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For the hydrogen atom, if $$\int |u|^2 ~dx = 1,$$ at $t = 0$, I am trying to show that this is true at all later times.

What I need help is with differentiating the integral with respect to $t$, and taking care about the solution being complex valued. Except that my notation is getting me mixed up. I think this might get me there.

Following Ben's hint, here is what I have:

  1. Change $|u|^ 2$ into $u^* u$.
  2. Bring the derivative inside the integral.
  3. Apply the product rule.
  4. Apply the Schrödinger equation and try to show that the result is zero.

$$\int u^* u ~dx = 1 $$

and to bring the derivative inside the integral, isn't $dx$ already inside?

From Schrödinger equation I have:

$$-i\hslash u_t = \sum_{i=1}^n \frac{\hslash^2}{2m_i}(u_{x_i x_i} + u_{y_i y_i} + u_{z_i z_i}) + V(x_1,\ldots,z_n)u$$

for $n$ particles and the potential would $V$ depend on all of the $3n$ coordinates.

I'm not sure how to extend it to even 2 dimensions with the notation below

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Thanks Henry for the edit –  mary Sep 1 '12 at 22:49
    
Please define your notation. $u$ isn't the wavefunction, is it? Or if it is, then don't you want the integral to equal 1 rather than 0? As written, your equation would imply $u(x,0)=0$ for all $x$. I'm assuming this is homework, so I'm tagging it that way. –  Ben Crowell Sep 1 '12 at 23:34
    
u is the wave function. This isn't homework, but extra practice. Also, in the absolute value it should be absolute value(u)^2 dx –  mary Sep 2 '12 at 0:02
    
OK, and the integral equals 1, which makes more sense. This expresses the fact that the total probability of finding the electron somewhere must be 1. Physically, this isn't something specific to the hydrogen atom, so this suggests that mathematically, your method of solution shouldn't depend on any of the details of the specific potential that applies in this case. It's also true in any number of dimensions, so you might want to warm up by doing it in one dimension. –  Ben Crowell Sep 2 '12 at 0:11
2  
So in broad strokes: (1) Change $|u|^2$ into $u^*u$. (2) Bring the derivative inside the integral. (3) Apply the product rule. (4) Apply the Schrodinger equation and try to show that the result is zero. This should be easier in one dimension. Then you can try to generalize to three. –  Ben Crowell Sep 2 '12 at 0:21

5 Answers 5

Wave function and time: With a potential (V is the potential) that does not depend on time, a general technique is to see what one gets when one writes the wave function as a product of a function only dependent on time, and another function only dependent on place. The Schrodinger equation then gives the result that the complex conjugate of the time-dependent part (up to a constant) equals the reciprocal. One needs to know what the derivative of the exponential function is.

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I will do for you the case of one particle in one dimension. See if you can do the many particles in three dimension case yourself.

It suffices to show $$ \frac{d}{dt} \int u^*u \mathrm{d}x = 0 $$

Bringing the derivative inside the integral sign you need to compute $$ \int \partial_t(u^*) u + u^* \partial_t u \mathrm{d}x \tag{1}$$ Now we plug in Schroedinger's equation $$ \partial_t(u) = \frac{i \hbar}{2m} u_{xx} + \frac{i}{\hbar} V u $$ which implies, by taking the complex conjugate (we assume that $V$ is a real valued potential; otherwise the conservation law may not be true) $$ \partial_t (u^*) = - \frac{i \hbar}{2m} u^*_{xx} - \frac{i}{\hbar} V u^* $$ So expression (1) can be rewritten as $$ \int \frac{i}{\hbar}\left( V |u|^2 - V |u^2|\right) + \frac{i \hbar}{2m} \left(u^* u_{xx} - u^*_{xx} u\right) \mathrm{d}x \tag{2}$$ The terms inside the first set of parentheses clearly cancel. We treat the second set of parentheses by integration by parts. This shows that the terms inside the second parentheses cancel each other. From which we conclude that $$ (1) = (2) = 0 $$ as desired.


Now, one quick word on the idea behind the computation above: the magic all sits behind the $i$ factor in the $i\partial_t$ of Schroedinger's equation. Observe the right hand side is given by some real valued function times the wave-function $u$ itself (okay, there is also the Laplacian term; we need that the Laplacian is "self-adjoint", which means that as an operator it is "real-valued"). If we divide the $i$ over the the right hand side, the equation is schematically $$ \partial_t u = i \times \text{something real} \times u $$ Now using our elementary complex analysis, we know that "$i$ times a complex number $z$ is orthogonal to $z$ itself", and this is what, when all is said and done, allows you to say that the $L^2$ norm of $u$ is conserved under the Schroedinger flow.

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In (2) I think you need another set of parens around the whole integrand. –  Ben Crowell Sep 9 '12 at 3:13
    
@Ben huh? It is customarily understood that everything between $\int \ldots \mathrm{d}x$ is the integrand. Why complicate notations? –  Willie Wong Sep 10 '12 at 9:01

Wave function and time: we put a possible solution in the form $u(t, x_i,y_i,z_i)=w(t)S(x_i,y_i,z_i).$ If we write the Schrodinger equation as $$ -i \hbar u_t = f(u) + V(x_i, y_i, z_i) u$$ and put in this $u$, then $$- i \hbar w_t(t) S(x_i,y_i,z_i) = w(t) f(S(x_i,y_i,z_i))+ V(x_i,y_i,z_i)w(t)S(x_i,y_i,z_i)$$ now "almost everywhere" where $u \not= 0$ we may divide by $u$ on both sides, and get $$ -i \hbar \frac{w_t(t)}{w(t)} = \frac{f(u)}{S(x_i,y_i,z_i)} + V(x_i,y_i,z_i)=E(x_i,y_i,z_i)$$ As RHS is independent of $t$, LHS is constant, and that differential equation has the exponential function as solution, $$w(t) = \exp^{i \frac{E}{\hbar} t}$$ Among other things we get the result that $|w(t)|^2$ = 1, and then $|u(t,x_i,y_i,z_i)|^2= |S(x_i,y_i,z_i)|^2$ and the integral is likewise independent of time. (The constant $\hbar$ has dimension energy $\times$ time, so $E$ is actually an energy).

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This answer addresses the case of wave functions with separate variables: $u(t,\mathbf{x})=w(t)S(\mathbf{x}).$ What about the general case? –  Giuseppe Negro Sep 4 '12 at 9:56

For a beginning, define $3n$ orthonormal basis vectors with respect to an inner product "$\cdot$", $\hat x_j, \hat y_j, \hat z_j$ in a real vector space, and define $$ \nabla = \sum_j \frac{\hbar}{\sqrt{2 m_j}}\left(\hat x_j \frac{\partial}{\partial x_j}+ \hat y_j \frac{\partial}{\partial y_j}+ \hat z_j\frac{\partial}{\partial z_j} \right) $$ Then $$\nabla \cdot \nabla u = \sum_j \frac{\hbar^2}{2 m_j}\left(u_{x_i x_i} + u_{y_i y_i} + u_{z_i z_i} \right) $$ The Schrodinger equation and its complex conjugate, multiplied with $u^*$ and $u$ respectively gives
$$-i \hbar u_t u^* = (\nabla \cdot \nabla u ) u^* + V u u^* $$ and $$ i\hbar u_t^* u = (\nabla \cdot \nabla u^*)u + V u^* u $$. From this, $$u_t u^* + u_t^* u = \frac{1}{i \hbar} \left( -(\nabla \cdot \nabla u)u^* + (\nabla \cdot \nabla u^*)u \right). $$ Right hand side can be written $$\frac{1}{i \hbar} \nabla \cdot ((\nabla u)u^* + (\nabla u^*)u) =\nabla \cdot F(u).$$ When $W$ is a volume in $3n$-dimensional space independent of time, then $$ \frac{\partial}{\partial t}\int_Wu^* u \;\; d w = \int_W u_t u^* + u u_t^* \;\; dw = \int_W \nabla \cdot F(u) \;\; dw .$$ From the divergence theorem, assuming that the surface $A$ of the volume $W$ is so far out that the wavefunction $u$ is zero there, and then also $F(u),$ the last expression equals $$ \int_A F(u) \cdot da = 0.$$ That is, $ \int_W |u|^2 dw$ is independent of time.

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As I understand this question, it really has nothing to do with the Hydrogen atom or even the dimensionality of the space involved. The Schrodinger equation is

$$H|\psi\rangle=i\hbar\frac{\partial|\psi\rangle}{\partial t}$$

And your question is "how do I know that my wave function $|\psi\rangle$ remains normalized as time goes on?" We have

$$\frac{d}{dt}\langle\psi|\psi\rangle=\frac{\partial}{\partial t}\langle\psi|\psi\rangle=\frac{\partial\langle\psi|}{\partial t}|\psi\rangle+\langle\psi|\frac{\partial|\psi\rangle}{\partial t}$$

The quantity $\partial\langle\psi|/\partial t$ is the complex conjugate of $\partial|\psi\rangle/\partial t$. Due to the $i$ in Schrodinger's equation and the fact that $H$ is hermitian (specifically that it acts equally well on the left as on the right), we find

$$\frac{d}{dt}\langle\psi|\psi\rangle=-\frac{1}{i\hbar}\langle\psi|H|\psi\rangle+\frac{1}{i\hbar}\langle\psi|H|\psi\rangle=0,$$

so that the wave function remains normalized at all times. This is a fundamental result, relying only on the Schrodinger equation and the hermiticity of $H$.

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I agree with this point of view. I would explicitly mention self-adjointness of $H$, which is what makes things work: as you know, if $H$ were not self-adjoint, we would have $$\frac{d}{dt}\langle \psi|\psi\rangle = \frac{1}{i \hbar}\left(\langle \psi|H|\psi\rangle - \langle\psi|H|\psi\rangle^\star\right),$$ and this, in general, would be different from $0$. –  Giuseppe Negro Sep 6 '12 at 19:28
    
True, I guess I assumed it was implicit in a discussion of QM, but I will add it the post. Thanks. –  Jonathan Sep 6 '12 at 19:45

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