## A Conversation for Quantum Mechanics

### Erm....have I misunderstood........?

Stephen Started conversation Aug 14, 2003

In the normalization process shouldnt the imaginary part of the function be multplied by i, sqroot -1, rather that -1 as stated in note 6?

Also, can the integral of a function be 1? My dim memories of A level maths are that the integral of any function includes c, an unspecified constant..and if the integral is just c, must the funtion itself not be a first degree function? (sounds simple to my untutored brain but probably isn't!)

Im confused.

Hey! This is a great article though! Fascinating and really clear!

### Erm....have I misunderstood........?

Cefpret Posted Aug 14, 2003

You are wrong in both cases. (That's typical of questions about QM )

The complex conjugate is yielded by multiplying its imaginery part -- and only it -- by -1. Thus the product of a complex number and its conjugate is real. And the conjugate of a real number is itself.

As far as the integral is concerned, we have limits here (although they are infinite). In this case the constant c is eliminated. So the primitive (do you call it like this?) of f(x)=x^2 is x^3/3+c, but the integral of f(x) from 0 to 2 is 2^3/3+c-0^3/3-c = 8/3.

### Erm....have I misunderstood........?

U195408 Posted Aug 14, 2003

I think you nailed both on the head Cefpret. Just to make in explicit, there are two operations. The first is taking the complex conjugate, the second is normalization/integration, both of which are exactly described by Cefpret.

thanks for reading! glad you enjoyed it.

dave

### Erm....have I misunderstood........?

U195408 Posted Aug 14, 2003

I think you nailed both on the head Cefpret. Just to make in explicit, there are two operations. The first is taking the complex conjugate, the second is normalization/integration, both of which are exactly described by Cefpret.

thanks for reading! glad you enjoyed it. BTW - I lost track of the number of times that I was told "both are wrong" when I was posing a question or hypothesis while I was taking QM.

dave

### Erm....have I misunderstood........?

Stephen Posted Aug 15, 2003

Thank you both.

The integral being 1 I now understand...I'm familiar with integration between limits of course but did not understand the statement that way. All is now clear.

The multipilcation by -1 bit I'm still less sure about...

Let the imaginary part of the number be xi

xi*(-1)=x(i)^3= x(-i) which is still imaginary isn't it?

Am I being thick here?

### Erm....have I misunderstood........?

Cefpret Posted Aug 15, 2003

The complex conjugate of an imaginery number is purely imaginery again. That's a consequence of the definition.

Notice that there are two important mathematical expressions in wave mechanics that are totally equal:

1. |psiĀ²| and

2. psi*(complex conjugate of psi).

Both are *real*, and must be, because they denote a measurable quantity (probability density of spatial position). The latter seems to be more complicated, but it's important for quantities other than spatial position, because then an operator stands between the factors.

### Erm....have I misunderstood........?

U195408 Posted Aug 15, 2003

Stephen - you are right on about the complex conjugate. But what you then have to do is multiply your complex conjugate by the original wavefunction. Let me give you an example:

a*i is our imaginary number

-a*i is the complex conjugate of our imaginary number

(-a*i)*(a*i) is the complex conjugate times the original

=-(a^2)*(i^2)

=+a^2

Here's a more complicated example:

b + a*i is our complex number (with a real & imaginary part)

b - a*i is the complex conjugate of our imaginary number

(note how just the imaginary part was made negative)

(b + a*i)*(b - a*i) is the complex conjugate times the original

=b^2 + (-b*a*i) + a*i*b - (a^2)*(i^2) multiply it out - the middle

=b^2 + a^2 terms cancel out

Does this help? The second part may be more than you wanted!

dave

### Erm....have I misunderstood........?

U195408 Posted Aug 15, 2003

ps. I forgot to add, that the end result of each calculation is a real (no imaginary parts!) number.

dave

### Erm....have I misunderstood........?

Stephen Posted Aug 18, 2003

Thanks Dave and Cefpret. Yes I now understand the maths! I still have to come to terms with the concept as a whole but I am working on it.

Stephen

### Erm....have I misunderstood........?

U195408 Posted Aug 19, 2003

That's where the real fun begins! Have fun!

dave

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