# Integration by Parts To Derive Expectation Value of Velocity

1. Jul 3, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Why can't you do integration-by-parts directly on the middle expression in equation 1.29--pull out the time derivative over onto x, note that $\displaystyle \frac{\partial x}{\partial t} = 0$, and conclude that $\displaystyle \frac{d \langle x \rangle }{dt} = 0$

2. Relevant equations

Equation 1.29 $\displaystyle \frac{d \langle x \rangle }{dt} = \int x \frac{\partial}{\partial t} | \psi |^2 dx = \frac{i \hbar}{2m} \int x \frac{\partial }{\partial x} \left( \psi^* \frac{\partial \psi}{\partial x} - \frac{\partial \psi^*}{\partial x} \right) dx$

3. The attempt at a solution

Here is my solution, which is not consistent with the one found in the answer key:

The notation $\displaystyle \frac{\partial x}{\partial t} = 0$ expresses two things: one is that the position of the particle depends on things other than time; the second is that the position of the particle does not change with time, which is not true, in general. One could make such an assumption, but it would be a rather restricting one; the result you would derive from such an assumption would be that the particle never changes its position, and we would not even need such sophisticated mathematics and physics to predict/describe the behavior of the system.

For integration by parts, the differentiation has to be with respect to the integration variable – in this case the differentiation is with respect tot, but the integration variable is x. It’s true that

$\displaystyle \frac{\partial }{\partial t} (x |\psi |^2 )= \frac{\partial x}{\partial t} |\psi |^2 + x \frac{\partial }{\partial t} |\psi |^2$

but this does not allow us to perform the integration:

$\displaystyle \int_a^b x \frac{\partial }{\partial t} {\psi }^2 dx = \int_a^b (x |\psi |^2)dx = (x |\psi|^2)|_a^b$

Is my reasoning incorrect?

2. Jul 3, 2014

### dauto

The answer key is correct. Your answer makes no sense. ∂x/∂t = 0 is just an expression of the fact that x and t are independent variables. That's what's meant by a partial derivative. You keep x constant while deriving with respect to t. Well, if x is constant than its derivative with respect to t is zero by definition. x is not the position of the particle. It is a coordinate.

3. Jul 3, 2014

### Bashyboy

Yes, but can't the position be a function of time?

4. Jul 3, 2014

### Bashyboy

I have another question. In equation 1.29, how come the author slipped the partial derivative operator $\frac{\partial}{\partial t}$ past x? As I said, couldn't x be a function of time, $x = x(t)$, thereby requiring the use of the product rule on $x(t) | \psi (x,t)|^2$?

5. Jul 7, 2014

6. Jul 7, 2014

### Fredrik

Staff Emeritus
The function $x\mapsto\psi(x,t)$ represents the procedure that the particle has been put through until time t. It's a square-integrable complex-valued function on $\mathbb R$. Measuring devices are represented by self-adjoint linear operators on the vector space of such functions. A time-dependence of such an operator would mean that we're dealing with different measuring devices at different times. It's not obvious that (for example) a particle detector located near position x and time t, should be considered the same device and therefore have the same mathematical representation as the same detector once it's been moved to position x' at time t' (or equivalently, if we use a coordinate system with a different origin). But we assume that it's appropriate to represent them by the same operator, because it would be very hard to do science if "the same experiment" has different results at different locations or at different times. As long as there are no experiments that contradict this, there's no reason to view this as a flaw in the theory.

In particular, the position operator Q is defined by $(Qf)(x)=xf(x)$ for all square-integrable f and all x in its domain. There's no t in its definition. However, if we use the preparation procedure reprented by f and then wait a time t, this should be viewed as a different preparation procedure that should be represented by a different function $f_t$. So for all x and all t, we have $(Qf_t)(x)=xf_t(x)$. As you can see, the only time-dependence on the right-hand side is that $f_t$ is a different function for each t.

The relationship between your $\psi$ and my $f_t$ is that $\psi(x,t)=f_t(x)$ for all x and t. In particular, we have $\psi(x,0)=f_0(x)=f(x)$.

Most QM books use the notation $\hat x$ instead of $Q$. This is of course only a convention. They also write things like $\hat x\psi(x,t)$. This should be viewed as an abuse of notation. The notation really means $(\hat x f_t)(x)$.

A more direct answer to your question is that $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial t}$ are supposed to denote the partial derivatives with respect to the first and second variables respectively. This choice of notation alone is a very strong hint that x and t are meant to be treated as independent variables.