Inner product of functions help

[Niles]

Homework Statement

Hi all.

Say that I have two functions g(x) and h(y). Is it correct that their inner product is given by:

$$<g,h> = \int\int {g(x)^*h(y)}dxdy,$$

where the asterix denotes complex conjugation?

Thanks in advance.Niles.

 

[quantumdude]

It should be:

$$\left<g,h\right>=\int g(t)h^*(t)dt$$

 

[Niles]

Thanks for replying.

But also when they are functions of different variables?

 

[quantumdude]

That shouldn't pose any difficulty. For instance if $g(x)=3x^2-2x+1$, then $g(t)=3t^2-2t+1$. You just plug in $x=t$. You can do that for $h(y)$ also. What's important is that the two functions have the same domain.

 

[Niles]

My question arises from quantum mechanics, and I know that luckily you are good at that. This is where my problem comes from:

E.g. when we have $\psi(x_1)$ and $\psi(x_2)$, and we want to find $<\psi_1|H_1|\psi_2>$, where H₁ acts only on x₁. Then do we also rewrite as above? I personally don't think so, but I am having my doubts.

 

[quantumdude]

I see. First thing to mention is that $<\psi_1|H_1|\psi_2>$ is not an inner product. I'm having trouble imagining how this quantity even arises in quantum mechanics. Typically if you have a 2-body problem you have something like $H=H_1+H_2$ for a Hamiltonian, where $H_i$ acts on the coordinates of the $i^{th}$ particle. Then the wavefunction is expressed as a product $\psi(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)$ whose factors satisfy the following equations separately.

$H_i\phi_i(x_i)=E_i\phi_1(x_i)$, $i=1,2$

Can you briefly explain why you even need to compute this thing?

 

[Niles]

First thing to mention is that $<\psi_1|H_1|\psi_2>$ is not an inner product.

When an operator works on a ket, it gives a new ket. So we have a bra and a ket - why is it not an inner product then?

You are absolutely correct. The Hamiltonian is H = H₁ + H₂, and I have to find the following quantity:

$$<\psi | H | \psi>,$$

where one of the parts of this expression is $<\psi_1|<\psi_2|H_1|\psi_1>|\psi_2>$. But what I do normally is do look at this as an integral, because then it becomes very obvious to me which parts I equal to 1 and which parts I have to calculate directly (if any). But suddenly when I was doing this, I was wondering whether I am even allowed to view this as an integral of two variables or not.

 

[quantumdude]

When an operator works on a ket, it gives a new ket. So we have a bra and a ket - why is it not an inner product then?

Because $|\psi_1>$ and $|\psi_2>$ are in different Hilbert spaces, call them $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively. In order to accommodate a 2-body problem the appropriate Hilbert space is the tensor product of the two, $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. The state ket for the system is now written as $|\psi>=|\psi_1>\otimes|\psi_2>$.

Let $|\psi^A>=|\psi^A_1>\otimes|\psi^A_2>$, $|\psi^B>=|\psi^B_1>\otimes|\psi^B_2>\in\mathcal{H}$. Inner products are defined as follows.

$$<\psi^A|\psi^B>=(<\psi^A_2|\otimes<\psi^A_1|)(|\psi^B_1>\otimes|\psi^B_2>)$$

$$<\psi^A|\psi^B>=<\psi^A_1|\psi^B_1><\psi^A_2|\psi^B_2>$$

Does that help?

 

[Niles]

It helped a little, but I don't know what a tensorproduct is. But the part of your post that I can use it this:

$$<\psi^A|\psi^B>=<\psi^A_1|\psi^B_1><\psi^A_2|\psi^ B_2>$$

So this means that I am allowed to look at the quantity $<\psi_1|<\psi_2|H_1|\psi_1>|\psi_2>$ as an integral over the two variables?

 

[quantumdude]

It helped a little, but I don't know what a tensorproduct is. But the part of your post that I can use it this:

$$<\psi^A|\psi^B>=<\psi^A_1|\psi^B_1><\psi^A_2|\psi^ B_2>$$

That's all you really need.

So this means that I am allowed to look at the quantity $<\psi_1|<\psi_2|H_1|\psi_1>|\psi_2>$ as an integral over the two variables?

It means that you can look at it as a product of two integrals, each over a different single variable. Because $|\psi_1>\notin\mathcal{H}_2$, it passes right through operators on $\mathcal{H}_2$. Likewise $|\psi_2>$ passes through operators on $\mathcal{H}_1$. So you shouldn't even see quantities such as $<\psi_2|H_1|\psi_1>$ (I don't even think such a thing is defined).

So we get the following.

$$<\psi_2|\otimes<\psi_1|(H_1+H_2)|\psi_1>\otimes|\psi_2>=<\psi_1|H_1|\psi_1><\psi_2|\psi_2>+<\psi_1|\psi_1><\psi_2|H_2|\psi_2>$$

$$<\psi_2|\otimes<\psi_1|(H_1+H_2)|\psi_1>\otimes|\psi_2>=E_1+E_2$$

 

[Niles]

It means that you can look at it as a product of two integrals, each over a different single variable.

Very nice, that is what I wanted to know. I have a final question, which I actually think is a little interesting. Let's say that we are looking at a state containing also spin:

$$\psi = \varphi (x_1 )\varphi (x_2 )\left| \uparrow \right\rangle _1 \left| \downarrow \right\rangle _2$$

Now we wish to find the following quantity: $<\psi| H_1 | \psi>$, where we assume that H₁ has nothing to do with the spin. Now if we insist on writing the above mentioned integral, then we get:

$$\left[ {\int {\varphi ^* (x_1 )} H_1 \varphi (x_1 )dx_1 } \right]\left[ {\int {\varphi ^* (x_2 )} \varphi (x_2 )dx_2 } \right]\left[ {\int {''\left\langle { \uparrow _1 } \right|\left| \uparrow \right\rangle _1 ''d1} } \right]\left[ {\int {''\left\langle { \downarrow _2 } \right|\left| { \downarrow _2 } \right\rangle ''d2} } \right].$$

First of all, the reason why I have used quotation marks around the spin-part is that I know it is very bad to write it this way, but I hope you get the idea.

My question is, what "d1" and "d2" in this case represent?

 

[quantumdude]

$d1$ and $d2$ don't represent anything. Spin doesn't have a coordinate representation. (Contrast this with orbital angular momentum which does have such a representation). In Sakurai's book Modern Quantum Mechanics he gives two arguments for this. One, spherical harmonics aren't defined for half-integer values of $l$. And second, he appeals to $2\pi$ rotations of spin-1/2 particles (are you familiar with how that works out?).

However, in 2005 a paper entitled "Explicit Spin Coordinates" was uploaded to the LANL arXiv (authors Hunter and Schlifer). I'm refraining from posting a link because the paper was never accepted by a peer-reviewed journal, but it should be easy to find if you are so inclined. I don't know if it's worthwhile, just thought I'd mention a counterpoint. But the orthodox view is the one expressed in the first paragraph.

 

[Niles]

And second, he appeals to $2\pi$ rotations of spin-1/2 particles (are you familiar with how that works out?).

I haven't heard of this, but I will ask my professor about it.

Thanks for helping. It was very useful for me.

 

[quantumdude]

I haven't heard of this,

Then allow me!

Let $|\alpha>$ be a spin state of a spin-1/2 particle. Expanded in the $S_z$ basis we have:

$$|\alpha>=c_+|+>+c_-|->$$

If the particle has no orbital angular momentum then the rotation operator by an angle $\phi$ about an axis parallel to unit vector $\hat{n}$ is given by the following.

$$\mathcal{D}(\phi,\hat{n})=exp\left(-\frac{\vec{S}\cdot\hat{n}\phi}{\hbar}\right)$$

Let's rotate our spin-1/2 particle about the z axis by an angle of $2\pi$.

$$\mathcal{D}(2\pi,\hat{k})|\alpha>=exp\left(-\frac{2S_z\pi}{\hbar}\right)(c_+|+>+c_-|->)$$

$$\mathcal{D}(2\pi,\hat{k})|\alpha>=c_+exp\left(-\frac{2S_z\pi}{\hbar}\right)|+>+c_-exp\left(-\frac{2S_z\pi}{\hbar}\right)|->$$

Since $S_z|+>=-S_z|->=\hbar /2$ we get the following.

$$\mathcal{D}(2\pi,\hat{k})|\alpha>=c_+exp(-i\pi)|+>+c_-exp(i\pi)|->$$

$$\mathcal{D}(2\pi,\hat{k})|\alpha>=-|\alpha>$$

See what happened? If you take a spin-1/2 particle and rotate it through a full revolution you don't return to the original state. You have to rotate it through two full revolutions before coming back to the original state. For this reason, spin cannot possibly be based on coordinates.

 

[Niles]

Wow, very nice indeed. And I actually understood the whole proof.

I found the article you referred to earlier: http://arxiv.org/abs/quant-ph/0507008. I'll take a look at it.

Thanks again; it was very useful.

 

[quantumdude]

No problem. This thread was interesting to me, and in fact I've just higlighted it. I've solved plenty of 2D/2-body QM problems before, but all I ever did was find the eigentstates and eigenvalues. I've never concerned myself with finding expectation values on these problems before. I had to do some digging and thinking before I wrote post #8.

 

[Anglea]

Hi guys, I am really sorry if Ia m not allowed to post in this thread, but I added a subscription and I am struggling with a similar question, so I though I might get a help from a generous people in this thread...it is QM context and I am new to this area, ...here I am trying to understand how they calculate the ensemble variance...<\psi \psi*> there is no comma in between... .I add the equation as in LaTex...
[tex] \eta[\psi]=\int\{\mid\psi(r)\mid\}^2 dr[\tex]<br /> [tex]Var(\eta)=2\iint\{\langle\mid\langle\psi(r_1)\psi^*(r_2)\rangle\rangle\mid}^2 dr_1 dr_2[\tex]<br /> then it is easy to obtain the following composition rule<br /> $$\int\langle\psi(r_1)\psi^*(r)\rangle\langle\psi(r)\psi^*(r_2)\rangle dr=1\N\langle\psi(r_1)\psi^*(r_2)\rangle [\tex]$$[/tex][/tex]

 

[P.j.S]

I am not too familiar with calculus as you may readily soon see.

fx ((g(x)=h(y)) >
but I feel the operator for multiplycation should be on both sides of what ia indicative. Not just the right hand side of the notation.

Thank-you! from P.j.S


1. Homework Statement
Hi all.

Say that I have two functions g(x) and h(y). Is it correct that their inner product is given by:




where the asterix denotes complex conjugation?

Thanks in advance.


Niles.

 

[P.j.S]

I am not too familiar with calculus as you may readily soon see.

fx ((g(x)=h(y)) >
but I feel the operator for multiplycation should be on both sides of what ia indicative. Not just the right hand side of the notation.

rather fx (g(x) x (h(y)) = niles solution




Thank-you! from P.j.S


1. Homework Statement
Hi all.

Say that I have two functions g(x) and h(y). Is it correct that their inner product is given by:




where the asterix denotes complex conjugation?

Thanks in advance.


Niles.