Orthonormal spin functions (Szabo and Ostlund problem 2.1)

[HeavyMetal]

Homework Statement

[/B]
Taken straight out of Szabo and Ostlund's "Quantum Chemistry" problem 2.1:

Given a set of K orthonormal spatial functions, $\{\psi_{i}^{\alpha}(\mathbf{r})\}$, and another set of K orthonormal functions, $\{\psi_{i}^{\beta}(\mathbf{r})\}$, such that the first set is not orthogonal to the second set, i.e.,
$$\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r}) = S_{ij}$$
where $\mathbf{S}$ is an overlap matrix, show that the set $\{\chi_{i}\}$ of 2K spin orbitals, formed by multiplying $\{\psi_{i}^{\alpha}(\mathbf{r})\}$ by the $\alpha$ spin function and $\{\psi_{i}^{\beta}(\mathbf{r})\}$ by the $\beta$ spin function, i.e.,
$$\left.\begin{aligned} \chi_{2i-1}(\mathbf{x}) = \psi_{i}^{\alpha}(\mathbf{r})\alpha(\omega)\\ \chi_{2i}(\mathbf{x}) = \psi_{i}^{\beta}(\mathbf{r})\beta(\omega) \end{aligned} \right\} \qquad \text{i = 1, 2, ..., k}$$
is an orthonormal set.

Homework Equations

The above given data, and [possibly] these identities:
$$\int d\omega\ \alpha^*(\omega)\alpha(\omega) = \langle\alpha\|\alpha\rangle = 1$$
$$\int d\omega\ \alpha^*(\omega)\beta(\omega) = \langle\alpha\|\beta\rangle = 0$$
$$\int d\mathbf{x}\ \chi_{i}^*(\mathbf{x})\chi_{j}(\mathbf{x}) = \langle\chi_{i}\|\chi_{j}\rangle = \delta_{ij}$$

The Attempt at a Solution

First, I insert two 1's stealthily into the first equation to get:
$$\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})*1*1*\psi_{j}^{\beta}(\mathbf{r})\\ = \int\int d\mathbf{r}\ d\omega\ \psi_{i}^{\alpha*}(\mathbf{r})\alpha^*(\omega)\alpha(\omega)\beta^*(\omega)\beta(\omega)\psi_{j}^{\beta}(\mathbf{r})$$

Then, I insert the definition of the spin orbitals:
$$\int \int d\mathbf{x}\ d\omega\ \chi_{i}^{\alpha*}(\mathbf{x}) \alpha(\omega)\beta^*(\omega) \chi_{j}^{\beta}(\mathbf{x})$$

I then get lost and, searching for a solution equal to a Kronecker delta, I say that it rearranges to:
$$\int \int d\mathbf{x}\ d\omega\ \chi_{i}^{\alpha*}(\mathbf{x})\chi_{j}^{\beta}(\mathbf{x})\alpha(\omega)\beta^*(\omega)\\ = \int d\omega\ \langle\chi_{i}^{\alpha}\|\chi_{j}^{\beta}\rangle\alpha(\omega)\beta^*(\omega)\\ = \int d\omega\ \delta_{ij}\ \alpha(\omega)\beta^*(\omega)\\ = \delta_{ij}\ * 0$$

But this answer gives me zero, not just the Kronecker delta that I am searching for...

I am very new to index notation, so I am sorry if this is trivial!

As always, thanks in advance!
-HeavyMetal

 

[DrClaude]

The above given data, and [possibly] these identities:
$$\int d\omega\ \alpha^*(\omega)\alpha(\omega) = \langle\alpha\|\alpha\rangle = 1$$
$$\int d\omega\ \alpha^*(\omega)\beta(\omega) = \langle\alpha\|\beta\rangle = 0$$
$$\int d\mathbf{x}\ \chi_{i}^*(\mathbf{x})\chi_{j}(\mathbf{x}) = \langle\chi_{i}\|\chi_{j}\rangle = \delta_{ij}$$

That last identity is what you are trying to prove. Start from the left-hand side and use the definition of χ to try and get the Kronecker delta. As an additional hint, I can tell you that you will have to consider different cases.

 

[HeavyMetal]

That last identity is what you are trying to prove. Start from the left-hand side and use the definition of χ to try and get the Kronecker delta. As an additional hint, I can tell you that you will have to consider different cases.

Thanks! I think I have it. Does this look correct?

$$\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\\ = \langle\psi^{\alpha}\|i\rangle\langle j\|\psi^{\beta}\rangle\\ = \sum_{i}\langle\psi^{\alpha}\|i\rangle\langle j \|i \rangle \langle i\|\psi^{\beta}\rangle\\ = \langle\psi^{\alpha}\|i\rangle\delta_{ji}\langle i\|\psi^{\beta}\rangle\\ = \langle\psi_{i}^{\alpha}|\delta_{ji}|\psi_{i}^{\beta}\rangle\\ = \langle\psi_{i}^{\alpha}\|\psi_{i}^{\beta}\rangle\delta_{ji}$$

-HeavyMetal

 

[DrClaude]

Thanks! I think I have it. Does this look correct?

$$\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\\ = \langle\psi^{\alpha}\|i\rangle\langle j\|\psi^{\beta}\rangle\\ = \sum_{i}\langle\psi^{\alpha}\|i\rangle\langle j \|i \rangle \langle i\|\psi^{\beta}\rangle\\ = \langle\psi^{\alpha}\|i\rangle\delta_{ji}\langle i\|\psi^{\beta}\rangle\\ = \langle\psi_{i}^{\alpha}|\delta_{ji}|\psi_{i}^{\beta}\rangle\\ = \langle\psi_{i}^{\alpha}\|\psi_{i}^{\beta}\rangle\delta_{ji}$$

That's not correct. Already in the second line, I do not understand what the states $|i\rangle$ and $|j\rangle$ are. Also, I suggest you start with $\langle \chi_i | \chi_j \rangle$.

On a minor note, when writing a bra and a ket together, there is is single vertical line inbetween.

 

[HeavyMetal]

That's not correct. Already in the second line, I do not understand what the states $|i\rangle$ and $|j\rangle$ are.

I do not mean these to be states, they're just basis vectors. When I write $\big|\psi_{j}\rangle$, I take it to mean the scalar product of a basis vector, $\langle j\big|$, and the wavefunction, $\big|\psi\rangle$. This gives $\langle j\big|\cdot \big|\psi\rangle = \big|\psi_{j}\rangle$. I used this step because I can simultaneously do a resolution of identity and get a Kronecker delta.

On a minor note, when writing a bra and a ket together, there is is single vertical line inbetween.

Yes, I know. I didn't know about the \big function in latex and so I was escaping the '|', which doubles it for some odd reason.

Also, I suggest you start with $\langle \chi_i | \chi_j \rangle$.

I don't know what you mean by this. If I start there the solution would just be to convert it into a Kronecker delta from the definition. Done. If i = j, then the integral is one, and if i $\neq$ j, the integral is zero. I want to know how to get to there from what is given! I'm sorry if I'm missing something, I'm just a bit confused as to what you're asking.

 

[DrClaude]

I do not mean these to be states, they're just basis vectors.
When I write $\big|\psi_{j}\rangle$, I take it to mean the scalar product of a basis vector, $\langle j\big|$, and the wavefunction, $\big|\psi\rangle$. This gives $\langle j\big|\cdot \big|\psi\rangle = \big|\psi_{j}\rangle$. I used this step because I can simultaneously do a resolution of identity and get a Kronecker delta.

You are confusing many things here. There is no such thing as "just a basis vector." States that form a basis will be eigenfunctions of some observable. Also, you are already given two basis sets for the spatial part, $\left| \psi_i^\alpha \right\rangle$ and $\left| \psi_i^\beta \right\rangle$, and one for the spin part, $\{ \left| \alpha \right\rangle, \left| \beta \right\rangle\}$. There are no other basis states to be used here.

Also, $\langle j\big|\cdot \big|\psi\rangle = \big|\psi_{j}\rangle$ doesn't make sense. First, there is no $\cdot$ in Dirac notation. Second, $\left\langle j \right. \left| \psi \right\rangle$ is a complex number, not a ket. Third, the $j$ in $\left| \psi_j^\alpha \right\rangle$ is just an index.

If i = j, then the integral is one, and if i $\neq$ j, the integral is zero.

How do you know that? You have been given the definition of $\left| \chi_i \right\rangle$, use it to prove that $\left\langle \chi_i \right| \left. \chi_j \right\rangle = \delta_{ij}$.

You seem to be assuming that $\left\langle \phi_i \right| \left. \phi_j \right\rangle = \delta_{ij}$ is a general statement for any $\left| \phi_i \right\rangle$. It is not. It is only true if the set of $\left| \phi_i \right\rangle$ is orthonormal, which is what you have to prove here.

 

[HeavyMetal]

Also, $\langle j\big|\cdot \big|\psi\rangle = \big|\psi_{j}\rangle$ doesn't make sense. First, there is no $\cdot$ in Dirac notation. Second, $\left\langle j \right. \left| \psi \right\rangle$ is a complex number, not a ket. Third, the $j$ in $\left| \psi_j^\alpha \right\rangle$ is just an index.

You're right, I'm confusing a few things. The math review for this book starts at basic vectors and gradually builds. They start with dot products in Euclidean space, but once they've introduced complex vector spaces they continue to call inner products "scalar products." This is confusing, and as you pointed out, the result of an inner product is a complex number. The following equation has also caused some confusion:
$$\langle j|a\rangle=\sum_{i}\langle j|i\rangle a_{i}=\sum_{i}\delta_{ji}a_{i}=a_{j}$$
which is equation 1.48a in S&O. I am tempted to just rename the vector 'a' as '$\psi$' and go in the reverse direction, which is what I attempted to do earlier.

You seem to be assuming that $\left\langle \phi_i \right| \left. \phi_j \right\rangle = \delta_{ij}$ is a general statement for any $\left| \phi_i \right\rangle$. It is not. It is only true if the set of $\left| \phi_i \right\rangle$ is orthonormal, which is what you have to prove here.

Okay. I know that I'm given two sets of spatial orbitals (with K total orbitals each) and a set of spin functions, and together they make a single set of spin orbitals (for a total of 2K spin orbitals). I think they're looking for me to prove that the alpha spin orbitals are orthonormal to the beta spin orbitals by finding that they are equal to the Kronecker delta.

Maybe you could help to steer me in the right direction? This isn't an assigned homework problem, I'm just trying to grasp this material. I can imagine that $\left\langle \chi_i \right|$ represents the alpha spin orbitals and that $\left| \chi_j \right\rangle$ represents the beta spin orbitals. So then how do I go about proving that the inner product is $\delta_{ij}$? I'm obviously a bit lost.

 

[DrClaude]

$$\langle j|a\rangle=\sum_{i}\langle j|i\rangle a_{i}=\sum_{i}\delta_{ji}a_{i}=a_{j}$$
which is equation 1.48a in S&O. I am tempted to just rename the vector 'a' as '$\psi$' and go in the reverse direction, which is what I attempted to do earlier.

It is important to recognize the different types of mathematical "objects" that can appear in equations in QM. a_j is a complex number, while ψ_j is a wave function, which is a particular case of a state vector that could be written |ψ_j> or more simply |j>.

I can imagine that $\left\langle \chi_i \right|$ represents the alpha spin orbitals and that $\left| \chi_j \right\rangle$ represents the beta spin orbitals. So then how do I go about proving that the inner product is $\delta_{ij}$? I'm obviously a bit lost.

Looking back at the definition, you have 2k χ orbitals. The indices i and j have to run over all possible values, so you can't take χ_i to be an α orbital and take χ_j to be β. (In other words, I'm free to chose i = 3 and j = 5, for instance.)

So, start from <χ_i|χ_j> and substitute using ψ^α and ψ^β. This immediately poses another question: which one should you take? You have to consider all four possibilities: i even/j even, i even/j odd, i odd/j even, i odd/j odd.

 

[HeavyMetal]

Okay, I think I know what you're saying now. I was trying to start with the individual spatial orbitals and go to the spin orbitals. So then I think we can arrange these 16 possibilities into a matrix:

$$\begin{bmatrix} \alpha^*\psi_i^{\alpha*}\psi_i^{\alpha}\alpha&\alpha^*\psi_i^{\alpha*}\psi_i^{\beta}\beta&\alpha^*\psi_i^{\alpha*}\psi_j^{\alpha}\alpha&\alpha^*\psi_i^{\alpha*}\psi_j^{\beta}\beta\\ \beta^*\psi_i^{\beta*}\psi_i^{\alpha}\alpha&\beta^*\psi_i^{\beta*}\psi_i^{\beta}\beta&\beta^*\psi_i^{\beta*}\psi_j^{\alpha}\alpha&\beta^*\psi_i^{\beta*}\psi_j^{\beta}\beta\\ \alpha^*\psi_j^{\alpha*}\psi_i^{\alpha}\alpha&\alpha^*\psi_j^{\alpha*}\psi_i^{\beta}\beta&\alpha^*\psi_j^{\alpha*}\psi_j^{\alpha}\alpha&\alpha^*\psi_j^{\alpha*}\psi_j^{\beta}\beta\\ \beta^*\psi_j^{\beta*}\psi_i^{\alpha}\alpha&\beta^*\psi_j^{\beta*}\psi_i^{\beta}\beta&\beta^*\psi_j^{\beta*}\psi_j^{\alpha}\alpha&\beta^*\psi_j^{\beta*}\psi_j^{\beta}\beta \end{bmatrix}$$

Whose values turn out to be:

$$\begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} = \delta_{ij}^{\alpha\beta}$$

 

[HeavyMetal]

Here is my train of thought: If i=j, then we are referring to the same spin orbital. Because both sets of spatial orbitals are orthonormal w.r.t. their own elements, the inner product of any spin orbital with itself must be one. If i $\neq$ j, then we are referring to any two different spin orbitals. If the spin part is the same, the inner product is zero because they were originally composed from a set of orthonormal spatial functions, yet they don't have the exact same spatial part. If the spin part is different, we know that they originally come from different sets of spatial orbitals that are not necessarily orthogonal...but imposing the fact that all alpha spin orbitals must be orthogonal to beta orbitals, we find that the inner product is zero.

 

[DrClaude]

Forget about matrices. I don't see were you are going with that.

There are not 16 possibilities, only the four I listed. There are two possibilities for χ_i, depending on whether i is even or odd.

Here is my train of thought: If i=j, then we are referring to the same spin orbital. Because both sets of spatial orbitals are orthonormal w.r.t. their own elements, the inner product of any spin orbital with itself must be one. If i $\neq$ j, then we are referring to any two different spin orbitals. If the spin part is the same, the inner product is zero because they were originally composed from a set of orthonormal spatial functions, yet they don't have the exact same spatial part. If the spin part is different, we know that they originally come from different sets of spatial orbitals that are not necessarily orthogonal...but imposing the fact that all alpha spin orbitals must be orthogonal to beta orbitals, we find that the inner product is zero.

This is all correct. You simply need to formalize it with a few lines of math.

 

[HeavyMetal]

Okay, sorry this took me so long...four situations:

$\left\langle \chi_i \big| \chi_j \right\rangle$

odd/odd: $= \int \int d\mathbf{r}\ d\omega\ \alpha^*(\omega)\psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\alpha}(\mathbf{r})\alpha(\omega) = 1$
odd/even: $= \int \int d\mathbf{r}\ d\omega\ \alpha^*(\omega)\psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\beta(\omega) = 0$
even/odd: $= \int \int d\mathbf{r}\ d\omega\ \beta^*(\omega)\psi_{i}^{\beta*}(\mathbf{r})\psi_{j}^{\alpha}(\mathbf{r})\alpha(\omega) = 0$
even/even: $= \int \int d\mathbf{r}\ d\omega\ \beta^*(\omega)\psi_{i}^{\beta*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\beta(\omega) = 1$

So if i=j, then the inner product is 1. If i $\neq$ j, then the inner product is 0. Therefore $\left\langle \chi_i \big| \chi_j \right\rangle = \delta_{ij}$. I see now why you reminded me that i and j are just indices. They are just labels referring to which spatial orbital you are talking about out of the set, or in other words, any member of $\{\chi_i\}$. It all seems obvious now .

 

[DrClaude]

Let me correct that a bit:

odd/odd: $= \int \int d\mathbf{r}\ d\omega\ \alpha^*(\omega)\psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\alpha}(\mathbf{r})\alpha(\omega) = \int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\alpha}(\mathbf{r}) = \delta_{ij}$
odd/even: $= \int \int d\mathbf{r}\ d\omega\ \alpha^*(\omega)\psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\beta(\omega) = 0$
even/odd: $= \int \int d\mathbf{r}\ d\omega\ \beta^*(\omega)\psi_{i}^{\beta*}(\mathbf{r})\psi_{j}^{\alpha}(\mathbf{r})\alpha(\omega) = 0$
even/even: $= \int \int d\mathbf{r}\ d\omega\ \beta^*(\omega)\psi_{i}^{\beta*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\beta(\omega) = \int d\mathbf{r}\ \psi_{i}^{\beta*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r}) = \delta_{ij}$

 

[HeavyMetal]

Yes, of course. I didn't properly account for the fact that inner products between spatial orbitals of the same set such as those in $\{\psi_i^{\alpha}\}$ or $\{\psi_i^{\beta}\}$ can be different, and most of the time are.

Thanks for the help! It exposed a few holes I had in notation and understanding.

-HeavyMetal