Some problems with... mathematical operators

[math.op.]

Hi!

 

I'm Patrick, and I'm new to this forum.

However, I'm studying Applied Natural Science in germany and at the moment I'm trying to recapture my knowledge about quantum physics and quantum theory. Well, since my QT lecture was in 3rd semester I can barely remember anything and I did not do a great job in understanding the mathematics behind the theory.

Right now I'm trying to work through the lessons from lectures of the University of Oxford (I discovered they are offered in iTunes U for free) about QT. The lecturer does a great job in teaching the mathematic basics, but sometimes its not enough (for me...).

At the moment I'm stuck at the following problem:

 

We just discovered, that any operator, which is defined like the Hamiltonian operator: Sum over k qk |qk><qk| (is there any way to insert a latex code or something, because if I always insert formula like this I will spend a lot of time explaining what I mean ^^). We just covered that, if squeezed between a bra and a ket the result is the expectation value. And further we said that, if that Operator Qk works on the bra <qi| it gives the Eigenvalues times the Eigenket of the Operator Q. Thats what I understood, at least...

 

However, later we worked with this Operator Qk and "bra'd through" from the left with a bra (of course). And now my problem: How does one calculate or simplify the following:

<j|Q|i>, where Q is the Operator described above and j and i are normal bras and kets. If I substitute Q I know what happens to <qk|i> (Kroneka delta and so on), but what happens if a ket meets a bra? so |qk><qi|?

 

I hope someone can help me and give me more information about operators or can recommend a book where the mathematical basics are described (best for dummies ^^) well...

 

I'm looking foreword to many, hopefully helpful answers.

 

Patrick

[timo]

Hi! I'm Patrick, and I'm new to this forum. [...] Is there any way to insert a latex code or something [...][?]

Hi Patrick, welcome to sfn. There is a code for latex which is [ math] latex here [ /math] (as you may guess: without the space in the tags).

Maybe you can use it to rewrite your question in the form of proper mathematical expressions, because this way it is hard to follow what you are trying to say/ask. For this, note that a proper definition of the terms used is an integral part of mathematical expressions.

[DrRocket]

....However, later we worked with this Operator Qk and "bra'd through" from the left with a bra (of course). And now my problem: How does one calculate or simplify the following:

<j|Q|i>, where Q is the Operator described above and j and i are normal bras and kets.

 

This is just <j |Qi> or in terms of inner products b<j, Qi>.

 

And there is a way to insert latex code: [math]\langle j |Q| i \rangle = \langle j, Q i \rangle [/math] (if you reply the codes are revealed iin the reply box).

 

I am a bit unsure of what it is that is troubling you. Perhaps this Wiki article can help.

http://en.wikipedia....ra-ket_notation

 

It relates the Dirac notation to more conventional notation for operators on a Hilbert space, and to the relationship between a Hilbert space and its dual, reflected by bra and ket vectors.

[math.op.]

Okay, at first I will try to read the important parts of the wikipedia article (thanks for that). However, just to be clear, here the Latex notation:

Q is defined by

[math]Q=\sum\limits_{k} q_k |q_k \rangle \langle q_k|[/math]

 

If squeezed between [math] \langle i|Q| j \rangle [/math] and I substitute Q I get:

[math]= q_k \langle i |q_k \rangle \langle q_k | j\rangle[/math]

 

is this correct? And now I start from the right, saying [math] q_k [/math] and [math] i [/math] only get values different from 0 if k = i, which leads to [math]q_i[/math]. And now?

Edited November 7, 2011 by math.op.

[timo]

Well, two comments here:

1) I didn't say "a proper definition of the terms used is an integral part of mathematical expressions" just for fun. Since I am familiar with standard QM lectures I have a good idea what [math]| q_k \rangle[/math] and [math]q_k[/math] are supposed to be, namely an orthonormal basis of eigenstates and their eigenvalues, respectively. But I think saying this should have been part of your post, not my reply.

 

2) [math] \langle i|Q| j \rangle = \langle i| \left( \sum_k q_k | q_k \rangle \langle q_k | \right) | j \rangle = \sum_k q_k \langle i | q_k \rangle \langle q_k | j \rangle[/math]. Or in other words: No your expression is not correct, you are missing a summation. Similarly, it is not correct to say that [math] \langle i | q_k \rangle [/math] is either one or zero, and especially "k=i" makes little to no sense (it would make sense if the expression had been [math]\langle q_i | q_k \rangle[/math], but it apparently isn't).

Edited November 7, 2011 by timo

[math.op.]

Okay, that's what I wanted to know. I already thought that the sum is missing, because later we said that this construction yields in a matrix [math]Q_{ij}[/math], with two sums this is only logical. Well sorry for my silly question, but in 6 semesters of studying there was no lecturer who could explain to me (us) the mathematical basics. And as I read ur answer I recognized that it was complete nonsense ^^

Well I'll try to figure out the meaning of that equation

 

Okay thanks for now.

[Roentgenquantum]

Hi!

Since I forgot my PW... and I was too lazy to get a new one, I simply connected with my FB account, and here I am, posting some more questions...

I just reviewed linear operators. And found out that they can have an adjoint.

Well. I asked myself a question while looking at this equation:

[math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math]

,where A is a linear operator, and [math]\phi[/math] and [math]\Psi[/math] are wave functions in the hilbert space. While trying to reproduce the steps I wondered if one is allowed to move bras and kets without forming there conjegate-complex.

Meaning:

[math]| \Psi \rangle \langle \Phi| = \langle \Phi| \Psi \rangle[/math]

 

Thanks for answering!

 

EDIT

 

I just realized that by definition [math] ( \Psi | \phi )= (\phi | \Psi )^* [/math].

Can this be applied to the bra-ket notation? And what happens if an operator is squeezed in between...?

Edited December 3, 2011 by Roentgenquantum

[Mystery111]

I will try and answer your questions... I little buisy at the mo, will be back on the comp soon.

 

Hi!

Since I forgot my PW... and I was too lazy to get a new one, I simply connected with my FB account, and here I am, posting some more questions...

I just reviewed linear operators. And found out that they can have an adjoint.

Well. I asked myself a question while looking at this equation:

[math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math]

 

,where A is a linear operator, and [math]\phi[/math] and [math]\Psi[/math] are wave functions in the hilbert space. While trying to reproduce the steps I wondered if one is allowed to move bras and kets without forming there conjegate-complex.

Meaning:

[math]| \Psi \rangle \langle \Phi| = \langle \Phi| \Psi \rangle[/math]

 

Thanks for answering!

 

EDIT

 

I just realized that by definition [math] ( \Psi | \phi )= (\phi | \Psi )^* [/math].

Can this be applied to the bra-ket notation? And what happens if an operator is squeezed in between...?

 

Right... well, when you come across something like this [math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math] the fact we can see [math]A^\dagger = A[/math] actually implies something important; if [math]A[/math] is a matrix, then this implies our matrix is hermitian, meaning all the diagonal elements in the matrix are real. Real things are associated to observables in quantum mechanics. This will take you into things like calculating matrices and the hermitian conjugate. You simply transponse your matrix but then complex conjugate it. I could write up some matrices, but I really can't be bothered unless you ask me to. Anyway... coming across something like [math]|\psi><\phi|[/math] is called an outer product, obviously the opposite name-meaning of the inner product. You may come across something like

 

[math]|\psi><\psi|\phi>[/math] this means that the eigenvector has been chosen, then this is an orthogonal vector [math]\phi[/math].

I would have written more, but am in a rush again!!! lol

Edited December 3, 2011 by Mystery111

[Roentgenquantum]

I will try and answer your questions... I little buisy at the mo, will be back on the comp soon.

 

 

 

Right... well, when you come across something like this [math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math] the fact we can see [math]A^\dagger = A[/math] actually implies something important; if [math]A[/math] is a matrix, then this implies our matrix is hermitian, meaning all the diagonal elements in the matrix are real. Real things are associated to observables in quantum mechanics. This will take you into things like calculating matrices and the hermitian conjugate. You simply transponse your matrix but then complex conjugate it. I could write up some matrices, but I really can't be bothered unless you ask me to. Anyway... coming across something like [math]|\psi><\phi|[/math] is called an outer product, obviously the opposite name-meaning of the inner product. You may come across something like

 

[math]|\psi><\psi|\phi>[/math] this means that the eigenvector has been chosen, then this is an orthogonal vector [math]\phi[/math].

I would have written more, but am in a rush again!!! lol

 

Hi and thanks for the quick answer.

My whole question is: How to get from [math]\langle \phi|A| \Psi \rangle ^* [/math] to [math] \langle \Psi|A^\dagger| \phi \rangle[/math]. Is this only per definition that way (I already understood that this is the case for matrix elements A, because this is the way for creating the adjoint operator - transpose and conjugate-complex.). Because I tried to review the single steps to get from the left side to the right side.

 

 

Or maybe my question is:

If I transpose the scalar product, do I just change the bra and the ket?

Sorry if this sounds confused, but in fact, I am o.O

 

 

Thanks in advance, Patrick

Edited December 3, 2011 by Roentgenquantum

[Mystery111]

Hi and thanks for the quick answer.

My whole question is: How to get from [math]\langle \phi|A| \Psi \rangle ^* [/math] to [math] \langle \Psi|A^\dagger| \phi \rangle[/math]. Is this only per definition that way (I already understood that this is the case for matrix elements A, because this is the way for creating the adjoint operator - transpose and conjugate-complex.). Because I tried to review the single steps to get from the left side to the right side.

 

 

Or maybe my question is:

If I transpose the scalar product, do I just change the bra and the ket?

Sorry if this sounds confused, but in fact, I am o.O

 

 

Thanks in advance, Patrick

 

You should check what I say in this following post. Prove to yourself for instance that the inner product of [math]<b|[/math] and [math]|a>[/math] is the complex conjugate of [math]<a|b>[/math]. Here is one that is less obvious. Take any matrix, hermitian or not hermitian as [math]<b|M|a>[/math]

 

Take M and multiply it onto the ket will give you a new vector. Then you take the inner product of b. This is related to

 

[math]<a| M^{\dagger}|b>^{*}[/math]

 

where the <a| is acting like a complex conjugate, in fact, this has been complex conjugated where all rows and columns have been interchanged. Incidently, M on (a) will give you a vector, but M and (a) after this onto (b) will give you a number. It won't give you a vector in this case, it's just a number.

 

So you get from [math]<b|M|a>[/math] to the expression [math]<a| M^{\dagger}|b>^{*}[/math] by complex conjugating it. In fact, if it is hermitian you can now state it as

 

[math]<b|M|a> = <a| M|b>^{*}[/math]

 

where we have just erased the conjugation dagger sign. This means it is equal to its own hermitian conjugate. That is just the definition of Hermitian. So the <b| and |a> matrix element of M is the same as the <a| and |b> matrix element of M.

 

Don't forget also you can have the case where a=b so

 

<a|M|a> = <a|M|a>*

 

Is real. It also has a special name. It is called the expectation value.

Edited December 3, 2011 by Mystery111

[Roentgenquantum]

Okay thank u

 

This helped a lot I'm still trying to get familiar to the whole notation and concept of working with operators

Thanks a lot

Hopefully this will be my last dumb question

Edited December 3, 2011 by Roentgenquantum

[Mystery111]

Don't be silly. No question is a dumb question if you don't know the answer