Integrals over operators.


by hello45
Tags: integrals, operators
hello45
hello45 is offline
#1
Feb24-12, 03:41 PM
P: 4
Greetings chaps,

This will probably be old hat to most of you, but I'm beginning to start Quantum mech. so that I can develop a deeper understanding of its application in Chemistry ( I'm a Chemistry undergrad -gauge my level from that if you will!)

i.) First of all, would I be right in saying that an operator acts upon a function that gives us an observable, an observable being anything that can be measured i.e. the position operator acts upon a given function, f(x), and returns a position?

ii.) The first part is just to make sure I ask the right question for the second part. In a text that I am following, namely 'Molecular Quantum Mechanics' by Atkins, a passage states (p.15):

'When we want to make contact between a calculation done using operators and the actual outcome of an experiment, we need to evaluate certain integrals. These integrals have the form:

∫f* Ω f δv where f* is the complex conjugate of f and Ω an operator. '

What does this part actually mean? For example does it mean that in order for our position operator to really give a position when acting upon a function we have to use the above integral to get anything meaningful? An extended question that stems from this is what does this mean when I have a non-complex function - how can I can take the complex conjugate of a function which doesn't involve any terms based on i?

Thanks in advance, I know it's quite a heavy question and fairly generic. As I say I'm new to this so please, be gentle.

Regards.
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M Quack
M Quack is offline
#2
Feb24-12, 04:51 PM
P: 640
Hi there,

the easier answer first:

the complex conjugate of a real function is just the function itself. Think of it as a complex function that happens to have imaginary part=0.

Operators can be tricky.

If you are lucky, then the operator applied to a function will give the function multiplied by a number. The famous example is
[itex]\hat{H}\Psi=E\Psi[/itex],
where H is the Hamiltonian operator, and E the energy. This equation holds if the state psi has a nicely defined energy.

In general this is not the case, and the operator applied to the function gives ... a mess, a sum of different functions, all multiplied by different values.
In that case you have to calculate the expectation value of the operator as you have shown. Position is such an example, as in QM wave functions are often better characterized by different quantum numbers, such as angular momentum etc.
hello45
hello45 is offline
#3
Feb24-12, 04:55 PM
P: 4
Hi M Quack,

Thanks for that!

Could you elaborate perhaps on the 'expectation value' that you mentioned? It's only that I have no idea what the integral represents, and only hazarded a guess as shown above.

Thanks.

M Quack
M Quack is offline
#4
Feb24-12, 05:08 PM
P: 640

Integrals over operators.


The expectation value is the average result you would obtain from a number of measurements of this quantity, e.g. position. The expectation value is the result of the integral.

[itex]\left<\Omega\right> = \int f^* \hat{\Omega} f \ \mathrm{d}v [/itex]

<Ω> is the expectation value of the operator Ω for the wave function f. The wave function describes the quantum state. The expectation value is the value you expect to measure.

This is the basis of all QM and occurs so frequently that it can be expressed in many different forms. There are also a few shorthand notations that are extremely common. For example

[itex]\left<\Omega\right> = \left<f\left|\hat{\Omega}\right|f\right>[/itex]

and instead of the wave function f, only the relevant quantum numbers are put inside the "ket" |...>
hello45
hello45 is offline
#5
Feb25-12, 06:09 AM
P: 4
So back to my previous question on operators; is an operator a mathematical construction which operates on a function to return position (for example with a position operator)?

In that is the case if I'm dealing with expectation values I'm not dealing with eigenfunctions? Does this then mean that the expectation value is kind of the homologue of the eigenvalue?

Thanks for all this, I've learnt more on this forum than I have reading a few of the chapters in my book!
ibazulic
ibazulic is offline
#6
Feb25-12, 06:23 AM
P: 2
Quote Quote by hello45 View Post
Greetings chaps,

This will probably be old hat to most of you, but I'm beginning to start Quantum mech. so that I can develop a deeper understanding of its application in Chemistry ( I'm a Chemistry undergrad -gauge my level from that if you will!)

i.) First of all, would I be right in saying that an operator acts upon a function that gives us an observable, an observable being anything that can be measured i.e. the position operator acts upon a given function, f(x), and returns a position?
if i'm reading this correctly, then yes. observables are indeed physical properties that can be measured. for example, if we seek total energy of some eigen-state [tex]|\psi>[/tex] then we will use a Hamiltonian operator to masure that total energy:

[tex]H|\psi>=E|\psi>[/tex]

which gives us a Schroedinger equation.

ii.) The first part is just to make sure I ask the right question for the second part. In a text that I am following, namely 'Molecular Quantum Mechanics' by Atkins, a passage states (p.15):

'When we want to make contact between a calculation done using operators and the actual outcome of an experiment, we need to evaluate certain integrals. These integrals have the form:

∫f* Ω f δv where f* is the complex conjugate of f and Ω an operator. '

What does this part actually mean? For example does it mean that in order for our position operator to really give a position when acting upon a function we have to use the above integral to get anything meaningful? An extended question that stems from this is what does this mean when I have a non-complex function - how can I can take the complex conjugate of a function which doesn't involve any terms based on i?

Thanks in advance, I know it's quite a heavy question and fairly generic. As I say I'm new to this so please, be gentle.

Regards.
the integral you used here is the expected value of the observable [tex]\Omega[/tex] in the state f. you can write that in operator form like

[tex]<f^*|\Omega|f>[/tex]

which is the operator form for the above mentioned integral.

i think my answers are correct, but i'm still only a student :-) so i hope someone else will confirm my post.
hello45
hello45 is offline
#7
Feb25-12, 10:12 AM
P: 4
Thanks ibazulic.

Could you clarify the definition of an 'eigenstate', I could quite follow the definition from the material I'm working with. Also, with Dirac notation, do we perform the operation on both the function and its complex conjugate separately, and then integrate?
ibazulic
ibazulic is offline
#8
Feb25-12, 11:32 AM
P: 2
Quote Quote by hello45 View Post
Thanks ibazulic.

Could you clarify the definition of an 'eigenstate', I could quite follow the definition from the material I'm working with. Also, with Dirac notation, do we perform the operation on both the function and its complex conjugate separately, and then integrate?
an eigenstate is a vector whose purpose is to describe the (complete) state of the system. that's because pure wave functions do not have the whole information of the system. for example, one could look at the same system using different wave functions: position, impulse, angular momentum etc. when you're working with eigenstates, you could easily just apply different operators to the eigenstate to get any information you want out: [tex]\hat{x}[/tex] for position, [tex]\hat{p}_x[/tex] for impulse, [tex]\hat{H}[/tex] for energy, [tex]\hat{L}[/tex] for angular momentum etc.

the best thing is working with operators is much more easier than working with integrals in the form of wave functions. one only needs to know (to define) operators of "destruction" and creation to in fact calculate everything you need.

as for the 2nd question: no. you act with your operator on the ket vector and then use the rule of scalar product to multiply bra and ket vector.

EDIT: it is possible to use the operator on the bra vector as well, but the possibility of making a mistake is much higher. so it's better to just act on the ket vector and then use the scalar product to obtain the result.


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