Undergrad Complex Conjugate of Wave Function

[Icycub]

TL;DR

Question about taking the complex conjugate of functions

I've been studying quantum mechanics this semester in school and have ran into an issue I can't find an answer for. I understand why we take the complex conjugate of the wave function, such as when calculating expectation values. I'm a little confused though as to why we take the complex conjugate of a function whenever multiplying any two functions, such as ψ*_nψ_m. What would be the benefit of this? Does this still remove all imaginary numbers?

 

[PeterDonis]

I'm a little confused though as to why we take the complex conjugate of a function whenever multiplying any two functions, such as ψ*_nψ_m.

You don't take a complex conjugate whenever you multiply any two functions in QM. You only do it in specific contexts. In what context are you seeing a product like $\psi_n^* \psi_m$?

 

[PeterDonis]

Normally it’s because $z z^* = |z|^2$, which is just a real number, i.e. the modulus $z$ squared. If $z$ is the wave function, then its modulus squared is a probability amplitude!

Note that the OP isn't talking about taking the squared modulus of a single wave function. The OP is talking about expressions that involve two different wave functions.

 

[Icycub]

The text uses that notation when discussing the orthogonality of the spatial solutions to the infinite square well. It also uses it when introducing the notation for the inner product of two functions.

 

[etotheipi]

Note that the OP isn't talking about taking the squared modulus of a single wave function. He's talking about expressions that involve two different wave functions.

Yeah, I re-read the post and then noticed the subscripts were different. But in that case I don’t really understand the question, since it’s perfectly fine mathematically to multiply $\psi_n \psi_m$, or $\psi_n ^* \psi_m^*$ or whatever combination you like, since they’re just complex numbers. Whether or not it’s a useful quantity is a different question!

 

[Icycub]

I may be overthinking it, but I'm just confused about the purpose of it? Why do the complex conjugate of one of the functions? The text seems to do this whenever multiplying any two functions.

 

[PeterDonis]

The text

What text? Can you give a reference?

 

[Icycub]

Yes, it's Introduction to Quantum Mechanics Third Edition by Griffiths.

 

[etotheipi]

The text uses that notation when discussing the orthogonality of the spatial solutions to the infinite square well. It also uses it when introducing the notation for the inner product of two functions.

Ah okay, I think I understand the question. Basically a complex inner product space has an inner product which satisfies $\langle x, y \rangle = \overline{\langle y, x \rangle}$, is linear in the second argument $\langle x, c_1y_1 + c_2 y_2 \rangle = c_1 \langle x, y_1 \rangle + c_2 \langle x, y_2 \rangle$ [and by extension conjugate linear in the first argument], and it's positive definite. This inner product in $\mathbb{C}^n$ looks like$$\langle x, y \rangle = \sum_i \overline{x_i} y_i$$You need the complex conjugate in there, so that the norm makes sense ($|v| = \sqrt{\langle v , v \rangle}$); to see why, try considering the vector $iv$ and its inner product $\langle iv, iv \rangle$ (credit to @Infrared from a previous thread, for this example!)!

You can also consider things like a space of complex-valued continuous functions which has an inner product like$$\langle f, g \rangle = \int \overline{f(x)} g(x) dx$$In exactly the same way, if your spatial solutions to the infinite square well are orthogonal then they'll satisfy$$\langle \psi_n | \psi_m \rangle = \int \psi_n^*(x) \psi_m(x) dx = 0$$

 

[Icycub]

Sorry, I don't think I fully understand the reasoning. I've just now started Chp. 3 in the text which begins to discuss linear algebra in QM. In general, in QM, what would be the benefit of doing the complex conjugate whenever multiplying any two functions?

 

[Icycub]

So using the complex conjugate ensures that the square root of the inner product of two vectors is positive?

 

[etotheipi]

It's how the inner product in a complex inner product space is defined, specifically because it must be positive definite. If we didn't conjugate one of the terms in the inner product, then $$\langle iv, iv \rangle = \sum_a i^2 {v_a}^2 = \sum_a - {v_a}^2 < 0$$Instead, if we define the inner product so that one of the terms is conjugated,$$\langle iv, iv \rangle = \sum_a -i^2 {v_a}^2 = \sum_a {v_a}^2 > 0$$And now we can relax, since $|iv| = \sqrt{\langle iv, iv \rangle}$ makes sense.

In general, in QM, what would be the benefit of doing the complex conjugate whenever multiplying any two functions?

You don't, you can multiply whatever you want, e.g. $\psi_n(x) \psi_m(x)$, or $\psi^*_n(x) \psi_m(x)$, or $\psi_n(x) \psi^*_m(x)$, or $\psi^*_n(x) \psi^*_m(x)$ are all just complex numbers.

It's just that a lot of structures are defined in the pattern you mentioned (how the inner product is defined), because a lot of structures are related to the inner product. E.g. if the state of the system is $\psi$, then the expectation of an observable $A$ is$$\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \langle \psi | \hat{A} \psi \rangle$$or $$\langle A \rangle = \int \psi^* (\hat{A} \psi) dx = \int \psi^* \hat{A} \psi dx$$

 

[Icycub]

Okay, thanks!

 

[PeroK]

So using the complex conjugate ensures that the square root of the inner product of two vectors is positive?

For complex valued square-integrable functions:
$$\langle f|g \rangle = \int_{-\infty}^{+\infty} f(x)^*g(x)dx$$
Defines an inner product. Exercise: check the properties of an inner product. Whereas:
$$\langle f|g \rangle = \int_{-\infty}^{+\infty} f(x)g(x)dx$$
Does not define an inner product. Exercise: which properties does it fail to have?

 

[vanhees71]

For complex functions you want to define a complex Hilbert space, i.e., a complex vector space with a scalar product. In complex vector spaces you have to define
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x),$$
so that the "square"
$$\langle \psi|\psi \rangle \geq 0.$$
You also want to have it positive definite, i.e.,
$$\langle \psi|\psi \rangle=0 \; \Leftrightarrow \psi=0.$$
This implies that we have to define a function already as 0, if
$$\int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \psi(x)=0.$$