis are called "eigenstates" (or eigenvectors) in state vector language, or "eigenfunctions" in wave function language (though the latter is used much less often in this connection, even if you are using the wave function representation). So, if we assume a collapse interpretation, then the measurement makes the quantum state "jump" to one of the eigenstates, i.e., to one of the elements of the basis picked out by the observable being measured.
The fact that a particular measurement results in a value does not mean a particular basis, or a particular element of that basis, "has a value" in a general sense. At most, it means that state is associated with that value (the "eigenvalue") if we are making that particular measurement. But that is a very restricted association.

You said: "For a given observable, the Hermitian operator associated with that observable picks out one basis from the infinite number of possible ones". I understand this. But when I'll share this in a lecture to complete newbies introducing QM and don't want to complicate it by mentioning basis or vectors. Can I just say "For a given observable, the Hermitian operated picks out one eigenvalue". Wasn't this the language we heard from Schroedinger before Von Neumann formalize it. Or what other words can I describe it that is correct that doesn't use basis or vectors since the newbies won't understand this.

 

[bhobba]

]So "no" was a mistake and it should be "yes"? If still no. Please explain. Would appreciate it a lot.

Well I have read a number of books on QFT and am totally unaware of how to do that.

But Peter does know more QFT than me - he has corrected some misconceptions I had on a few occasions. But as another science adviser, Stragerep, says - you find humility in QFT - I certainly have.

That said I doubt it can be explained except at at least the I level - probably A.

Had a quick look at the Wikipedia link Peter gave - its at the A level and would take me a serious amount of time to digest. If you want it to help beginner students then sorry - I don't think it would help,

Thanks
Bill

 

[PeterDonis]

Can I just say "For a given observable, the Hermitian operated picks out one eigenvalue".

I don't think so, because the Hermitian operator itself just picks the basis; it doesn't pick which of the basis elements determines the result of the measurement. The result of the measurement is a random choice among the possible values, with the probability of each value being equal to the squared modulus of the complex amplitude associated with the corresponding basis element.

I would also observe that your suggested language here has nothing whatever to do with wave functions, which I thought was what you were trying to use as your "newbie" version of QM.

what other words can I describe it that is correct that doesn't use basis or vectors since the newbies won't understand this.

I don't know. I'm not sure this is even doable, or worth doing. To me, trying to teach QM without using basis or vectors is like trying to teach arithmetic without using addition.

A key advantage of the state vector formalism in QM is that it avoids having to commit to any specific choice of basis or representation of states; you're just using the underlying Hilbert space structure that is there no matter what basis or representation you choose. That makes it much more general and much more useful than the wave function formalism, which commits you to a specific choice of basis (in wave function language, this would be the variable or variables that the wave function is a function of, like $x$ in the position representation) and a specific representation.

 

[fanieh]

I don't think so, because the Hermitian operator itself just picks the basis; it doesn't pick which of the basis elements determines the result of the measurement. The result of the measurement is a random choice among the possible values, with the probability of each value being equal to the squared modulus of the complex amplitude associated with the corresponding basis element.

I would also observe that your suggested language here has nothing whatever to do with wave functions, which I thought was what you were trying to use as your "newbie" version of QM.
I don't know. I'm not sure this is even doable, or worth doing. To me, trying to teach QM without using basis or vectors is like trying to teach arithmetic without using addition.

A key advantage of the state vector formalism in QM is that it avoids having to commit to any specific choice of basis or representation of states; you're just using the underlying Hilbert space structure that is there no matter what basis or representation you choose. That makes it much more general and much more useful than the wave function formalism, which commits you to a specific choice of basis (in wave function language, this would be the variable or variables that the wave function is a function of, like $x$ in the position representation) and a specific representation.

Why, wave function language without committing to any choice of basis and without any specific representation doesn't make sense? but the schroedinger equation can still be written without any basis.. is it not.

 

[PeterDonis]

the schroedinger equation can still be written without any basis.. is it not

No. The Schrodinger equation is explicitly written in the position basis. It takes derivatives with respect to $x$ (and $t$, but time in non-relativistic QM is a parameter, not a basis choice); that means you've chosen the position basis.

 

[fanieh]

No. The Schrodinger equation is explicitly written in the position basis. It takes derivatives with respect to $x$ (and $t$, but time in non-relativistic QM is a parameter, not a basis choice); that means you've chosen the position basis.

reading about wave function in Wikipedia I reached the last paragraph and it was written: https://en.wikipedia.org/wiki/Wave_function

"Whether the wave function really exists, and what it represents, are major questions in the interpretation of quantum mechanics. Many famous physicists of a previous generation puzzled over this problem, such as Schrödinger, Einstein and Bohr. Some advocate formulations or variants of the Copenhagen interpretation (e.g. Bohr, Wigner and von Neumann) while others, such as Wheeler or Jaynes, take the more classical approach[42] and regard the wave function as representing information in the mind of the observer, i.e. a measure of our knowledge of reality. Some, including Schrödinger, Bohm and Everett and others, argued that the wave function must have an objective, physical existence. Einstein thought that a complete description of physical reality should refer directly to physical space and time, as distinct from the wave function, which refers to an abstract mathematical space.[43]"

Why do we heard it asked whether wave function really exist or have an objective, physical existence. Yet we never heard it asked whether the state vectors really exist or have an objective, physical existence? This is the point of this thread.

 

[PeterDonis]

Why do we heard it asked whether wave function really exist or have an objective, physical existence. Yet we never heard it asked whether the state vectors really exist or have an objective, physical existence?

Yes, we have. You are reading that paragraph way too literally. As the term "wave function" is used in that particular paragraph in the Wiki article, it just means "whatever mathematical object we are using to describe the quantum state of the system"--i.e., it could be a wave function in the stricter sense of $\psi(x)$, or it could be a state vector. If you look at the actual literature on the subject, you will see people asking the question both ways--"does the wave function really exist" and "does the state vector really exist". The question is the same either way.

This is the point of this thread.

Then this whole thread has been based on a simple misunderstanding on your part, and is therefore closed.