# Time-independence of seperable solutions of the Schrödinger eqn

1. Feb 6, 2014

### Runei

Hey there clever folk,

So im taking a class in Quantum Mechanics, and i've just completed an assignment about seperable solutions to the schrödinger equation.

One of the questions was to show that seperable solutions give rise to time-independent probability densities.

So I was having the equation $$\left|{\psi(x,t)}\right|^2=\left|{T(t)}\right|^2\left|{\psi(x)}\right|^2$$

So the time dependence will be determined only by the function of t. What I did was simply to solve the differential equation I had earlier to find T, and then simply multiply T by its complex conjugate. However, one of the later questions were to actual derive T(t) so I began wondering if there were another way (perhaps quicker) way to show that the complex conjugate product of T(t) would be time independent.

What I arrived at was that since

$$i\hbar\frac{d}{dt}T(t) = E T(t)$$

$$T(t) = \frac{i\hbar}{E}\frac{d}{dt}T(t)$$

$$T^*(t) = \frac{-i\hbar}{E}\frac{d}{dt}T^*(t)$$

$$T(t) \cdot T^*(t) = \frac{\hbar^2}{E^2}\frac{d}{dt}T^*(t)\frac{d}{dt}T(t)$$

Therefore I was down to that the time dependence or independence had to be determined from the product

$$\frac{dT^*}{dt}\frac{dT}{dt}$$

Am I on a completely wild goose hunt? Or is there something about it? When I arrived here I was like "I don't think I have heard about that a product of complex conjugate derivatives cancel".

:grumpy:

Best regards,
Rune

2. Feb 6, 2014

### WannabeNewton

You almost had it but went in the wrong direction halfway through. Go back to the step where you had $\frac{dT}{dt} = \frac{-iE}{\hbar}T$ and solve the DE (it's trivial). The result will be a unitary time evolution.

3. Feb 6, 2014

### Bill_K

I'd suggest the obvious. Start with (d/dt)|T(t)|2 and see if you can show it is zero.

4. Feb 6, 2014