# Time independent schrödinger equation query

1. Feb 15, 2015

### Maylis

Hello, the TISE can be simplified
$$H \psi = E \psi$$
Where $H$ is the Hamiltonian, and $E$ is the eigenvalue, but why don't the $\psi$ terms cancel, leaving $H = E$?

Also, what the heck does the eigenvalue $E$ have to do with the eigenvalue that I have previously encountered in mathematics (linear algebra)

$A \textbf {v} = \lambda \textbf {v}$
where $A$ is a matrix, $\textbf {v}$ is a vector, and $\lambda$ is the eigenvalue. Where was linear algebra ever invoked when doing TISE to use such a name for E, Eigenvalue?

2. Feb 15, 2015

### Cruz Martinez

The Hamiltonian is a differential operator, not a mumber, you cant cancel the wave functions on both sides for this reason.

3. Feb 15, 2015

### Maylis

That was my suspicion

4. Feb 15, 2015

### Maylis

Well, I have some more issues with this. I don't understand why there are two ways of defining and then calling them equivalent, $\hat {H} \psi = E \psi$

So I think the quantum mechanical Hamiltonian is defined

$$\hat {H} = - \frac {\hbar^{2}}{2m} \frac {\partial^{2}}{\partial x^{2}} + V$$

So I will just take this for granted as a letter that will signify the sum of the kinetic and potential energy, with the kinetic energy having a fancy operator attached to it. Now if I multiply through by $\psi$, I will get

$$\hat {H} \psi = - \frac {\hbar^{2}}{2m} \frac {\partial^{2} \psi}{\partial x^{2}} + V \psi$$

But for some reason, in deriving the TISE, we set
$$i \hbar \frac {1}{f} \frac {\partial f}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {1}{\psi} \frac {d^{2} \psi}{dx^{2}} + V = E$$

Where $f$ is a function of time, taken from separating variables of the wave function $\Psi (x,t) = \psi (x) f(t)$

Why create this constant, $E$? At the end of it all, it becomes $\hat {H} \psi = E \psi$. I don't understand why this is necessary to even have a letter $E$.

Is it perhaps meaningless to write $\Psi (x,t) = \psi(x) e^{\frac {-i \hat {H} t}{\hbar}}$ since $\hat {H}$ contains the second derivative operator, and for that reason $E$ is useful?

Last edited: Feb 15, 2015
5. Feb 15, 2015

### vela

Staff Emeritus
E is a separation constant. The term involving $f$ is a function of $t$ only while the stuff in the middle is a function of $x$ only. The only way this equality can hold is if they're equal to a constant.

6. Feb 15, 2015

### Maylis

What the heck is the difference between E and the Hamiltonian?

7. Feb 15, 2015

### Staff: Mentor

The potential energy $V$ is also an operator and it may be a function of $x$. For example, if we're looking for the wave function of a particle in a harmonic oscillator, $V=\frac{1}{2}kx^2$ and the time-independent Schrodinger equation becomes
$$\frac {\hbar^{2}}{2m} \frac {\partial^{2} \psi}{\partial x^{2}}+\frac{1}{2}kx^2\psi=E\psi$$

Solutions (other than the trivial $\psi=0$) to this differential equation only exist for certain values of $E$.

This is the same concept of eigenvalue and eigenvector that you saw in linear algebra. All that's going on now is that the vector space on which the operator H is acting is the space of functions $\psi$. H maps the function $\psi$ onto the function $H\psi$, and the TISE is looking for functions that are mapped to scalar multiples of themselves by the operator H.

Last edited: Feb 15, 2015
8. Feb 15, 2015

### vanhees71

The fundamental equation of non-relativistic quantum theory in its formulation as "wave mechanics" (discovered by Erwin Schrödinger in 1926 when on skiing vacation) is the time-dependent Schrödinger equation
$$\mathrm{i} \hbar \partial_t \psi(t,x)=\hat{H} \psi(t,x),$$
where the Hamiltonian in the most simple case of a point particle moving in a potential is given by
$$\hat{H}=-\frac{\hbar^2}{2m} \partial_x^2 + V(x).$$
Given the wavefunction at $t=0$, you can calculate the wave function at any later time by solving this partial differential equation (with some boundary conditions, ensuring that $|\psi(t,x)|^2$ can be integrated over $x \in \mathbb{R}$.

One method to solve linear partial differential equations is a separation ansatz. This lets you look for solutions of the equation which factor in a product of functions of $t$ and $x$:
$$\psi(t,x)=f(t) \Psi(x).$$
Plugging this into the Schrödinger equation, you get
$$\mathrm{i} \hbar \Psi(x) \frac{\mathrm{d}}{\mathrm{d} t} f(t) =f(t) \left [-\frac{\hbar^2 \Psi''(x)}{2m} + V(x) \Psi(x) \right ].$$
Now you divide this equation by $f(t) \Psi(x)$. Then you get
$$\mathrm{i} \hbar \frac{1}{f(t)} \frac{\mathrm{d}}{\mathrm{d} t} f(t) =\frac{1}{\Psi(x)} \left [-\frac{\hbar^2 \Psi''(x)}{2m} + V(x) \Psi(x) \right ].$$
Now the left-hand side does not depend on $x$ and the right-hand side does not depend on $t$. This implies that it cannot depend on both $x$ and $t$, i.e., it's a constant, which we suggestively call $E$.

From the left-hand side this means
$$\mathrm{i} \hbar \frac{\dot{f}}{f}=E.$$
This you can immediately integrate to get
$$f(t)=f_0 \exp \left (-\frac{\mathrm{i} E t}{\hbar} \right )=u_E(t).$$
For the right-hand side you get
$$\hat{H} \Psi(x)=E \Psi(x).$$
Together with the approriate boundary conditions you get the possible set of eigenvalues $E$ of the Hamiltonian and the corresponding eigensolutions, $\Psi_E(x)$. Usually the possible eigensolutions involve a set of discrete energy eigenvalues $E_j \in \mathbb{R}$ ("bound states") and a continuous set $S(\hat{H}) \subseteq \mathbb{R}$. Then you can write any wavefunction in terms of these eigensolutions:
$$\psi(t,x)=\sum_n \exp \left(-\frac{\mathrm{i} E t}{\hbar} \right) A_n \Psi_{E_n}(x) + \int_{S(\hat{H})} \mathrm{d} E A(E) \Psi_{E}(x).$$

9. Feb 15, 2015

### Maylis

I know $\Psi (x,t)$ is the wave function, what does $\psi(x)$ by itself represent or mean?

10. Feb 15, 2015

### Staff: Mentor

Its the expansion of a pure state in terms of the position eigenfunctions.

Here are the two axioms of QM:

Axiom 1
Associated with each measurement we can find a Hermitian operator O, called the observations observable, such that the possible outcomes of the observation are its eigenvalues yi.

Axiom 2 - called the Born Rule
Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

Axiom 2 is to some extent, without going into the details, implied by axiom 1 via Gleason's theorem - but no need to go into that here - we will simply focus on axiom 1.

Given a Hermitian operator O by the spectral theorem it can be expanded in terms of its eigenvalues and eigenvectors O = ∑yi |yi><yi|. Since the |yi> form an orthonormal basis any vector |v> can be expanded in terms of that basis |v> = ∑vi |yi>. The vi is called the representation of the vector v in terms of the observable O. Ignoring the subtlety of a continuous basis whose rigorous resolution involves what's called a Rigged Hilbert Space the wave-function is the representation of the systems state in terms of the position observable.

Without proving it, it turns out, from the Born rule, given a wave-function ψ(x), the probability of finding a particle at position x if you observe it is |ψ(x)|^2.

To really understand this stuff get a copy of Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

Thanks
Bill

Last edited: Feb 15, 2015
11. Feb 15, 2015

### Staff: Mentor

If $H$ is not a function of $t$ (which is equivalent to saying that the potential is constant over time), then we can solve the time-dependent Schrodinger equation using separation of variables. The solutions will be $\Psi(x,t)=\psi(x)e^{-iEt / \hbar}$ where $\psi(x)$ is the solution of the time-independent Schrodinger equation $H\psi=E\psi$ with $E$ a scalar.

Because $H$ is Hermitian we know that $E$ is real. Therefore the magnitude of the exponential is equal to one, and $|\Psi(x,t)|^2$, the probability distribution of finding a particle with energy $E$ at position $x$ at time $t$, is equal to $|\psi(x)|^2$. More complicated states with more complicated and interesting probability distributions will be the square of linear combinations of different solutions of the TISE with different values of $E$.

Last edited: Feb 15, 2015