I Basis and Object

jlcd

It may be a valid argument that "A basis is not a property of an object. It's a choice humans make in the math for convenience. It makes no sense to say an object has or doesn't have a position basis or any other basis.".

So state vector basis could be the map.

object is the territory.

But then to describe an object, using wave function or state vectors are more accurate than anything to describe the object. What if its the only way?

Lets take the case of time dilation in special relativity. Object doesn't time dilate directly but it is the result of speed and time and generally derived from the temporal part the lorentz transformation. Here:

lorentz transformation = map

object = territory

The best way to derive at time dilation is using the math of lorentz transformation and the observation is a result of geometry. There seems to be no other way to explain time dilation.

Right now. We describe objects and physics by means of math model. It is our only way to describe reality.

Therefore would it be wrong to say the basis in state vector where particle is described by it the only way how object arises (observable for example)? Then state vector is more fundamental.

Mathematically is there another way you can describe the basis and observable without using the concept of state vector basis or wave function? If there is none. Then the particle behaving or taking part in the dynamics of state vector and basis is the object itself. Here a basis can be considered as the property of an object. What if state vector and basis describing object is the true reality, then can't we say basis is a property of object and not just a choice humans make in the math for convenience?

Is there other things in physics where the choice humans make in the math for convenience is really what the objects are (isn't it Dirac Equation is like this, the math being the only way to describe the object and no other way, here the particle and dirac equation is the object itself. In this case perhaps one mustn't use the analogy of map and territory because it is a hybrid)).

In QFT, what does the basis stand for (if it is not observable in QM)?

Related Quantum Physics News on Phys.org

vanhees71

Gold Member
Obviously you are very confused. You have to look at the fundamental structure of QM to understand it. Here we can concentrate on the "kinematical part".

A quantum system is described by a Hilbert space $\mathcal{H}$. The special case of a pure state of the system is described by a unit ray in Hilbert space, $[\psi]$, which is given by a unit vector $|\psi \rangle$ modulo a phase factor, i.e., all vectors which are different from $|\psi \rangle$ only by a phase factor, $|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle$ with $\varphi \in \mathbb{R}$ represent the same state. Equivalently you can also represent this pure state by a statistical operator $\hat{\rho}=|\psi \rangle \langle \psi|$, which is independent of the choice of the overall phase and thus provides a one-to-one mapping of mathematical objects (statistical operator) and pure states.

Any observable of the quantum system is represented by a self-adjoint operator $\hat{O}$. The possible values the observable can take is an eigenvalue (or more precisely spectral value) of its representing operator. Then there exists a orthonormal set $|o,\alpha \rangle$ of (generalized) eigenvectors, where $\alpha$ labels some set of parameters numbering the orthonormal eigenvectors with the same eigenvalue $o$.

The physical meaning of this construction is given by Born's rule. If the system is prepared in the pure state $\hat{\rho}=|\psi \rangle \langle \psi |$, then the probability to find the value $o$ when measuring the observable $O$, represented by the self-adjoint operator $\hat{O}$ is
$$P(o)=\sum_{\alpha} \langle o,\alpha|\hat{\rho}|o,\alpha \rangle = \sum_{\alpha} |\langle o,\alpha|\psi \rangle|^2.$$
If there are continuous parts of possible values for $\alpha$ the sum is substituted by an integral over these values. In the following I always write sums to simplify the notation.

A complete set of observable $O_j$ is defined such that all the representing operators commute $[\hat{O}_j,\hat{O}_k]=0$, and the common eigen-spaces are one-dimensional. Then one defines the wave function with respect to the corresponding orthonormal eigenbasis $|o_1,o_2,\ldots,o_n \rangle$ as
$$\psi(o_1,\ldots,o_n)=\langle o_1,\ldots,o_n|\psi \rangle.$$
One also says, this is the wave function in $(o_1,\ldots,o_n)$-representation.

Since the eigenbasis is complete, i.e.,
$$\sum_{o_j} |o_1,\ldots,o_j \rangle \langle o_1,\ldots,o_j|=\hat{1}$$
there's a one-to-one mapping between the Hilbert space of wave functions (which is the Hilbert space of square-summable sequences if all eigenvalues are discrete or the Hilbert space of square-Lebesgue-integrable functions or a mixture thereof).

If the pure state is given by the wave function (modulo an overall phase), then the probability that the observables $O_j$ take the values $o_j$ respectively is
$$P(o_1,\ldots,o_n)=|\psi(o_1,\ldots,o_n)|^2.$$
The scalar product of two vectors is given with help of the above completeness relation
$$\langle \phi|\psi \rangle=\sum_{o_j} \langle \phi|o_1,\ldots,o_n \rangle \langle o_1,\ldots,o_n |\psi \rangle = \sum_{o_j} \phi^*(o_1,\ldots,o_n) \psi(o_1,\ldots,o_n).$$

• DrClaude and weirdoguy

jlcd

Obviously you are very confused. You have to look at the fundamental structure of QM to understand it. Here we can concentrate on the "kinematical part".

A quantum system is described by a Hilbert space $\mathcal{H}$. The special case of a pure state of the system is described by a unit ray in Hilbert space, $[\psi]$, which is given by a unit vector $|\psi \rangle$ modulo a phase factor, i.e., all vectors which are different from $|\psi \rangle$ only by a phase factor, $|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle$ with $\varphi \in \mathbb{R}$ represent the same state. Equivalently you can also represent this pure state by a statistical operator $\hat{\rho}=|\psi \rangle \langle \psi|$, which is independent of the choice of the overall phase and thus provides a one-to-one mapping of mathematical objects (statistical operator) and pure states.

Any observable of the quantum system is represented by a self-adjoint operator $\hat{O}$. The possible values the observable can take is an eigenvalue (or more precisely spectral value) of its representing operator. Then there exists a orthonormal set $|o,\alpha \rangle$ of (generalized) eigenvectors, where $\alpha$ labels some set of parameters numbering the orthonormal eigenvectors with the same eigenvalue $o$.

The physical meaning of this construction is given by Born's rule. If the system is prepared in the pure state $\hat{\rho}=|\psi \rangle \langle \psi |$, then the probability to find the value $o$ when measuring the observable $O$, represented by the self-adjoint operator $\hat{O}$ is
$$P(o)=\sum_{\alpha} \langle o,\alpha|\hat{\rho}|o,\alpha \rangle = \sum_{\alpha} |\langle o,\alpha|\psi \rangle|^2.$$
If there are continuous parts of possible values for $\alpha$ the sum is substituted by an integral over these values. In the following I always write sums to simplify the notation.

A complete set of observable $O_j$ is defined such that all the representing operators commute $[\hat{O}_j,\hat{O}_k]=0$, and the common eigen-spaces are one-dimensional. Then one defines the wave function with respect to the corresponding orthonormal eigenbasis $|o_1,o_2,\ldots,o_n \rangle$ as
$$\psi(o_1,\ldots,o_n)=\langle o_1,\ldots,o_n|\psi \rangle.$$
One also says, this is the wave function in $(o_1,\ldots,o_n)$-representation.

Since the eigenbasis is complete, i.e.,
$$\sum_{o_j} |o_1,\ldots,o_j \rangle \langle o_1,\ldots,o_j|=\hat{1}$$
there's a one-to-one mapping between the Hilbert space of wave functions (which is the Hilbert space of square-summable sequences if all eigenvalues are discrete or the Hilbert space of square-Lebesgue-integrable functions or a mixture thereof).

If the pure state is given by the wave function (modulo an overall phase), then the probability that the observables $O_j$ take the values $o_j$ respectively is
$$P(o_1,\ldots,o_n)=|\psi(o_1,\ldots,o_n)|^2.$$
The scalar product of two vectors is given with help of the above completeness relation
$$\langle \phi|\psi \rangle=\sum_{o_j} \langle \phi|o_1,\ldots,o_n \rangle \langle o_1,\ldots,o_n |\psi \rangle = \sum_{o_j} \phi^*(o_1,\ldots,o_n) \psi(o_1,\ldots,o_n).$$
Thanks you surely are a textbook writer who can write down everything in one stroke.

So a particle can be represented as wave function or state vector and it's eigenbasis is the observable like position, momentum, energy, right?

My point is. Can you treat a basis as a property of a particle?

Can you produce an observable without using the formalism of state vector in Hilbert space or wave function? How is basis represented in Heisenberg Matrix Mechanics for example? Does it still use the concept of basis? I want other synonyms for basis. I want to use other words beside say "the Preferred basis problem of MWI". What words can you use besides "Preferred basis problem". I'm not trying to argue for interpretations but just want another mathematical synonym for it. Thanks.

vanhees71

Gold Member
Again, you have to be very pedantic in the foundations to understand quantum mechanics. You cannot simply shorten what I wrote above (and what's written in textbooks more or less equivalently to what I wrote)!

Pure states: represented by unit rays in a Hilbert space (unit vectors modulo a phase factor)
possible values observables can take: eigenvalues of their representing self-adjoint operators
Common eigenbasis of a complete set of observables:$|o_1,\ldots,o_n\rangle$
meaning of the state (vector): $P(o_1,\ldots,o_n)=|\langle o_1,\ldots,o_n|\psi \rangle|^2=|\psi(o_1,\ldots,o_n)|^2$ is the probability for the complete set of observables taking the values $(o_1,\ldots,o_n)$.

There's no preferred basis problem in the standard interpretation. Don't bother about interpretations at all at your stage of knowledge. Learn the minimal interpretation first. Then you can, in your free time, dig into the socalled "interpretational problems" of quantum theory. That's a purely philosophical subject, amusing sometimes to think about in your spare time, and then coming to the conclusion that it's irrelevant for physics, but it helps to understand the full physical meaning of the formalism to think about them.

All there is, is the following: A system is prepared in some state (e.g., a pure state as discussed here), and then you know the probabilities for the outcome of measurements of all observables of this system. For this you need to choose the basis of the operator representing the measured observable. The only sense the idea of "preferred basis" makes is that it's the basis to choose to calculate these probabilities. Which basis to choose simply depends on which probability you want to know, i.e., which observable your experimental setup is supposed to measure.

The big problem with interpretational problems is that people try to discuss them without even knowing what the formal physical part of the theory really is, and that leads to even more unnecessary confusions than this murky unsharply defined philosophical pseudo-problems produce anyway.

For the best treatment of "interpretations of quantum theory", see

S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press

His conclusion is that the question of the right interpretation is open. That makes his treatment so valuable in not trying to convince you of one interpretation. He just describes the facts about the various interpretations and provides also a convincing argument for the independence of Born's rule from the other postulates (particularly the dynamical, not touched yet in this thread).

Note that I don't agree with Weinberg on the issue that there is a interpretational problem, and that I think that the minimal statistical interpretation (i.e., the Copenhagen flavor that does not contain a collapse postulate) is all there is in terms of QT as a physical theory, but that's of course a personal opinion. Nevertheless I think Weinberg's exhibition of the "interpretation problem" is the best there is in the textbook literature.

Concerning the philsophical part of scientific theories it finally always boils down to a personal choice, which however has not much impact on what the theory means concerning the purely objective scientific content: That's just about describing what's observed and quantitatively measured in nature as completely and as accurately you can, and this is provided by the minimal statistical interpretation. This view is very well represented in Ballentine, Quantum Mechanics, Addison Wesley.

The true remaining physics problem, in my opinion, of QT is not its interpretation but its lack of a satisfactory description of the gravitational interaction (and closely related to that of spacetime itself). It may well be that one day, a more comprehensive theory is found, including the description of gravity and spacetime, deviating from todays QT but containing QT as a certain limit (like non-relativistic physics is a limiting case of the more comprehensive relativistic physics for small velocities and weakly interacting particles/bodies).

jlcd

Again, you have to be very pedantic in the foundations to understand quantum mechanics. You cannot simply shorten what I wrote above (and what's written in textbooks more or less equivalently to what I wrote)!

Pure states: represented by unit rays in a Hilbert space (unit vectors modulo a phase factor)
possible values observables can take: eigenvalues of their representing self-adjoint operators
Common eigenbasis of a complete set of observables:$|o_1,\ldots,o_n\rangle$
meaning of the state (vector): $P(o_1,\ldots,o_n)=|\langle o_1,\ldots,o_n|\psi \rangle|^2=|\psi(o_1,\ldots,o_n)|^2$ is the probability for the complete set of observables taking the values $(o_1,\ldots,o_n)$.

There's no preferred basis problem in the standard interpretation. Don't bother about interpretations at all at your stage of knowledge. Learn the minimal interpretation first. Then you can, in your free time, dig into the socalled "interpretational problems" of quantum theory. That's a purely philosophical subject, amusing sometimes to think about in your spare time, and then coming to the conclusion that it's irrelevant for physics, but it helps to understand the full physical meaning of the formalism to think about them.

All there is, is the following: A system is prepared in some state (e.g., a pure state as discussed here), and then you know the probabilities for the outcome of measurements of all observables of this system. For this you need to choose the basis of the operator representing the measured observable. The only sense the idea of "preferred basis" makes is that it's the basis to choose to calculate these probabilities. Which basis to choose simply depends on which probability you want to know, i.e., which observable your experimental setup is supposed to measure.

The big problem with interpretational problems is that people try to discuss them without even knowing what the formal physical part of the theory really is, and that leads to even more unnecessary confusions than this murky unsharply defined philosophical pseudo-problems produce anyway.

For the best treatment of "interpretations of quantum theory", see

S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press

His conclusion is that the question of the right interpretation is open. That makes his treatment so valuable in not trying to convince you of one interpretation. He just describes the facts about the various interpretations and provides also a convincing argument for the independence of Born's rule from the other postulates (particularly the dynamical, not touched yet in this thread).

Note that I don't agree with Weinberg on the issue that there is a interpretational problem, and that I think that the minimal statistical interpretation (i.e., the Copenhagen flavor that does not contain a collapse postulate) is all there is in terms of QT as a physical theory, but that's of course a personal opinion. Nevertheless I think Weinberg's exhibition of the "interpretation problem" is the best there is in the textbook literature.

Concerning the philsophical part of scientific theories it finally always boils down to a personal choice, which however has not much impact on what the theory means concerning the purely objective scientific content: That's just about describing what's observed and quantitatively measured in nature as completely and as accurately you can, and this is provided by the minimal statistical interpretation. This view is very well represented in Ballentine, Quantum Mechanics, Addison Wesley.

The true remaining physics problem, in my opinion, of QT is not its interpretation but its lack of a satisfactory description of the gravitational interaction (and closely related to that of spacetime itself). It may well be that one day, a more comprehensive theory is found, including the description of gravity and spacetime, deviating from todays QT but containing QT as a certain limit (like non-relativistic physics is a limiting case of the more comprehensive relativistic physics for small velocities and weakly interacting particles/bodies).
vanheez71. I really understood the minimum interpretation and even the math you shared. I'm just asking whether one can just treat a particle as something that has state vector as part of its makeup. Or is it ok to think an object or particle is simply something that really has state vector and basis inherent in it? What do you think. Sorry for this short reply but I just want your answer on it.

About true remaining physics problem of quantum gravity only. I'm writing a paper or document to prove to you and Neumaier and Demystifier and the rest that there is a new physics the world has never seen before (at least in the mainstream physics community). And I will even include scientific evidence and demonstration. However. In the conceptual parts. I need to write about state vectors and basis so I need your comment on the first paragraph above whether you think it is ok to state that a particle has inherent state vector and basis in its behavior as part of its makeup something along the thought of math have objective existence as proposed by Penrose and company (like Tegmark The Mathematical Universe).

Nugatory

Mentor
I'm writing a paper or document to prove to you and Neumaier and Demystifier and the rest that there is a new physics the world has never seen before (at least in the mainstream physics community).
But be mindful of the forum rule about new theories - it’ll have to be published elsewhere before it can be discussed here.

• weirdoguy and vanhees71

microsansfil

Or is it ok to think an object or particle is simply something that really has state vector and basis inherent in it?
In the context of quantum information theory, there is always a basis for which the qubit is well defined.

/Patrick

vanhees71

Gold Member
Aehm, can you specify this further? As an observable a q-bit is well-defined and independent of any basis. I don't think that's different in quantum information theory, which is just usual QT applied to multi-qbit systems anyway ;-).

microsansfil

It seems that Ballentine says the same thing in a different way :
From "The Statistical Interpretation of Quantum Mechanics" /Patrick

Mentz114

Gold Member
[]
. I'm writing a paper or document to prove to you and Neumaier and Demystifier and the rest that there is a new physics the world has never seen before (at least in the mainstream physics community). And I will even include scientific evidence and demonstration.
[..]
Great ! But you should learn some maths and physics first.

vanhees71

Gold Member
HI,

there is always a basis against which the qubit is well defined. This not possible with an incoherent mixture!

/Patrick
I still don't understand what you mean by that. Of course, what you quoted from Ballentines paper is correct. For any statistical operator $\hat{\rho}$, which is a self-adjoint operator there's an eigenbasis,
$$\hat{\rho}=\sum_{n} \rho_n |\phi_n \rangle \langle \phi_n|.$$
Since $\hat{\rho}$ is positive semidefinite and of trace 1 you have $0 \leq \rho_n \leq 1$ and $\sum_n \rho_n=1$.

$\hat{\rho}$ thus represents a pure state if and only if for one $n=j$ you have $\rho_j=1$ and $\rho_n=0$ for all $n \neq j$. Then and only then it's a projection operator as it must be for a pure state.

Now take a q-bit. What do you mean by "the q-bit is well defined against some basis"? Do you mean that then there's a spin component in some direction which has a determined value? But that can be only the case if $\hat{\rho}$ represents a pure state. For a q-bit this can be seen elegantly as follows.

Here the Hilbert space is two-dimensional and in an arbitrary basis (usually chosen such that it's the eigenbasis of $\hat{s}_3=\hat{\sigma}_3/2$) you can write it in terms of hermitean $2 \times 2$ matrices, of which a basis is $\hat{1}$, $\hat{\sigma}_j$ with $j \in \{1,2,3 \}$, where $\hat{\sigma}_j$ are the three Pauli matrices. Then any statistical operator can be written in the form
$$\hat{\rho}=\frac{1}{2} (\hat{1} + c_j \hat{\sigma}_j)$$
with real $c_j$'s.

Now by a unitary transformation, representing a rotation $\hat{D}_{1/2}(\vec{n} \varphi)$, you can diagonalize the statistical operator. Redefining our spatial directions accordingly then the matrix of the statistical operator in this basis gets
$$\hat{\rho}'=\frac{1}{2} (\hat{1} + r \hat{\sigma}_3)=\frac{1}{2} \mathrm{diag}(1+r,1-r).$$
We have $|r|=\sqrt{c_1^2+c_2^2+c_3^2}$ and the properties of the statistical operator implies that $r \in [-1,1]$.

This implies that $\hat{\rho}$ represents a pure state with a determined value $r/2$ for the spin-component in 3'-direction iff $\rho \in \{1,-1 \}$. This means that one spin component for a statistical operator $\hat{\rho}$ takes a determined value if and only if $\sum_j c_j^2=1$ and thus $\hat{\rho}$ represents the corresponding pure state, and the direction of the then determined spin component is given by $(c_1,c_2,c_3)$.

In an arbitrary basis thus all (pure as well as mixed) states of the q-bit can be represented by three real values $(c_j)$ or, written as a vector $\vec{c} \in \mathbb{R}^3$, with $|\vec{c}| \leq 1$. That's known as the Bloch sphere. Any pure state is represented by the surface of this Bloch sphere, i.e., by a unit vector $\vec{c}$. All other states with a $\vec{c}$ with $|\vec{c}|=r<1$ are mixed.

• weirdoguy

microsansfil

I still don't understand what you mean by that. Of course, what you quoted from Ballentines paper is correct.
The paragraph "1.2 Qubits" of the book "The Physics of Quantum Information" would not be correct when it is write : ?

The essential point here is that for a coherent superposition there is always a basis in which the value of the qubit is well defined, while for an incoherent mixture it is a mixture whatever way we choose to describe it.
Also presented in CNRS courses : http://paristech.institutoptique.fr/site.php?id=334&fileid=1114

Now take a q-bit. What do you mean by "the q-bit is well defined against some basis"? Do you mean that then there's a spin component in some direction which has a determined value? But that can be only the case if $\hat{\rho}$ represents a pure state.
That's what I also understand.

/Patrick

weirdoguy

In QFT, what does the basis stand for (if it is not observable in QM)?
I think that first and foremost you should understand that basis is a mathematical concept used in linear algebra and functional analysis:
And since these branches of mathematics are used to formulate quantum theories, we talk about bases also when we talk about quantum mechanics or QFT. We use basis to assign to every vector a set of real or complex numbers that will represent our vector, just for our convenience. Even in maths we sometimes need to use numbers instead of abstract quantities. Different basis - different set of numbers, and which basis we will use is up to us. But we always have to remember that these numbers that basis gives us and abstract vectors are not the same thing.

• Klystron

jlcd

I think that first and foremost you should understand that basis is a mathematical concept used in linear algebra and functional analysis:
And since these branches of mathematics are used to formulate quantum theories, we talk about bases also when we talk about quantum mechanics or QFT. We use basis to assign to every vector a set of real or complex numbers that will represent our vector, just for our convenience. Even in maths we sometimes need to use numbers instead of abstract quantities. Different basis - different set of numbers, and which basis we will use is up to us. But we always have to remember that these numbers that basis gives us and abstract vectors are not the same thing.
There are many mathematical methods or tools. So for these word "Preferred basis". What synonym can you use to replace "basis" if you want to describe QM or QFT without using linear algebra or functional analysis or any concept of basis?

Nugatory

Mentor
if you want to describe QM or QFT without using linear algebra or functional analysis or any concept of basis?
That's like trying to describe your ancestry while not using the concept that people have parents.

And as we are now far from the Physics Forums mission of helping people understand physics, this thread is closed.

• DrClaude

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