i try to finished my thesis, (Just have a problem with the wave mechanics) this is wave function: $$\Psi(\vec x, t)=A\exp{i(\phi+\vec k.\vec x-\omega t)}$$

In mathematics, the symbol $i$ is conventionally used to represent the square-root of minus one: i.e., $i^2=-1$

In addition, a general complex number is written: $$z=x+iy$$ $$z=r|\cos\theta + i\sin\theta|=r\exp{i\theta}$$

that which means: $$\Psi(\vec x, t)=A|\cos(\phi+\vec k.\vec x-\omega t) + i\sin(\phi+\vec k.\vec x-\omega t)|$$

ok i want to know that:

  1. $$A\cos(\phi+\vec k.\vec x-\omega t)=?$$
  2. $$A\sin(\phi+\vec k.\vec x-\omega t)=?$$

in the case of complex numbers is: 1. $$r\cos\theta=x$$ 2. $$r\sin\theta=y$$

and also i want to know that what is unit of $A$ ?

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    $\begingroup$ It's not clear what you are asking. $A \cos(\phi + \vec{k} \cdot \vec{x} - \omega t)$ and $A \sin(\phi + \vec{k} \cdot \vec{x} - \omega t)$ are simply functions of space and time, what more do you want to know about them? Also, you seem to be using symbols to mean multiple things. For example, you use $\phi$ to denote the generic phase of a complex number, but then you also use it as a phase offset in your particular wavefunction. Additionally, you might be confusing the $x$ and $y$ in your expression for a generic complex number with the spatial variables in your wave function. $\endgroup$ Commented May 25, 2012 at 5:58
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    $\begingroup$ Does $cis\phi$ mean $cos\phi + isin\phi$? $\endgroup$ Commented May 25, 2012 at 6:36
  • $\begingroup$ @John Rennie yes $\endgroup$
    – user8784
    Commented May 25, 2012 at 9:15
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    $\begingroup$ Good riddance... $\endgroup$
    – user2963
    Commented May 25, 2012 at 23:01
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    $\begingroup$ @BadBoy: The "answer" was in point of fact a followup question, which is not how we do things around here. $\endgroup$ Commented May 25, 2012 at 23:34

1 Answer 1


Re the units of $A$: $A$ has dimensions of $L^{-3/2}$. The modulus of $\Psi^2(\vec{x},t)$ is a probability density so it has units of probability per unit volume, i.e. it's dimension is $L^{-3}$, so the dimensions of $\Psi$ and therefore $A$ must be $L^{-3/2}$. I don't think there is an SI unit corresponding to $\sqrt{P/m^3}$, but then it's hard to see when you would need such a unit.

Re the other question: the function:

$$\Psi(\vec x, t)=Ae^{i(\phi+\vec k\vec x-\omega t)}$$

is indeed a solution to the Schrodinger equation for a free particle, but you need to be careful about it's physical meaning. As mentioned above, the modulus of $\Psi^2$ gives you the probablity of finding the particle per unit volume, but $\Psi$ itself doesn't have a physical interpretation. You're quite correct that the equation can also be written:

$$\Psi(\vec x, t)=Acos(\phi+\vec k\vec x-\omega t) + Aisin(\phi+\vec k\vec x-\omega t)$$

but you can't attach any physical meaning to the two terms on the right.

P = probability

Re the comment: I suppose the SI unit would be the metre$^{3/2}$, but the value of the wavefunction has no physical significance so no such unit has ever been defined.

This is why the value has no physical significance: suppose $\phi$ is zero, then what is $\Psi(0, 0)$. The answer is of course $A$. But the phase $\phi$ has no absolute meaning, it's a phase relative to some reference point that can be freely selected. So suppose I use a reference point that is $\pi/2$ different to yours then I'd calculate $\Psi(0, 0)$ to be $iA$ not zero.

So you think $\Psi(0, 0) = A$ while I think $\Psi(0, 0) = iA$. Does this make any difference in the real world. Well the probability deniity is $\Psi^*\Psi$ so you calculate the probablility density to be $A^2$, and I calculate the probability density to be, erm, $A^2$ and we get the same answer even though we have different values for $\Psi$.

  • $\begingroup$ can you more explain about its unit!? $$A \to \frac {1}{L^{\frac {3}{2}}}$$ $$A^2 \to \frac {1}{L^ {3}}$$ $\endgroup$
    – user8784
    Commented May 25, 2012 at 12:25

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