The derivative of the complex conjugate of the wave function

[Tony Hau]

TL;DR Summary

Given a wavefunction $\Psi$, why is it that $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$?

It is a rather simple question:
In my textbook it writes something like: $$\frac {\partial \Psi} {\partial t}= \frac{i\hbar}{2m}\frac {\partial^2 \Psi} {\partial x^2}- \frac{i}{\hbar}V\Psi$$
$$\frac {\partial \Psi^*} {\partial t}= -\frac{i\hbar}{2m}\frac {\partial^2 \Psi^*} {\partial x^2}+\frac{i}{\hbar}V\Psi^* $$

I don't understand why, because I have never learned the topic of complex function. My educated guess is as follows:
Assume that $\Psi^{*}(x,t) = |Z|e^{i\omega t}$, and $\Psi(x,t) = |Z|e^{-i\omega t}$. Obviously $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$, which then can be generalized into any case. Am I right?

 

[Tony Hau]

That's not quite right in general. You actually need the properties that $(v+w)^* = v^* + w^*$, $(vw)^* = v^* w^*$, $i^* = -i$ and finally that the conjugate commutes with the partial derivative.

So how does it apply to the wave function case? Maybe I shouldn't generalize that $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$, but that happens in my textbook...

 

[martinbn]

So how does it apply to the wave function case? Maybe I shouldn't generalize that $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$, but that happens in my textbook...

No, that is not what happens. You take complex conjugate of both sides of the first equation, that gives you the second.

 

[vanhees71]

Just write down the Schrödinger equation. It's much better to keep the usual form, where you have the self-adjoint Hamiltonian on the right-hand side, i.e.,
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi(t,\vec{x})+V(\vec{x}) \psi(t,\vec{x}).$$
Now take the conjugate complex. Since $t$ and $\vec{x}$ are real you immediately get
$$-\mathrm{i} \hbar \partial_t \psi^*(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi^*(t,\vec{x}) + V(\vec{x}) \psi^*(t,\vec{x}).$$
The equation in the summary of #1 is obviously wrong!

 

[Tony Hau]

Just write down the Schrödinger equation. It's much better to keep the usual form, where you have the self-adjoint Hamiltonian on the right-hand side, i.e.,
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi(t,\vec{x})+V(\vec{x}) \psi(t,\vec{x}).$$
Now take the conjugate complex. Since $t$ and $\vec{x}$ are real you immediately get
$$-\mathrm{i} \hbar \partial_t \psi^*(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi^*(t,\vec{x}) + V(\vec{x}) \psi^*(t,\vec{x}).$$
The equation in the summary of #1 is obviously wrong!

How do you take the complex conjugate of a function? I know how to take a complex conjugate of a complex number $z$. For example, for $z= 1 + 2i$, its conjugate is $z^* = 1-2i$. Forgive me but my complex number knowledge stops there.

 

[PeroK]

How do you take the complex conjugate of a function? I know how to take a complex conjugate of a complex number $z$. For example, for $z= 1 + 2i$, its conjugate is $z^* = 1-2i$. Forgive me but my complex number knowledge stops there.

A complex-valued function, $f(x)$, of a real variable can be written:$$f(x) = a(x) + ib(x)$$ The complex conjugate of the function is simply: $$f^*(x) = a(x) - ib(x)$$

 

[strangerep]

I have never learned the topic of complex function.

Mate, trust me when I tell you that if you're trying to study QM without prior knowledge of complex variables, you're pretty-much stuffed.

Get thee to a complex variables textbook NOW. I suggest "Schaum Outline of Complex Variables", which can be had quite cheaply (new and used) via Amazon. This textbook covers a lot of material quickly and efficiently, with many examples.

 

[Tony Hau]

Mate, trust me when I tell you that if you're trying to study QM without prior knowledge of complex variables, you're pretty-much stuffed.

Get thee to a complex variables textbook NOW. I suggest "Schaum Outline of Complex Variables", which can be had quite cheaply (new and used) via Amazon. This textbook covers a lot of material quickly and efficiently, with many examples.

Good idea. You know my school's education sucks. It even doesn't teach Fourier Series and Fourier transform, which I am self-learning during this winter break. I am using the Mathematical methods for physical science of Boas, which my school uses for teaching ODEs.

 

[strangerep]

Good idea. You know my school's education sucks. It even doesn't teach Fourier Series and Fourier transform,

which I am self-learning during this winter break.

Good, but be aware you'll need knowledge of complex variables (and contour integration) in order to make good progress with the Fourier transform (which is also crucial in many areas of physics).

 

[vanhees71]

For Fourier transformations and generalized functions, I recommend

M. Lighthill, Introduction to Fourier analysis and generalised
functions, Cambridge University Press (1958).

 

[Thomas Michael]

Summary:: Given a wavefunction $\Psi$, why is it that $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$?

It is a rather simple question:
In my textbook it writes something like: $$\frac {\partial \Psi} {\partial t}= \frac{i\hbar}{2m}\frac {\partial^2 \Psi} {\partial x^2}- \frac{i}{\hbar}V\Psi$$
$$\frac {\partial \Psi^*} {\partial t}= -\frac{i\hbar}{2m}\frac {\partial^2 \Psi^*} {\partial x^2}+\frac{i}{\hbar}V\Psi^* $$

I don't understand why, because I have never learned the topic of complex function. My educated guess is as follows:
Assume that $\Psi^{*}(x,t) = |Z|e^{i\omega t}$, and $\Psi(x,t) = |Z|e^{-i\omega t}$. Obviously $\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}$, which then can be generalized into any case. Am I right?

Normally, it would look something like: $$i\hbar \frac {\partial \Psi} {\partial t}= \frac{\hbar^2}{2m}\frac {\partial^2 \Psi} {\partial x^2}+ V\Psi$$

And then the complex conjugate of $$i\hbar$$

would just be $$-i\hbar$$

thereby changing all the signs on the right hand side as well.

 

[vanhees71]

The correct equation is
$$\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2}{2m} \Delta \psi + V \psi.$$
Here all quantities except $\mathrm{i}$ and $\psi$ are real. Thus complex conjugation gives no sign change on the right-hand side but just $\psi$ becomes $\psi^*$. So you have
$$-\mathrm{i} \hbar \partial_t \psi^*=-\frac{\hbar^2}{2m} \Delta \psi^* + V \psi^*.$$

 

[Tony Hau]

The correct equation is
$$\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2}{2m} \Delta \psi + V \psi.$$
Here all quantities except $\mathrm{i}$ and $\psi$ are real. Thus complex conjugation gives no sign change on the right-hand side but just $\psi$ becomes $\psi^*$. So you have
$$-\mathrm{i} \hbar \partial_t \psi^*=-\frac{\hbar^2}{2m} \Delta \psi^* + V \psi^*.$$

I have asked one of my frds from math department and obtained the following proof. Forgive me for uploading only two piece of paper because I would like to spare my effort in typing latex.

.

I do not understand why taking the complex conjugate seems so intuitive and obvious to you guys.

 

[PeroK]

I have asked one of my frds from math department and obtained the following proof. Forgive me for uploading only two piece of paper because I would like to spare my effort in typing latex.
I do not understand why taking the complex conjugate seems so intuitive and obvious to you guys.

That's far too long winded. All you really need to show (if you don't already know this result) is that: $$(\frac{\partial \Psi}{\partial t})^* = (\frac{\partial \Psi^*}{\partial t}) \ \ \text{and} \ \ (\frac{\partial^2 \Psi}{\partial x^2})^* = (\frac{\partial^2 \Psi^*}{\partial x^2})$$ And the rest follows from taking the complex conjugate of an equation.

The point being that by the time you come to study QM, you shouldn't need to prove that, say, $(z_1z_2 + w)^* = z_1^*z_2^* + w^*$ from first principles. That's a basic result from complex numbers that you should know and apply without further ado.

If you are learning complex numbers for the first time, then it's a different matter, but Griffiths will expect you to know at least the basics.

 

[Tony Hau]

That's far too long winded. All you really need to show (if you don't already know this result) is that: $$(\frac{\partial \Psi}{\partial t})^* = (\frac{\partial \Psi^*}{\partial t}) \ \ \text{and} \ \ (\frac{\partial^2 \Psi}{\partial x^2})^* = (\frac{\partial^2 \Psi^*}{\partial x^2})$$ And the rest follows from taking the complex conjugate of an equation.

The point being that by the time you come to study QM, you shouldn't need to prove that, say, $(z_1z_2 + w)^* = z_1^*z_2^* + w^*$ from first principles. That's a basic result from complex numbers that you should know and apply without further ado.

If you are learning complex numbers for the first time, then it's a different matter, but Griffiths will expect you to know at least the basics.

Thanks for your hints. That's much much better. Honestly I don't know what it means by learning complex number. I have taken a course(maths for physics student I, you can say) where I was taught the Euler's formula, the complex equation for natural logarithm, sine and cosine. But obviously I have never seen useful identities such as these: $(z_1z_2 + w)^* = z_1^*z_2^* + w^*$, $(\frac{\partial \Psi}{\partial t})^* = (\frac{\partial \Psi^*}{\partial t})$, etc. Where can you find a very quick reference to these identities? I really think that Griffiths should include a chapter on complex number, just like what he does for his EM book by introducing a chapter on the vector calculus.

 

[PeroK]

Where can you find a very quick reference to these identities?

You could do something extravagent like googling for "complex conjugate identities".

 

[vanhees71]

It's just a nice exercise to prove for yourself that, e.g.,
$$(a b)^*=a^* b^*$$
by decomposing both sides of the equation in real and imaginary parts.

 

[Tony Hau]

It's just a nice exercise to prove for yourself that, e.g.,
$$(a b)^*=a^* b^*$$
by decomposing both sides of the equation in real and imaginary parts.

That's easy. $$ (e^{i\theta_1}\cdot e^{i\theta_2})^* = (e^{i(\theta_1 +\theta_2)})^*$$
$$= e^{-i(\theta_1 +\theta_2)}$$
$$=e^{-i\theta_1}\cdot e^{-i\theta_2}$$
$$=e^{i\theta_1*} \cdot e^{i\theta_2 *}$$
What really surprises me is the complex conjugate on the partial derivatives.

 

[vanhees71]

The derivative is taken wrt. a real quantity like $t$ or $x$. from this you get
$$\left (\frac{\partial \psi}{\partial t} \right)^*=\frac{\partial \psi^*}{\partial t}.$$
This follows from the general rules by thinking of the derivative in terms of the limit of a difference coefficient, and of course again
$$\left (\frac{a}{b} \right)^*=\frac{a^*}{b^*}.$$

 

[PeroK]

What really surprises me is the complex conjugate on the partial derivatives.

We have $$\psi(x, t) = \psi_1(x, t) + i \psi_2(x, t)$$ and $$\frac{\partial \psi}{\partial t} = \frac{\partial \psi_1}{\partial t} + i\frac{\partial \psi_2}{\partial t}$$ and I'll leave the rest to you.

 

[PeterDonis]

Forgive me for uploading only two piece of paper because I would like to spare my effort in typing latex.

Sorry, but "I would like to spare my effort" is not a valid excuse. PF rules require equations to be typed directly in the post using LaTeX. If you want discussion of the equations, you need to type them in.

 

[Tony Hau]

We have $$\psi(x, t) = \psi_1(x, t) + i \psi_2(x, t)$$ and $$\frac{\partial \psi}{\partial t} = \frac{\partial \psi_1}{\partial t} + i\frac{\partial \psi_2}{\partial t}$$ and I'll leave the rest to you.

$$(\frac {\partial \psi} {\partial t})^* = (\frac {\partial \psi_{1}} {\partial t} + i \frac {\partial \psi_2} {\partial t})^* $$
$$=(\frac {\partial \psi_{1}} {\partial t})^* + ( i \frac {\partial \psi_2} {\partial t})^*$$
$$=\frac {\partial \psi_{1}} {\partial t} - i\frac {\partial \psi_2} {\partial t}$$, because $\frac {\partial \psi_{1}} {\partial t}$ is just a real number, and and the complex conjugate of $i\frac {\partial \psi_2} {\partial t}$ is the negative of itself.
$$=\frac {\partial \psi^*} {\partial t}$$

 

[Tony Hau]

We have $$\psi(x, t) = \psi_1(x, t) + i \psi_2(x, t)$$ and $$\frac{\partial \psi}{\partial t} = \frac{\partial \psi_1}{\partial t} + i\frac{\partial \psi_2}{\partial t}$$ and I'll leave the rest to you.

By the way, where did you guys learn these from during your ug level study? Did you guys actually take a math course on complex analysis?

 

[strangerep]

By the way, where did you guys learn these from during your ug level study? Did you guys actually take a math course on complex analysis?

In Australia, I started to learn the basics of complex numbers at the start of year 11 in high school! More advanced complex variables theory followed in early university math courses.

I'm actually kinda shocked by your situation, which is why I made the urgent textbook suggestion in post #7. (Have you obtained, or ordered it yet?)

And I'm starting to understand why wealthy Asian parents often pay vast sums of money to send their kids to western universities (where the unfortunate kids often have to grind out a few months doing intensive catch-up courses before the main semester begins).

But, more than that, I'm actually sympathetic with your situation because I experienced something similar regarding basic group theory: I was allowed to drift into an unsuitable (i.e., too-easy) ug math course at my university, which didn't cover group theory, and then later it suddenly became essential in certain physics courses. When I realized this, I was furious. But there was nothing to be done, except to grab some textbooks and go into head-down tail-up mode for a while in order to catch up.

 

[Tony Hau]

In Australia, I started to learn the basics of complex numbers at the start of year 11 in high school! More advanced complex variables theory followed in early university math courses.

I'm actually kinda shocked by your situation, which is why I made the urgent textbook suggestion in post #7. (Have you obtained, or ordered it yet?)

And I'm starting to understand why wealthy Asian parents often pay vast sums of money to send their kids to western universities (where the unfortunate kids often have to grind out a few months doing intensive catch-up courses before the main semester begins).

But, more than that, I'm actually sympathetic with your situation because I experienced something similar regarding basic group theory: I was allowed to drift into an unsuitable (i.e., too-easy) ug math course at my university, which didn't cover group theory, and then later it suddenly became essential in certain physics courses. When I realized this, I was furious. But there was nothing to be done, except to grab some textbooks and go into head-down tail-up mode for a while in order to catch up.

Yes, I have found the online version of your book. I will work through it later because I have just finished the Fourier series and transform.

The problem of Hong Kong's educational system is not at the university level, but at the secondary school level. At secondary school, except for a few who are really good at maths, it is not required to study calculus, integration etc, let alone complex number arithmetics. Although I was lucky enough to enroll into an extended maths class, where I learned the the advanced maths like calculus, I didn't pay much effort into it and eventually got a near pass. The point is that this advanced maths course is an extention to the normal maths class and most of the universities do not consider it as a valid subject. So there is no incentive for us to score high.

Admitting students a majority of whom do not understand advanced maths, university at year 1 can only teach students basic calculus and integration without going deep into other maths. So it is not really the university that is to blame; it is the secondary school system to blame.

Yea many wealthy kids in Hong Kong go overseas to study, mainly the UK and Australia. But I really doubt if they can articulate well to those universities, since these students have never learned the A2 level before. At most Hong Kong's syllabus is equivalent to the AS level. Anyway, I sometimes envy those kids who have a wealthy and kind family sending them to foreign countries...

 

[strangerep]

Yes, I have found the online version of your book. I will work through it later

No! Start working through it now, in parallel with your other courses.
I know that might seem like a lot of work all at once, but you will never have more energy (nor brain plasticity, i.e., ability to learn) than you do now in your youth. (I often wish I could reach back through time and give my teenage self a good kick up the butt.)

because I have just finished the Fourier series and transform.

Hmmm, so you think you've "finished" the Fourier transform?

The problem of Hong Kong's educational system is not at the university level, but at the secondary school level. At secondary school, except for a few who are really good at maths, it is not required to study calculus, integration etc, let alone complex number arithmetics.

 

[PeterDonis]

where did you guys learn these from during your ug level study? Did you guys actually take a math course on complex analysis?

I took one as a sophomore in college--and realized that I had made the mistake of taking it from the math department when I wasn't a math major. I ended up having to take a different course in complex variables as a junior, which was much more suitable for physics and engineering majors--in other words, it taught you the methods you needed to know without getting bogged down in a lot of advanced and esoteric theory which was no doubt very interesting to math majors but which I couldn't make head or tail of.

 

[Tony Hau]

No! Start working through it now, in parallel with your other courses.
I know that might seem like a lot of work all at once, but you will never have more energy (nor brain plasticity, i.e., ability to learn) than you do now in your youth. (I often wish I could reach back through time and give my teenage self a good kick up the butt.)Hmmm, so you think you've "finished" the Fourier transform?

At least the transform covered in the Boas book...

 

[vanhees71]

Well, I think the high school-math education went down the hill worldwide. If I read this forum in this connection everything sounds familiar from the experience we have in Germany with the incoming 1st-semester students. It's clear that usually people start to study physics who liked math and physics at school and were good at it.

The problem is that there was a change in attitude by didactics people beginning around the mid-1990ies and then amplified by the socalled "PISA shock". This was the realization that the German level of education in STEM was below the average. The response to this, however, was in my opinion a complete desaster. In Germany the didactics people call it "competence learning", which however is an euphemism for the opposite of what's desired from the point of view of the STEM people, i.e., one propagated rote learning to solve standard problems in math and the sciences without providing a true understanding let alone enabling the students to be creative and versatile in using mathematics as a language to express real-world phenomena and a way of thinking logically to build up knowledge by understanding the connections between different subjects of the natural sciences.

So you are not alone with this problem, and you have to keep going in learning all the needed math for your physics studies either in lectures (usually the problem is that you need math in the physics courses that have not been covered yet in the standard math courses).

For quantum theory for sure you need a good deal of (multi-variable) calculus, including partial differential equations, linear algebra, and some probability theory.

Lie groups and algebras are very helpful to understand why the various operators representing observables look as they look in quantum mechanics. Behind this is Noether's theorem, which should also be covered in the classical-mechanics theory lecture. Closest to quantum mechanics is the formalism in the Hamiltonian version of the least-action principle in terms of Poisson brackets (which can be used as a realization of the Lie algebra of the Galilei group in terms of phase-space functions, explaining on a deeper level why the dynamical laws of Newtonian mechanics look as they look because of the underlying geometry of Galilean spacetime).

 

[Nugatory]

(I often wish I could reach back through time and give my teenage self a good kick up the butt.)

I feel the same way about my younger self.
Youth is wasted on the young. They aren't old enough to appreciate it.