Time-independent Schrodinger equation 01

— 1. Stationary states —

In the previous posts we’ve normalized wave functions, we’ve calculated expectation values of momenta and positions but never at any point we’ve made a quite logical question:

How does one calculate the wave function in the first place?

The answer to that question obviously is:

You have to solve the Schroedinger equation.

The Schroedinger equation is

\displaystyle i\hbar\frac{\partial \Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+V\Psi(x,t)

Which is partial differential equation of second order. Partial differential equations are very hard to solve whereas ordinary differential equations are easily solved.

The trick is to to turn this partial differential equation into ordinary differential equation.

To do such a thing we’ll employ the separation of variables technique.

We’ll assume that {\Psi(x,t)} ca be written as the product of two functions. One of the functions is a function of the position alone whereas the other function is solely a function of {t}.

\displaystyle  \Psi(x,t)=\psi(x)\varphi(t)

This restriction might seem as overly restrictive to the class of solutions of the Schroedinger Equations, but in this case appearances are deceiving. As we’ll see later on more generalized solutions of the Schroedinger Equation can be constructed with separable solutions.

Calculating the appropriate derivatives for {\Psi(x,t)} yields:

\displaystyle  \frac{\partial \Psi}{\partial t}=\psi\frac{d\varphi}{dt}

and

\displaystyle  \frac{\partial^2 \Psi}{\partial x^2} = \frac{d^2 \psi}{d x^2}\varphi

Substituting the previous equations into the Schroedinger equation results is:

\displaystyle  i\hbar\psi\frac{d\varphi}{dt}=-\frac{\hbar^2}{2m}\frac{d \psi^2}{d x^2}\varphi+V\psi\varphi

Dividing the previous equality by {\psi\varphi}

\displaystyle  i\hbar\frac{1}{\varphi}\frac{d\varphi}{dt}=-\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d \psi^2}{d x^2}+V

Now in the previous equality the left-hand side is a function of {t} while the right-hand side is a function of {x} (remember that by hypothesis {V} isn’t a function of {t}).

These two facts make the equality expressed in the last equation require a very fine balance. For instance if one were to vary {x} without varying {t} then the right-hand side would change while left-hand side would remain the same spoiling our equality. Evidently such a thing can’t happen! Te only way for all equality to hold is that both sides of the equation are in fact constant. That way there’s no more funny business of changing one side while the other remains constant.

For reasons that will become obvious in the course of this post we’ll denote this constant (the so-called separation constant) by {E}.

\displaystyle  i\hbar\frac{1}{\varphi}\frac{d\varphi}{dt}=E \Leftrightarrow \frac{d\varphi}{dt}=-\frac{i E}{\hbar}\varphi

and for the second equation

\displaystyle  -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d^2 \psi}{d x^2}+V=E \Leftrightarrow -\frac{\hbar^2}{2m}\frac{d^2 \psi}{d x^2}+V\psi=E\psi

The first equation of this group is ready to be solved and a solution is

\displaystyle  \varphi=e^{-i\frac{E}{\hbar}t}

The second equation, the so-called time-independent Schroedinger equation can only be solved when a potential is specified.

As we can see the method of separable solutions had lived to my promise. With it we were able to produce two ordinary equations which can in principle be solved. In fact one of the equations is already solved.

At this point we’ll state a few characteristics of separable solutions in order to better understand their importance (of one these characteristics was already hinted before):

  1. Stationary states

    The wave function is

    \displaystyle  \Psi(x,t)=\psi(x)e^{-i\frac{E}{\hbar}t}

    and it is obvious that it depends on {t}. On the other hand the probability density doesn’t depend on {t}. This result can easily be proven with the implicit assumption that {E} is real (in a later exercise we’ll see why {E} has to be real).

    \displaystyle \Psi(x,t)^*\Psi(x,t)=\psi^*(x)e^{i\frac{E}{\hbar}t}\psi(x)e^{-i\frac{E}{\hbar}t}=|\psi(x)|^2

    If we were interested in calculating the expectation value of any dynamical variable we would see that those values are constant in time.

    \displaystyle  <Q(x,p)>=\int\Psi^*Q\left( x,\frac{\hbar}{i}\frac{\partial}{\partial x} \right)\Psi\, dx

    In particular {<x>} is constant in time and as a consequence {<p>=0}.

  2. Definite total energy

    As we saw in classical mechanics the Hamiltonian of a particle is

    \displaystyle H(x,p)=\frac{p^2}{2m}+V(x)

    Doing the appropriate substitutions the corresponding quantum mechanical operator is (in quantum mechanics operators are denoted by a hat):

    \displaystyle \hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}+V

    Hence the time-independent Schroedinger equation can be written in the following form:

    \displaystyle  \hat{H}\psi=E\psi

    The expectation value of the Hamiltonian is

    \displaystyle  <\hat{H}>=\int\psi ^*\hat{H}\psi\, dx=E\int|\psi|^2\, dx=E

    It also is

    \displaystyle  \hat{H}^2\psi=\hat{H}(\hat{H}\psi)=\hat{H}(E\psi)=E\hat{H}\psi=EE\psi=E^2\psi

    Hence

    \displaystyle  <\hat{H}^2>=\int\psi ^*\hat{H}^2\psi\, dx=E^2\int|\psi|^2\, dx=E^2

    So that the variance is

    \displaystyle \sigma_{\hat{H}}^2=<\hat{H}^2>-<\hat{H}>^2=E^2-E^2=0

    In conclusion in a stationary state every energy measurement is certain to return the value {E} since that the energy distribution has value {E}.

  3. Linear combination

    The general solution of the Schroedinger equation is a linear combination of separable solutions.

    We’ll see in future examples and exercises that the time-independent Schroedinger equation holds an infinite number of solutions. Each of these different wave functions is associated with a different separation constant. Which is to say that for each allowed energy level there is a different wave function.

    It so happens that for the time-dependent Schroedinger equation any linear combination of a solution is itself a solution. After finding the separable solutions the task is to construct a more general solution of the form

    \displaystyle \Psi(x,t)=\sum_{n=1}^{+\infty}c_n\psi_n(x)e^{-i\frac{E_n}{\hbar}t}=\sum_{n=1}^{+\infty}c_n\Psi_n(x,t)

    The point is that every solution of the time-dependent Schroedinger equation can be written like this with the initial conditions of the problem being being studied fixing the values of the constants {c_n}.

    I understand that all of this may be a little too abstract so we’ll solve a few exercises to make it more palatable.

As an example we’ll calculate the time evolution of a particle that starts out in a linear combination of two stationary states:

\displaystyle \Psi(x,0)=c_1\psi_1(x)+c_2\psi_2(x)

For the sake of our discussion let’s take {c_n} and {\psi_n} to be real.

Hence the time evolution of the particle is simply:

\displaystyle  \Psi(x,t)=c_1\psi_1(x)e^{-i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{-i\frac{E_2}{\hbar}t}

For the probability density it is

{\begin{aligned} |\Psi(x,t)|^2 &= \left( c_1\psi_1(x)e^{i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{i\frac{E_2}{\hbar}t} \right) \left( c_1\psi_1(x)e^{-i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{-i\frac{E_2}{\hbar}t} \right)\\ &= c_1^2\psi_1^2+c_2^2\psi_2^2+2c_1c_2\psi_1\psi_2\cos\left[ \dfrac{E_2-E_1}{\hbar}t \right] \end{aligned}}

As we can see even though {\psi_1} and {\psi_2} are stationary states and hence their probability density is constant the probability density of the final wave function oscillates sinusoidally with angular frequency {(E_2-E_1)/t}.

Prove that for for normalizable solutions the separation constant {E} must be real.

Let us write {E} as

{E=E_0+i\Gamma}

Then the wave equation is of the form

\displaystyle  \Psi(x,t)=\psi(x)e^{-i\frac{E_0}{\hbar}t}e^{\frac{\Gamma}{\hbar}t}

{\begin{aligned} 1 &= \int_{-\infty}^{+\infty}|\Psi(x,t)|^2\, dx \\ &= \int_{-\infty}^{+\infty} \psi(x,t)^*\psi(x,t)e^{-i\frac{E_0}{\hbar}t}e^{i\frac{E_0}{\hbar}t}e^{\frac{\Gamma}{\hbar}t}e^{\frac{\Gamma}{\hbar}t}\, dx \\ &= e^{\frac{2\Gamma}{\hbar}t}\int_{-\infty}^{+\infty}|\psi(x,t)|^2\, dx \end{aligned}}

The final expression has to be equal to {1} to all values of {t}. The only way for that to happen is that we set {\Gamma=0}. Thus {E} is real.

Show that the time-independent wave function can always be taken to be a real valued function.

We know that {\psi(x)} is a solution of

\displaystyle  -\frac{\hbar^2}{2m}\frac{d^2 \psi}{d x^2}+V\psi=E\psi

Taking the complex conjugate of the previous equation

\displaystyle  -\frac{\hbar^2}{2m}\frac{d^2 \psi^*}{d x^2}+V\psi^*=E\psi^*

Hence {\psi^*} is also a solution of the time-independent Schroedinger equation.

Our next result will be to show that if {\psi_1} and {\psi_2} are solutions of the time-independent Schroedinger equation with energy {E} then their linear combination also is a solution to the time-independent Schroedinger equation with energy {E}.

Let’s take

\displaystyle \psi_3=c_1\psi_1+c_2\psi_2

as the linear combination.

{\begin{aligned} -\frac{\hbar^2}{2m}\frac{d^2 \psi_3}{d x^2}+V\psi_3 &= -\frac{\hbar^2}{2m}\left( c_1\dfrac{\partial ^2\psi_1}{\partial x^2}+c_2\dfrac{\partial ^2\psi_2}{\partial x^2} \right)+ V(c_1\psi_1+c_2\psi_2)\\ &= c_1\left( -\frac{\hbar^2}{2m}\dfrac{\partial ^2\psi_1}{\partial x^2}+V\psi_1 \right)+c_2\left( -\frac{\hbar^2}{2m}\dfrac{\partial ^2\psi_2}{\partial x^2}+V\psi_2 \right)\\ &= c_1E\psi_1 + c_2E\psi_2\\ &= E(c_1\psi_1+c_2\psi_2)\\ &= E\psi_3 \end{aligned}}

After proving this result it is obvious that {\psi+\psi^*} and that {i(\psi-\psi^*)} are solutions to the time-independent Schroedinger equation. Apart from being solutions to the time-independent Schroedinger equation it is also evident from their construction that these functions are real functions. Since they have same value for the {E} as {\psi} we can use either one of them as a solution to the time-independent Schroedinger equation

If {V(x)} is an even function than {\psi(x)} can always be taken to be either even or odd.

Since {V(x)} is even we know that {V(-x)=V(x)}. Now we need to prove that if {\psi(x)} is a solution to the time-independent Schroedinger equation then {\psi(-x)} also is a solution.

Changing from {x} to {-x} in the time-independent Schroedinger equation

\displaystyle  -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d (-x)^2}+V(-x)\psi(-x)=E\psi(-x)

in order to understand what’s going on with the previous equation we need to simplify

\displaystyle  \dfrac{d^2}{d (-x)^2}

Let us introduce the variable {u} and define it as {u=-x}. Then

\displaystyle \frac{d}{du}=\frac{dx}{du}\frac{d}{dx}=-\frac{d}{dx}

And for the second derivative it is

\displaystyle  \frac{d^2}{du^2}=\frac{dx}{du}\frac{d}{dx}\frac{dx}{du}\frac{d}{dx}=\left(-\frac{d}{dx}\right)\left(-\frac{d}{dx}\right)=\frac{d^2}{dx^2}

In the last expression {u} is a dummy variable and thus can be substituted by any other symbol (also see this post Trick with partial derivatives in Statistical Physics to see what kind of manipulations you can do with change of variables and derivatives). For convenience we’ll change it back to {x} and it is

\displaystyle  \dfrac{d^2}{d (-x)^2}=\dfrac{d^2}{d x^2}

So that our initial expression becomes

\displaystyle  -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d x^2}+V(-x)\psi(-x)=E\psi(-x)

Using the fact that {V(x)} is even it is

\displaystyle  -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d x^2}+V(x)\psi(-x)=E\psi(-x)

Hence {\psi(-x)} is also a solution to the time-independent Schroedinger equation.

Since {\psi(x)} and {\psi(-x)} are solutions to the time-independent Schroedinger equation whenever {V(x)} is even we can construct even and odd functions that are solutions to the time-independent Schroedinger equation.

The even functions are constructed as

\displaystyle  h(x)=\psi(x)+\psi(-x)

and the odd functions are constructed as

\displaystyle  g(x)=\psi(x)-\psi(-x)

Since it is

\displaystyle  \psi(x)=\frac{1}{2}(h(x)+g(x))

we have showed that any solution to the time-independent Schroedinger equation can be expressed as a linear combination of even and odd functions when the potential function is an even function.

Show that {E} must exceed the minimum value of {V(x)} for every normalizable solution to the time-independent Schroedinger equation.

Rewriting the time-independent Schroedinger equation in order of its second x derivative

\displaystyle \frac{d^2\psi}{dx^2}=\frac{2m}{\hbar ^2}(V(x)-E)\psi

Let us proceed with a proof by contradiction and assume that we have {V_{\mathrm{min}}>E}. Using the previous equation this implies that {\dfrac{d^2\psi}{dx^2}} and {\psi } have the same sign. This comes from the fact that {\frac{2m}{\hbar ^2}(V(x)-E)} is positive.

Let us suppose that {\psi} is always positive. Then {\dfrac{d^2\psi}{dx^2}} is also always positive. Hence {\psi} is concave up. In the first quadrant the graph of the function is shaped like

Since by hypothesis our function is normalizable it needs to go to {0} as {x\rightarrow -\infty}. in order for the function to go to {0} the plot needs to do something like this

Such a behavior would imply that there is region of space where the function is positive and its second derivative is negative (in our example such a region is delimited in {-0.5\leq x \leq 0.5}).

Such a behavior is in direct contradiction with the conclusion that {\psi} and {\dfrac{d^2\psi}{dx^2}} always have the same sign. Since this contradiction arose from the hypothesis that {V_{\mathrm{min}}>E} the logical conclusion is that {V_{\mathrm{min}}<E}.

With {V_{\mathrm{min}}<E}, {\psi} and {\dfrac{d^2\psi}{dx^2}} no longer need to have the same sign at all times. Hence {\psi} can turn over to {0} and {\psi} can go to {0}.

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