Undergrad Griffiths' Intro to Quantum Mechanics - primed variables

[Sparky_]

Hello,I have been going through Griffiths’ Intro To Quantum Mechanics in an attempt to self-teach myself some quantum mechanics.

I am currently in section 3.4 (Generalized Statistical Interpretation)

I am seeing a “step” occur more than once. When I first came across it, while I didn’t understand “why”, I ignored it and moved on. I am now seeing it again. The deal is an inner product with a “prime variable” and a “non-primed” variable.
I believe it was first introduced in example 3.2 (second edition) page 103 where the eigenfunctions and eigenvalues of the momentum operator are calculated.

This is where the Dirac Delta Function first shows up.

Inner product of < fp’ | fp > = Dirac Delta

Within this example, I just went with it, not fully understanding why the need for p’

Now in section 3.4 pgs 106-109 Griffiths (and some others I found googling) just take it as an obvious step to have a prime variable and non-primed variable within the inner product
Page 107 – inner product of psi (wave function) with itself = 1

< psi | psi > =1

Next step show the linear combination formula with the index variable n. The left side of the inner product now has n’ and right side has n. (Equation 3.48)
Down a couple of lines, the expectation of Q: inner product of psi and Qpsi – again a n’ and n show up. (Equation 3.50)Bottom line, as you can see, I do not understand what’s going on and why it should be obvious to the reader to have a 2nd variable n’ or p’ ….Help?

Thanks

Sparky_

 

[PeroK]

Hello,I have been going through Griffiths’ Intro To Quantum Mechanics in an attempt to self-teach myself some quantum mechanics.

I am currently in section 3.4 (Generalized Statistical Interpretation)

I am seeing a “step” occur more than once. When I first came across it, while I didn’t understand “why”, I ignored it and moved on. I am now seeing it again. The deal is an inner product with a “prime variable” and a “non-primed” variable.
I believe it was first introduced in example 3.2 (second edition) page 103 where the eigenfunctions and eigenvalues of the momentum operator are calculated.

This is where the Dirac Delta Function first shows up.

Inner product of < fp’ | fp > = Dirac Delta

Within this example, I just went with it, not fully understanding why the need for p’

Now in section 3.4 pgs 106-109 Griffiths (and some others I found googling) just take it as an obvious step to have a prime variable and non-primed variable within the inner product
Page 107 – inner product of psi (wave function) with itself = 1

< psi | psi > =1

Next step show the linear combination formula with the index variable n. The left side of the inner product now has n’ and right side has n. (Equation 3.48)
Down a couple of lines, the expectation of Q: inner product of psi and Qpsi – again a n’ and n show up. (Equation 3.50)Bottom line, as you can see, I do not understand what’s going on and why it should be obvious to the reader to have a 2nd variable n’ or p’ ….Help?

Thanks

Sparky_

It's a bit of a limitation only to be allowed one variable! For example, in the discrete case, you might have energy levels $E_n$ and $E_m$. And, in general, you might have something of the form $\Sigma_n \Sigma_m$.

With continuous variables, it's common to use $x, x'$ (although, personally I'd prefer $x_1, x_2$), or $p, p'$. Then you have things like:

$\int dx dx'$ or $\int dp dp'$

 

[Sparky_]

I understand that p and p' are different variables and are continuous instead of discrete.

I don't understand "why" the need to introduce a 2nd variable. p' or x'

I'm missing a "big picture" thing here.

One can do inner products without introducing a 2nd variable.

I have reviewed Griffiths and I do not see him stating why the introduction of a 2nd variable within an inner product. He just seems to take it for granted and does it.

-Sparky_

 

[PeroK]

I understand that p and p' are different variables and are continuous instead of discrete.

I don't understand "why" the need to introduce a 2nd variable. p' or x'

I'm missing a "big picture" thing here.

One can do inner products without introducing a 2nd variable.

I have reviewed Griffiths and I do not see him stating why the introduction of a 2nd variable within an inner product. He just seems to take it for granted and does it.

-Sparky_

You can't do inner products with only one variable. You need two, as in $\langle u, v \rangle$ or $\langle u_1, u_2 \rangle$ or $\langle u, u' \rangle$.

If you limit yourself to one variable, you can only ever do $\langle u, u \rangle$.

To turn the question round. Using any notation you like, fill in the gaps:

Let $p$ and ? be eigenvalues of the momentum operator.

It can't be "$p$ and $p$".

 

[PeroK]

PS One reason for using the prime is (taking an example of spatial coordinates):

You don't want to use $x, y$, because when you move to 3D, you'll need $x, y, z$ for the spatial components of a single position variable.

You don't want to use $x_1, x_2$, because when you consider more than one particle you might want $x_1$ to be the x-coordinate of the first particle and $x_2$ to be the x-coordinate of the second particle etc.

So, perhaps $x, x', x'', x'''$ are as good as any.

 

[Sparky_]

I am starting to get there. I did as an exercise review the inner product and I agree that the (simple) examples are with different values and variables.

I now see that my question / confusion is regarding the inner product of functions.

What I am seeing is : < f | f > or < psi | psi >

so my mind erroneously says it's the same "thing "f" or "psi" - (I know f and psi are not variables.)

I see within the text: "taking the inner product with itself" regarding functions

So since you say you can't take an inner product with the one variable (need two),

am i correct ... you can do an inner product with one function but the dependent variable must be different ?

< f(x) | f(x') >

??
Sparky_

 

[PeroK]

am i correct ... you can do an inner product with one function but the dependent variable must be different ?

< f(x) | f(x') >

??
Sparky_

There are a few things here. First, functions generally don't need a variable to be introduced. You can talk about functions $f, g$ etc. In physics, however, authors tend to put the variable in and say $f(x), g(x)$.

As you know, you can define an inner product on functions:

$\langle f | g \rangle = \int f^*(x)g(x)dx$

Where $x$ has been introduced as a dummy variable to satisfy the notational requirement of the integral - if that makes sense.

Note that functions are variables too. In fact, the whole idea of an "Operator" is that it acts on sets of functions to give other functions. In that sense, the functions act just like vectors. In particular, you have a momentum eigenfunction for every real number. Griffiths notation for that is:

$f_p$ is the eigenfunction associated with eigenvalue $p$. Here $p$ can be any real number.

Now, what if you want to take the inner product of two momentum eigenfunctions? Griffiths notation for this is:

$f_{p'}$ is the eigenfunction associated with eigenvalue $p'$ and we have:

$\langle f_{p'}| f_{p} \rangle = \int f_{p'}^*(x) f_p(x) dx = \delta(p-p')$

Now, I think using $p_1$ and $p_2$ instead of $p, p'$ might be simpler to understand, but the prime notation is quite common.

 

[Sparky_]

thank you for working with me on this. I do not fully understand but I am going back to review with your comments now ... thanks again!

I may have additional questions if the light bulb does not come on later
Sparky_

 

[Sparky_]

There are a few things here. First, functions generally don't need a variable to be introduced. You can talk about functions $f, g$ etc. In physics, however, authors tend to put the variable in and say $f(x), g(x)$.

As you know, you can define an inner product on functions:

$\langle f | g \rangle = \int f^*(x)g(x)dx$

Where $x$ has been introduced as a dummy variable to satisfy the notational requirement of the integral - if that makes sense.

Note that functions are variables too. In fact, the whole idea of an "Operator" is that it acts on sets of functions to give other functions. In that sense, the functions act just like vectors. In particular, you have a momentum eigenfunction for every real number. Griffiths notation for that is:

$f_p$ is the eigenfunction associated with eigenvalue $p$. Here $p$ can be any real number.

Now, what if you want to take the inner product of two momentum eigenfunctions? Griffiths notation for this is:

$f_{p'}$ is the eigenfunction associated with eigenvalue $p'$ and we have:

$\langle f_{p'}| f_{p} \rangle = \int f_{p'}^*(x) f_p(x) dx = \delta(p-p')$

Now, I think using $p_1$ and $p_2$ instead of $p, p'$ might be simpler to understand, but the prime notation is quite common.

PeroK -

Follow-up question for Eigenvalues ... unfortunately I am referencing Griffith's 2nd edition (if you don't have it) ...

In Chap 2, Section 2.2 - The Infinite Square Well -

Describing 3 properties of the solution of ψ_n = ...

the 3rd property is "They are mutually orthogonal ..."

The discussion following this statement is showing that the integral of ∫ψ_m^*ψ_n dx yields the "Kronecker" Delta Function

In reviewing our discussion in this thread I found myself back in this section ...

My question, can this subscripts m and n in this orthogonal / delta function discussion be thought of as eigenvalues?

Thanks
Sparky_

 

[PeroK]

PeroK -

Follow-up question for Eigenvalues ... unfortunately I am referencing Griffith's 2nd edition (if you don't have it) ...

In Chap 2, Section 2.2 - The Infinite Square Well -

Describing 3 properties of the solution of ψ_n = ...

the 3rd property is "They are mutually orthogonal ..."

The discussion following this statement is showing that the integral of ∫ψ_m^*ψ_n dx yields the "Kronecker" Delta Function

In reviewing our discussion in this thread I found myself back in this section ...

My question, can this subscripts m and n in this orthogonal / delta function discussion be thought of as eigenvalues?

Thanks
Sparky_

The eigenvalues are the energy levels. And the eigenfunctions are the stationary states. The subscript are simply labels.

 

[Sparky_]

well shoot, I'm still not getting it ... (not a question here just a comment of frustration) ...