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Vitani11
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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 said: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?
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 said:the probability amplitude of passing from the state ##A| \Phi \rangle## to ## |\Psi\rangle##
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 said: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
operator S is an observable ?Orodruin said: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 said:operator S is an observable ?
Obviously not. Yet it is a perfectly viable operator that is very important in QM. Your statement:Karolus said:operator S is an observable ?
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 said:So when we speak of states bra or ket, we make the implicit assumption that operators are hermitian
and that's what I wrote in post # 5Orodruin said: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##,.
blue_leaf77 pointed out that:Karolus said:the probability amplitude of passing from the state ##A| \Phi \rangle## to ## |\Psi\rangle##
to which you repliedblue_leaf77 said: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##".
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 said: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 said:It is unclear to me why you start involving observables .
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.Karolus said: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 said: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.
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 said: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 >.
blue_leaf77 said: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.
Basic bra-ket arithmetic is used in quantum mechanics to represent and manipulate quantum states and operators. It allows for the calculation of probabilities and expectation values for quantum measurements.
The fundamental rules include linearity, orthogonality, and normalization. Linearity means that the arithmetic operations can be applied to the entire state or operator, rather than just individual components. Orthogonality means that states that are perpendicular to each other have a dot product of zero. Normalization means that the length of a state vector is equal to 1.
In order to add or subtract bra-ket expressions, you must first ensure that they have the same bra or ket components. Then, you can simply add or subtract the coefficients of the matching components while keeping the non-matching components the same.
A bra represents a row vector, while a ket represents a column vector. Bra vectors are denoted by ⟨ ⟩, while ket vectors are denoted by ⟩ ⟨. In other words, a bra is the conjugate transpose of a ket vector.
Yes, you can multiply two bra-ket expressions together by applying the rules of bra-ket arithmetic. The result will be a scalar value. However, it is important to note that the order of multiplication matters, as the bra-ket notation is not commutative.