Basic question about Bra-Ket notation

[nomadreid]

This is so basic as to be embarrassing, but I haven't figured out my misunderstanding of some basic notation.

[1] If v and w are two vectors in a Hilbert space, then <v|w> is interpreted as the probability amplitude of w collapsing into v.

[2] However, if P is a projection, then <v|Pv> =<v|P|v> is the probability amplitude of v collapsing to the projection of v, that is, Pv.

These two seem to be contradicting each other. I suspect it is [2] that I have wrong, but I do not know why.

Thanks for the corrections.

 

[anuttarasammyak]

I am not sure of your projection operator P. On what P projects ?

 

[PeterDonis]

These two seem to be contradicting each other.

Why do you think so?

 

[gentzen]

Why do you think so?

Maybe because the first one looks like an amplitude, and the second like a probability? The second one looks like a probability to me, because
<v|Pv>=<vP|Pv>=||Pv||^2

The first one looks like an amplitude to me, because I image that v or w would be a canonical basis vector, and then this just tells me the corresponding amplitude.

 

[PeterDonis]

The second one looks like a probability to me

The first one looks like an amplitude to me

They're both inner products. An inner product can "look like" either an amplitude or a probability, depending on the context. Note that the OP uses the term "probability amplitude" to describe both. This makes me think it is unlikely that your conjecture is correct. But it would be best for the OP to answer the question.

 

[nomadreid]

Thank you for the replies. My responses

I am not sure of your projection operator P. On what P projects ?

Onto one of the eigenstates/eigenvectors, that is, one of the axes of my basis.

Why do you think so?

Er, I am not sure which part of my statement you are referring to.

Maybe because the first one looks like an amplitude, and the second like a probability? The second one looks like a probability to me, because
<v|Pv>=<vP|Pv>=||Pv||^2

The first one looks like an amplitude to me, because I image that v or w would be a canonical basis vector, and then this just tells me the corresponding amplitude.

Although I understand why <v|Pv> = <vP|v>,I'm afraid I don't understand why you write that <v|Pv> = <vP|Pv> .

They're both inner products. An inner product can "look like" either an amplitude or a probability, depending on the context. Note that the OP uses the term "probability amplitude" to describe both. This makes me think it is unlikely that your conjecture is correct. But it would be best for the OP to answer the question.

The remark that an inner product can be a probability as well as a probability amplitude is interesting. I was starting from the statement

"In quantum mechanics the expression ⟨φ|ψ⟩ is typically interpreted as the probability amplitude for the state ψ to collapse into the state φ. Mathematically, this means the coefficient for the projection of ψ onto φ. It is also described as the projection of state ψ onto state φ. "

in https://en.wikipedia.org/wiki/Bra–ket_notation#Overlap_of_states

Then, using "v" in place of φ in the above, and then letting Pv=ψ from the above, I get that <v|Pv> is interpreted as the probability amplitude for Pv to collapse into the state v. On the other hand, I thought that it was the state v being measured that collapsed into the projections of v onto the axes (eigenvectors). That is, the reverse. This is what I meant by contradictory usage.

Also from the above Wikipedia statement, I took the inner product to be a probability amplitude rather than a probability, but then on reflection, as <v|v> is a probability, so now I am even more confused.

Thanks for your patience, and I will be very grateful for further clarifications .

 

[PeterDonis]

I am not sure which part of my statement you are referring to.

Um, the part that I explicitly quoted? In any case, you respond to it further on in your post:

I thought that it was the state v being measured that collapsed into the projections of v onto the axes (eigenvectors). That is, the reverse. This is what I meant by contradictory usage.

Inner products are conjugate symmetric, meaning that $<b | a>$ is the complex conjugate of $<a | b>$. Since converting a probability amplitude into a probability means taking the squared modulus, that means $<b | a>$ and $<a | b>$ will give you the same probability.

Also, in an expression like $\bra{v} P \ket{v}$, the $P$ can be "attached" to either the bra or the ket; the two will give the same result.

So there is no contradiction.

The remark that an inner product can be a probability as well as a probability amplitude is interesting.

I said "look like", not "can be". And even that is really only the case when there is a projection operator $P$ sandwiched into the inner product.

I don't understand why you write that <v|Pv> = <vP|Pv>

Since $P$ is a projection operator, $P^2 = P$, so you can write $\bra{v} P \ket{v} = \bra{v} P^2 \ket{v} = \bra{vP} \ket{Pv}$.

 

[nomadreid]

Thanks a million, Peter Donis. Great explanations! Cleared the fog!

Um, the part that I explicitly quoted?

Oops. Scrolling too far hid that part while I was answering. Sorry about that, and thanks for your patience as well as the extremely helpful answers.

 

[Orodruin]

$<b | a>$

$\bra{v} P \ket{v}$,

The greatest of sins, knowing that $<$ and $>$ are relational operators and still using them instead of $\langle$ and $\rangle$

I am surprised this is not yet a bannable offence on PF …

 

[Nugatory]

I am surprised this is not yet a bannable offence on PF

It’s on my to-do list, but first we have to deal with the far more serious problem of photons in the relativity forum.

 

[H_A_Landman]

This is so basic as to be embarrassing, but I haven't figured out my misunderstanding of some basic notation.

[1] If v and w are two vectors in a Hilbert space, then <v|w> is interpreted as the probability amplitude of w collapsing into v.

[2] However, if P is a projection, then <v|Pv> =<v|P|v> is the probability amplitude of v collapsing to the projection of v, that is, Pv.

These two seem to be contradicting each other. I suspect it is [2] that I have wrong, but I do not know why.

Thanks for the corrections.

Let $e$ be a basis vector. Then $\langle e|v\rangle$ is the coefficient of $|e\rangle$ in the basis representation of $v$. One can view $|e\rangle\langle e|$ as the projection operator onto $e$, that is $|e\rangle\langle e|v\rangle$ is the $e$ component of $v$ in the basis. If the orthonormal basis vectors are $e_1,\ e_2,\ \ldots$ then $|v\rangle = |e₁\rangle\langle e₁|v\rangle + |e₂\rangle\langle e₂|v\rangle + \ldots$

Note that there is no collapse here. $\langle v|w\rangle$ is just a complex number (probability amplitude) that represents how much $v$ there is in $w$. If $v$ and $w$ are orthogonal, then it's 0. If $v = w$, then it's 1.

Let $P$ be one of the basis projections, that is $P = |e\rangle\langle e|$. Then $\langle v|P|v\rangle = \langle v|e\rangle\langle e|v\rangle$, and since $\langle v|e\rangle = \langle e|v\rangle^*$, $\langle v|P|v\rangle$ is a real number, the probability that if you measure/collapse $v$ into the basis, you will get $e$.

 

[nomadreid]

Thanks, H_A_Landman. This formulation is helping me to better understand why $\langle$ v|w $\rangle$ is a probability amplitude while $\langle$ v|v $\rangle$ is a probability.

 

[Orodruin]

Thanks, H_A_Landman. This formulation is helping me to better understand why $\langle$ v|w $\rangle$ is a probability amplitude while $\langle$ v|v $\rangle$ is a probability.

If $\ket v$ is a properly normalized state, then $\braket{v|v} = 1$ and the probability amplitude is equal to the probability.

 

[nomadreid]

Thanks, Ododruin. That point, in H_A_Landman's post, is good to emphasize.

 

[vanhees71]

This is so basic as to be embarrassing, but I haven't figured out my misunderstanding of some basic notation.

[1] If v and w are two vectors in a Hilbert space, then <v|w> is interpreted as the probability amplitude of w collapsing into v.

[2] However, if P is a projection, then <v|Pv> =<v|P|v> is the probability amplitude of v collapsing to the projection of v, that is, Pv.

These two seem to be contradicting each other. I suspect it is [2] that I have wrong, but I do not know why.

Thanks for the corrections.

First of all one must be very precise in formulating Born's rule. A pure state can be represented by a normalized vector in Hilbert space, $|\psi \rangle$. Then you have a self-adjoint operator $\hat{A}$ of some observable with real eigenvalues and a complete set of orthonormal eigenvectors $|u_a \rangle$. Then the probability to get the eigenvalue $a$ as a result of measuring this observable is $P(a)=\langle u_a|\psi \rangle|^2$. Here I simplified a bit, because I tacitly assumed that the eigenvalue $a$ is not degenerate, i.e., that there's only one linearly independent eigenvector with this eigenvalue.

The important point is that you have to distinguish the state ket, representing the state, and the eigenvectors of the operator that represents the observable measured. Only then the whole mathematical scheme of QT is consistent, particularly when it comes to the time evolution of state kets and operators representing observables and their eigenvectors in different "pictures of time evolution".

Further a pure state does not uniquely determine the state ket, because any state ket $|\psi' \rangle=\exp(\mathrm{i} \phi) |\psi \rangle$ describes the same state ($\phi \in \mathbb{R}$). The unique representant of a pure state thus is rather the projection operator, $\hat{\rho}=|\psi \rangle \langle \psi|$, the statistical operator representing this pure state. The most general state are then the socalled mixed states, represented by an arbitrary self-adjoint positive semidefinite operator $\hat{\rho}$ with $\mathrm{Tr} \hat{\rho}=1$, the statistical operator of the system. In this notation the probability (not the probability amplitude!) for getting $a$ as a measurement result when measuring $A$ is $P(a)=\langle u_a|\hat{\rho}|u_a \rangle$. For a pure state you get again $P(a)=\langle u_a |\psi \rangle \langle \psi|u_a \rangle=|\langle{u_a}|\psi \rangle|^2$, i.e., everything is indeed fully consistent.

 

[vanhees71]

It’s on my to-do list, but first we have to deal with the far more serious problem of photons in the relativity forum.

Photons belong to the relativity forum as well as to the QT forum. The more severe problem is that photons are the most misrepresented subjects by popular-science authors one can think of. That's why many people believe photons were in any sense like "massless classical particles", but that's not "bannable" but simply has to be explained, but that's another story.

 

[vanhees71]

If $\ket v$ is a properly normalized state, then $\braket{v|v} = 1$ and the probability amplitude is equal to the probability.

Once more, the "probability amplitudes", better known as "wave functions", in QT are always a scalar product of a state ket, representing a pure state and an eigenvector of the self-adjoint operator, representing the observable measured. Both have to be normalized, of course.

 

[nomadreid]

Many thanks, vanhees71. You are right, I was apparently a bit sloppy in my formulation, and from your summary I spot some mistakes that I have been making (not just in this post). I'll be working through that with the aid of your posts. Super!