Linked Questions

0
votes
2answers
202 views

Why is $i$ used in the equations for quantum mechanics? [duplicate]

Coming from someone who knows a tiny bit about the subject but who really wants to learn. I know it's the square root of -1 but I would like some insight as to why it's used at all.
-1
votes
2answers
106 views

On the spectrum of a quantum mechanical system [duplicate]

Can the spectrum of a quantum mechanical operator contain both real and complex numbers?
3
votes
0answers
60 views

Do imaginary measurements exists? [duplicate]

I'm sorry if this question is too metaphysical, but I will give it a try. My textbook in introductory quantum mechanics is basing a lot of its proof and derivations on the fact that the value of the ...
76
votes
13answers
29k views

About the complex nature of the wave function?

1. Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an ...
76
votes
12answers
19k views

QM without complex numbers

I am trying to understand how complex numbers made their way into QM. Can we have a theory of the same physics without complex numbers? If so, is the theory using complex numbers easier?
30
votes
7answers
3k views

Can one do the maths of physics without using $\sqrt{-1}$?

The use of imaginary and complex values comes up in many physics and engineering derivations. I have a question about that: Is the use of complex numbers simply to make the process of derivation ...
17
votes
4answers
12k views

Why do we use Hermitian operators in QM?

Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') ...
10
votes
2answers
2k views

Commutator not transitive

I noticed the following: $$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad [L^2,L_3]=0.$$ This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ ...
7
votes
2answers
2k views

Why hermitian, after all? [duplicate]

This question is going to look a lot like a duplicate, but I've read dozens of related posts and they don't touch the subject. Here we go. Why are observables represented by hermitian operators? ...
7
votes
3answers
2k views

What is the point of complex fields in classical field theory?

I see a lot of books/lectures about classical field theory making use of complex scalar fields. However why complex fields are used in the first place is often not really motivated. Sometimes one can ...
3
votes
5answers
322 views

Why is there a physical preference to real numbers?

In quantum mechanics, operators can only be observables if the eigenfunctions they operate on have real eigenvalues. If they are complex, I am told that, surely, some observable of a physical system ...
6
votes
2answers
3k views

Can expectation value be imaginary?

I was solving a problem and the result of the expectation value of an operator came out to be $-\frac{\hbar}{4}$ $i$. Is this result possible? It seems counter intuitive.
3
votes
5answers
409 views

Reality of the Wavefunction - Complex numbers and Degrees of Freedom in Configuration space

On the "reality" of the wavefunction, there seem to be two schools of thought on why treating $\psi$ as something more than a mathematical tool is erroneous: $\psi$ involves complex numbers. Only ...
5
votes
1answer
334 views

Energy scale dependence of coupling constants

I am trying to understand the meaning of the renormalization group equation and what i have understood is that, since observable (or physical?) quantities must not depend on arbitrary energy scales, ...
2
votes
1answer
489 views

Why are Hermitian operators linked to observables?

In Quantum Mechanics, why is it that a self-adjoint operator is linked to an observable? What makes it measureable? And why isn't a non-Hermitian operator linked to an observable? Also, what type of ...

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