-1

So in quantum mechanics, most of the people uses exponential equation to handle the problem and present the relationship. Another word, we often handle the classical wave function in trigonometric way by utilizing sine and cosine. How can these two be correlated?

Using Euler's formula the exponential equation will finally result in an excess imaginary term of $i\sin(kx-\omega t)$. What is the formal process to deal with this?

DanielSank
  • 24,439
user46899
  • 37
  • As I understand, you are asking about the correlation between the exponential-formalism and the trigonometric-formalism in QM. Actually, they are the same, by the means of the Euler's formula. You don't need to worry about the imaginary term: the coefficients in the expansion of wavefuntions into an orthornomal basis are, in general, complex; and measured physical quantities are in the modulus form (like $\left\langle\psi|H|\psi\right\rangle$ or $\left\langle\psi|\psi\right\rangle$). So, despite the fact that you should work with complex numbers in QM, physical values are always REAL. – Lê Dũng Jul 20 '17 at 16:59

1 Answers1

0

Imagine that you have a state in QM, which can describe any system you wish (e.g. Atom, Oscillator and so on). With the help of QM, the quantity which describes the entire system is called a $\textbf{state vector}$ (if you use Dirac approach, or a wave function if you use the Schrodinger approach) usually denoted with $|\Psi>$ or $\Psi(\textbf{r},t)$ for the latter approach. Your wave-function can have many forms including the exponential that you mentioned, so it can have both REAL and IMAGINARY parts.

The key thing here, as Le Dung already stated is that, the actual PHYSICAL quantity (which has a real meaning, and it is measurable) has always the form $|\Psi|^2$ if you want the probability distribution, or $E=<\Psi|H|\Psi>$ if you want to find the energies, $r=<\Psi|r|\Psi>$ if you want the expectation value of the particle's position(don't take this as it is though, it's just a poor and quick example, you can learn more here) and other relevant quantities. Because of the mathematics behind these operations (which usually just imply a modulus squared of the state vector), the imaginary part would not be "present" anymore, so we end up with a real, physical quantity. More on probability in QM here.

Hope this helps:)

PS: It's a little bit late for me, so I apologize in advance if I was not clear enough XD

Robert Poenaru
  • 269
  • So, essentially what really brings into my observation is the real part while the imaginary part of the exponential function can be utilized during the calculation of the probability density proposed by Born? – user46899 Jul 21 '17 at 02:44
  • I don't understand what you're saying. Essentially, when you try to calculate all the quantities which I enumerated in the answer, even the probability density proposed by Born, you take the entire wave-function as it is (with its real and imaginary parts) and apply a modulus squared. You don't make calculations only with the real part or only with the imaginary part. You always consider the entire wave-function. I hope this clarifies your question. Also, I added a link in the answer where you can find more details. Please vote this as an answer if you are pleased :) – Robert Poenaru Jul 21 '17 at 09:27
  • Thanks! Really appreciated. BTW could you upvote the question? – user46899 Jul 21 '17 at 10:25