Proof of time independence for normalization of wavefunction

[Jimmy87]

Hi pf, I am having trouble with understanding some of the steps involved in a mathematical proof that a normalized wavefunction stays normalized as time evolves. I am new to QM and this derivation is in fact from "An introduction to QM" by Griffiths. Here is the proof:

I am fine with most of the maths but I am struggling with a few points:

Equation 21. He refers to this later and says "Ordinarily we need to justify taking a derivative inside an integral, but we can console ourselves with the knowledge that for any wave function for which this is not a valid operation, that function is also not a possible description of a physical state (for one reason or another)".

Why do you need to justify taking a partial derivative inside an integral? Is this not normally a valid operation?

Equation 25. I really don't understand how you can set a second derivative of a function equal to a first derivative? He says further on that the expression in the brackets at the start of equation 25 can be expressed as a single derivative - that doesn't make any sense to me why you can do this?

Equation 26. Why is there no integral sign on the right hand side of the equation? The expression on the right hand side is the partial derivative of psi squared with respect to time (as shown in equation 22) whereas they are implying the right hand side of equation 26 is the integral of the partial derivative with respect to time of psi squared (i.e. the integration has been done hence no integral sign)?

Thanks for any help!

 

[Haborix]

21: By taking the derivative inside the integral you require certain assumptions on the continuity of the function and its derivatives. If the function didn't satisfy these conditions then changing the order of integration and differentiation would not hold.

25: It's basically inverting the product rule. Take the derivative on the RHS of 25 to convince yourself it is the same expression.

The reason to do the previous step is that you would like to appeal to the first fundamental theorem of calculus. By integrating a derivative of a function you are able to just evaluate the function at the end points. This leads to 26. The integral has been done using the first fundamental theorem (note the evaluation at the boundary).

 

[Jimmy87]

21: By taking the derivative inside the integral you require certain assumptions on the continuity of the function and its derivatives. If the function didn't satisfy these conditions then changing the order of integration and differentiation would not hold.

21. Thanks.

25: It's basically inverting the product rule. Take the derivative on the RHS of 25 to convince yourself it is the same expression.

The reason to do the previous step is that you would like to appeal to the first fundamental theorem of calculus. By integrating a derivative of a function you are able to just evaluate the function at the end points. This leads to 26. The integral has been done using the first fundamental theorem (note the evaluation at the boundary).

Thanks for your help.
25. I think I get what you are saying. On the RHS there is a d/dx and if you carry this out then you get back to the LHS. However, this d/dx seems to disappear in equation 26 and without the d/dx surely it is not equal to the LHS of equation 25?

26. I still don't follow. The RHS of this equation hasn't been integrated. The limits have just been stuck on the end. From how I see it, its like saying the integral of (x^2) evaluated at two limits is (x^2) with the limits added whereas the integral of x^2 is (x^3/3) evaluated at those limits.

Unless integrating the RHS of equation 25 makes the d/dx go away? Is that what you are saying? I am just making guesses based on the outcomes.

 

[Haborix]

The d/dx does disappear and that is because you have integrated the expression. Notice that the LHS of 26 is the integration of the LHS of 25. Similarly the integration of the RHS of 25 has already been done when it is written in 26. This was done using the following: $\int_a^bF'(x) dx=F(x)|_a^b$

 

[Jimmy87]

The d/dx does disappear and that is because you have integrated the expression. Notice that the LHS of 26 is the integration of the LHS of 25. Similarly the integration of the RHS of 25 has already been done when it is written in 26. This was done using the following: $\int_a^bF'(x) dx=F(x)|_a^b$

Thanks again. So on the RHS of 26 do you insert the limits where the curly d's are? So if you were to write it out fully when you evaluate the limits would it be:

I have called them a and b for plugging in the limits + and - infinity. And both of these equations would be equal to zero, is that right?

Could you also confirm the reason for having an integration and a partial derivative at the same time. I believe it is because the normalization needs an integration since it is the integral over all of space of psi squared. The reason for the partial derivative is because you want to see how the normalized wave function evolves with time. Is that the right way to think about it? Do you recommend any sources that discuss integrating derivatives because we haven't covered it and the book seems to assume the reader knows how to do this. All the other maths I have done but I have never come across integrals involving partial derivatives.

 

[vanhees71]

I don't understand what the infinity symbols should mean in the just quoted equations. I guess, it's just the usual proof of the conservation of the norm of the wave function. It follows from unitarity of the time evolution, i.e., the self-adjointness of the Hamiltonian. In the position representation ("wave mechanics"), you just use Schrödinger's equation
$$\hbar \mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H}(t,\vec{x}).$$
The norm is
$$\langle \psi|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(t,\vec{x})|^2.$$
The time derivative is
$$\frac{\mathrm{d}}{\mathrm{d} t} = \langle \psi|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} [\psi^{*} \partial_t \psi+(\partial_t \psi)^* \psi] =
(\partial_t \langle \psi|)|\psi \rangle + \langle \psi|\partial_t \psi \langle=\langle -\mathrm{i} \hat{H}/\hbar|psi \rangle + \langle \psi|-\mathrm{i} \hat{H}/\hbar \rangle = \langle \psi|+\mathrm{i} \hat{H}/\hbar \rangle+ \langle \psi|-\mathrm{i} \hat{H}/\hbar \rangle=0.$$