# Understanding Mixed vs Pure States in Quantum Mechanics

• msumm21
In summary, "In summary, the quote from the book refers to a wavefunction that determines a projection operator on the Hilbert space, taking it from the entire space to a smaller subspace. When the operators are restricted to this subspace, they act diagonally, meaning they are equivalent to simple multiplication by a scalar. This is a result of the Spectral Theorem. However, the wording in the book is confusing and may lead to misunderstandings for readers.
msumm21
Does the quote below (from a textbook) make sense to anyone? The context is that they are introducing mixed states, and the quote below is how they refer to the pure states discussed in previous chapters. Here is the quote:

"In quantum mechanics, such maximum information is contained in a wavefunction that simultaneously diagonalizes a complete set of commuting operators relevant to the system."

How does a wavefunction diagonalize an operator? I know some operators can be diagonalized with respect to certain basis sets of eigen functions, but I don't know what it means for a single wavefunction to diagonalize an operator. Maybe it should say the wave function is an eigenvector of every operator in a complete set of commuting operators?

The book is "Introductory Quantum Mechanics" forth Ed. and the author is Liboff.

Thanks

Let me ask this again in a compact and maybe better-worded form:

Does it make sense to say a (single) wave function diagonalizes a set of (many) operators? My textbook says this (see original post), but I think the author may be confused.

msumm21 said:
Does it make sense to say a (single) wave function diagonalizes a set of (many) operators? My textbook says this (see original post), but I think the author may be confused.

When the author says that a wave function "diagonalizes a set of operators" he means that the state vector corresponding to that wave function is a simultaneous eigenvector of the operators in that set.

Therefore the statement has content and can be interpreted, but it is also terribly sloppy to the point of being incorrect. This is a major problem with physics writing, and it gets worse as the topics become more advanced.

Maybe it should say the wave function is an eigenvector of every operator in a complete set of commuting operators?

That's also how I understand it. And I agree, current QM textbooks are junk.

But what does this have to do with pure vs. mixed states? Maybe we are talking about density matrices...

OK, thanks.

But what does this have to do with pure vs. mixed states? Maybe we are talking about density matrices...

The quote from the book occurred where Liboff was comparing pure states to mixed states, and it certainly reads as if he is saying a pure state is one which is an eigenvector of a complete set of commuting operators. If so, I believe again he is confused -- I believe what he is describing is a _DETERMINATE_ state with respect to these operators. The wording is so bad (and often wrong) that I really don't know what he is trying to say in a lot of these cases.

msumm21 said:
The quote from the book occurred where Liboff was comparing pure states to mixed states, and it certainly reads as if he is saying a pure state is one which is an eigenvector of a complete set of commuting operators.
I don't have that book, so I don't know the larger context. But from the small extract
you posted, I don't read it that way. Maybe if you post a larger extract I could try to say

The wording is so bad (and often wrong) that I really don't know what he is trying
to say in a lot of these cases.
If you ever write any educational material yourself, you'll find that it's surprisingly
hard to ensure that transmission from your mind to every reader's mind ends up clear.

strangerep said:
I don't have that book, so I don't know the larger context. But from the small extract
you posted, I don't read it that way. Maybe if you post a larger extract I could try to say

The section is titled "Pure and Mixed States" and the author is discussing the differences. Sorry I don't have the book with me now, but he is definitely discussing differences between pure and mixed states surrounding this quote.

strangerep said:
If you ever write any educational material yourself, you'll find that it's surprisingly
hard to ensure that transmission from your mind to every reader's mind ends up clear.

I agree that it's not easy, but I have read many math/science books and cannot recall any with near the amount of incorrect statements, math errors, nonsensical statements (like the vector "diagonalizing a set of operators" above), and misleading statements as this one. If I didn't know better I would think someone just learning QM and linear algebra wrote this book.

msumm21 said:
[...] nonsensical statements (like the vector "diagonalizing a set of operators" above), and misleading statements as this one.
That phrase should be understood in this way: the wavefunction determines a projection
operator on the Hilbert space. Think of a ket $|\psi\rangle$ and the associated
projection operator $|\psi\rangle\langle\psi|$. The crucial point is that projection operators
take you from the entire Hilbert space down to a subspace (in this case a very small subspace).
When the "set of operators" you mentioned (I'll just call them "A" collectively) are restricted
to this subspace, A acts diagonally -- meaning that A is equivalent to simple multiplication
by a scalar.

So that's what "diagonalizing an operator" means: when you restrict its domain to the
particular subspace specified by that wavefunction, the operator is equivalent to scalar
multiplication.

(This is actually a simple case of a deep mathematical result called the "Spectral Theorem".)

If I didn't know better I would think someone just learning QM and linear
algebra wrote this book.
It's the opposite, of course. People who know a great deal about a subject are not always
the clearest teachers, having forgotten what it's like to be learning the subject for the
first time.

strangerep said:
That phrase should be understood in this way: the wavefunction determines a projection
operator on the Hilbert space. Think of a ket $|\psi\rangle$ and the associated
projection operator $|\psi\rangle\langle\psi|$. The crucial point is that projection operators
take you from the entire Hilbert space down to a subspace (in this case a very small subspace).
When the "set of operators" you mentioned (I'll just call them "A" collectively) are restricted
to this subspace, A acts diagonally -- meaning that A is equivalent to simple multiplication
by a scalar.

So that's what "diagonalizing an operator" means: when you restrict its domain to the
particular subspace specified by that wavefunction, the operator is equivalent to scalar
multiplication.

The projection operator for a vector v maps any other vector w to a multiple of v (Pw=cv), so the subspace you mention is just Span{v}. Saying that each operator in A acts diagonally on Span{v} is equivalent to saying that v is an eigenvector of these operators, which is exactly what we were saying above.

strangerep said:
It's the opposite, of course. People who know a great deal about a subject are not always
the clearest teachers, having forgotten what it's like to be learning the subject for the
first time.

Well, this is just an opinion, but I disagree strongly. I've read a few things from Feynmann, Einstein, Bell, ... and it's all very clear, concise, and accurate. When I've tried to understand fellow students in the classroom I got nowhere because of the inaccuracies, confused language, ...

msumm21 said:
The projection operator for a vector v maps any other vector w to a multiple of v (Pw=cv), so the subspace you mention is just Span{v}. Saying that each operator in A acts diagonally on Span{v} is equivalent to saying that v is an eigenvector of these operators, which is exactly what we were saying above.
Yes.

[...] but I disagree strongly. I've read a few things from Feynmann, Einstein, Bell, ... and it's all very clear, concise, and accurate. When I've tried to understand fellow students in the classroom I got nowhere because of the inaccuracies, confused language, ...
Here we have an example of how what was in my mind failed to be transmitted into your mind
without distortion. I said that not every expert makes a good teacher, but you seem to have
heard "every expert is a bad teacher" -- which is not what I said. :-)

Anyway, you seem to have the answer you were looking for. (If not, post a followup question.)

strangerep said:
I said that not every expert makes a good teacher, but you seem to have
heard "every expert is a bad teacher" -- which is not what I said. :-)

Well I don't want to spend any more time on this, but for the record I said that when I see a poorly written book I assume it's written by someone that is not very familiar with the subject (poorlyExplained=>novice) and you said it's the opposite of that (poorlyExplained=>notNovice), so I said I disagree because I've observed plenty of poor explanations by novices new to a subject (not(poorlyExplained=>notNovice)).

Anyway, the main issue is that everyone agrees that the book was trying to say that the wave function was an eigenvector of all the operators.

## 1. What is a mixed state in quantum mechanics?

A mixed state in quantum mechanics refers to a state of a quantum system that cannot be described by a single pure state. It is a statistical combination of multiple pure states, each with a specific probability of occurring. This concept arises because in quantum mechanics, the state of a system is described by a wave function, which can be thought of as a probability distribution. So, a mixed state represents a combination of different probabilities of finding the system in different pure states.

## 2. What is a pure state in quantum mechanics?

A pure state in quantum mechanics refers to a state of a quantum system that can be described by a single wave function. It is a state in which the system has a definite set of properties and can be measured with certainty. In other words, the wave function of a pure state does not change with time and fully describes the quantum state of the system. This is in contrast to a mixed state, where the wave function is a combination of multiple pure states with different probabilities.

## 3. How is a mixed state different from a classical probability mixture?

A mixed state is different from a classical probability mixture in that it cannot be described by classical probability theory. In classical probability, all states are pure states, and a mixture of different states simply represents a combination of different probabilities. In quantum mechanics, however, a mixed state is a combination of different quantum states, each with its own wave function and set of probabilities. This difference arises due to the probabilistic nature of quantum mechanics and the concept of superposition, where a system can exist in multiple states simultaneously.

## 4. What is the significance of mixed vs pure states in quantum mechanics?

The distinction between mixed and pure states is crucial in understanding the behavior of quantum systems. Pure states are associated with definite properties and can be measured with certainty, while mixed states represent a combination of different probabilities. This has important implications for quantum measurements and the interpretation of quantum phenomena. Mixed states also play a crucial role in quantum information processing and quantum computing, where the manipulation and control of mixed states are essential for carrying out quantum operations.

## 5. How are mixed states and entanglement related in quantum mechanics?

Mixed states and entanglement are closely related in quantum mechanics. Entanglement is a phenomenon in which the quantum states of two or more particles become correlated, even when they are physically separated. This correlation is described by a mixed state, where the particles are in a superposition of all possible states. Entanglement is a key resource in quantum information processing, and mixed states are often used to describe entangled systems. This connection between mixed states and entanglement highlights the importance of understanding mixed states in quantum mechanics.

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