Understanding Scalar Quantity in Basic Bra-Ket Arithmetic

[Vitani11]

Say I have a vector product |x+a⟩⟨x| and I multiplied it by a ket vector |x'⟩. Can I pull the |x'⟩ into the ket vector |x+a⟩? also could you split up the ket vector |x+a⟩ into two ket vectors added together?

 

[Orodruin]

Say I have a vector product |x+a⟩⟨x| and I multiplied it by a ket vector |x'⟩. Can I pull the |x'⟩ into the ket vector |x+a⟩? also could you split up the ket vector |x+a⟩ into two ket vectors added together?

No and no.

 

[Vitani11]

Okay thank you

 

[mike1000]

Is every complete bra-ket result a scaler, no matter how complicated?

By every bra-ket I mean every operation that begins < and ends with >.

Also, how do we interpret something like this? What is that equation saying?

<Φ|A|Ψ> or <Ψ₂|A|Ψ₁>

 

[Karolus]

the probability amplitude of passing from the state $A| \Phi \rangle$ to $|\Psi\rangle$

 

[blue_leaf77]

the probability amplitude of passing from the state $A| \Phi \rangle$ to $|\Psi\rangle$

As some operators are not Hermitian, it's actually more quantum-mechanically intuitive to say instead "the probability amplitude of passing from the state $A| \Psi \rangle$ to $|\Phi\rangle$".

 

[Karolus]

for what I understand, quantum operators are hermitian, if an operator is not Hermitian, it is not associated to an observable. So when we speak of states bra or ket, we make the implicit assumption that operators are hermitian

 

[Orodruin]

for what I understand, quantum operators are hermitian, if an operator is not Hermitian, it is not associated to an observable. So when we speak of states bra or ket, we make the implicit assumption that operators are hermitian

What makes you think A is an observable? There are many operators in QM that are not Hermitian. The time propagation operator $S = e^{-iHt}$ comes to mind, or any other unitary operator.

 

[Karolus]

What makes you think A is an observable? There are many operators in QM that are not Hermitian. The time propagation operator $S = e^{-iHt}$ comes to mind, or any other unitary operator.

operator S is an observable ?

 

[hilbert2]

operator S is an observable ?

No it's not, because it doesn't have real eigenvalues except if for some value of t you have $e^{-iHt} = 1$ or $e^{-iHt}=-1$ by coincident.

 

[Karolus]

you have to read carefully .. I did not say that any operator in QM is hermitian, I said that the operator of an observable is hermitian. why are you unnecessarily complicating to a question about the elementary bases on the bra and ket. Go ahead and answer the question...

 

[Orodruin]

operator S is an observable ?

Obviously not. Yet it is a perfectly viable operator that is very important in QM. Your statement:

So when we speak of states bra or ket, we make the implicit assumption that operators are hermitian

is therefore wrong. For example, $\langle \phi \lvert S \rvert \psi \rangle$ would represent the probability amplitude of the state $\lvert \psi \rangle$ evolving into the state $\lvert \phi \rangle$, not the other way around. There is no implicit assumption that $A$ must be a Hermitian operator in an expression such as $\langle \phi \lvert A \rvert \psi \rangle$.

 

[Karolus]

Obviously not. Yet it is a perfectly viable operator that is very important in QM. Your statement:

is therefore wrong. For example, $\langle \phi \lvert S \rvert \psi \rangle$ would represent the probability amplitude of the state $\lvert \psi \rangle$ evolving into the state $\lvert \phi \rangle$,.

and that's what I wrote in post # 5
ok right, not in all generally, all operators are hermitian, but only associates to observable,

 

[Orodruin]

You said:

the probability amplitude of passing from the state $A| \Phi \rangle$ to $|\Psi\rangle$

blue_leaf77 pointed out that:

As some operators are not Hermitian, it's actually more quantum-mechanically intuitive to say instead "the probability amplitude of passing from the state $A| \Psi \rangle$ to $|\Phi\rangle$".

to which you replied

for what I understand, quantum operators are hermitian, if an operator is not Hermitian, it is not associated to an observable. So when we speak of states bra or ket, we make the implicit assumption that operators are hermitian

The last assertion in this post is wrong. We do not make the implicit assumption that operators are Hermitian when speaking of bras and kets. It is unclear to me why you start involving observables at all, this was not part of the original question and the first statement that you made generally needs the correction made by blue_leaf77. The operator $S$ is not an observable, but this does not matter, the bra-ket notation is perfectly capable of handling non-Hermitian operators and so blue_leaf77's correction of your post is perfectly warranted.

 

[Karolus]

It is unclear to me why you start involving observables .

because the concept of observable is really essential in quantum mechanics, and the formalism of the bra and ket, was introduced not on a whim, or because Dirac had nothing better to do but to clarify and mathematically formalize the concept of observable , among other things. I strongly advise you to read carefully "The principles of quantum mechanics" by Dirac.

 

[Orodruin]

because the concept of observable is really essential in quantum mechanics, and the formalism of the bra and ket, was introduced not on a whim, or because Dirac had nothing better to do but to clarify and mathematically formalize the concept of observable , among other things. I strongly advise you to read carefully "The principles of quantum mechanics" by Dirac.

Regardless of what Dirac intended, the A does not need to describe an observable and the braket notation is perfectly well suited to handle any linear operator on the Hilbert space. It is also done regularly in QM.

 

[mike1000]

Regardless of what Dirac intended, the A does not need to describe an observable and the braket notation is perfectly well suited to handle any linear operator on the Hilbert space. It is also done regularly in QM.

Thanks for the replies.

I did ask two questions in my post. I would really like to know the answer to the first question. Maybe it is so simple you think I should already know the answer. I suspect it is true.

Is every complete bra-ket result a scaler, no matter how complicated?

By every bra-ket I mean every operation that begins < and ends with >.

Basically, what I think I am asking is if every <...> is a statement of a probability result of some kind.

Also, based on the replies so far, I am seeing a similarity between the <| |> notations and conditional probability notation. In conditional probability notation, P(A|B) is interpreted to mean "the probability of A given that B has already occurred". Would I be wrong if I interpret the braket notation of, <Φ|A|Ψ> for examples as "the probability amplitude of being in state Φ given that the system is already in the state AΨ? (I was going to say observing instead of being but based on the previous answers I know that would be wrong)

 

[blue_leaf77]

Is every complete bra-ket result a scaler, no matter how complicated?

By every bra-ket I mean every operation that begins < and ends with >.

Depending on how we interpret the term scalar, yes it is always a "scalar". But sometimes there are some cases where the use of the word scalar can be ambiguous, consider the expectation value of position operator $\langle \psi | \mathbf r |\psi \rangle$. The operator $\mathbf r$ is a rank-1 tensor (vector) operator and thus its expectation value.

 

[mike1000]

Depending on how we interpret the term scalar, yes it is always a "scalar". But sometimes there are some cases where the use of the word scalar can be ambiguous, consider the expectation value of position operator $\langle \psi | \mathbf r |\psi \rangle$. The operator $\mathbf r$ is a rank-1 tensor (vector) operator and thus its expectation value.

Thanks.

By scaler I mean the quantity can be treated algebraically as a scaler.