# Wave function

1. May 16, 2015

### deependra1003

Suppose there is a wave function for a particle. What is the actual meaning of hermition conjugate of the wave function? Does the conjugate represent any other state of the wave function? What is the physical meaning of product of wave function and its conjugate? Please explain.

2. May 16, 2015

### Staff: Mentor

Different people have different ways of thinking about this stuff, but it may be safest (that is, least likely to mislead you) if think of the entire mathematical formalism as just a machine for calculating stuff with no physical significance of its own. I can use a calculator to calculate the kinetic energy of a moving object, but I don't ask how the calculator keys might be related to the object.

3. May 16, 2015

### The_Duck

This is just a convenient way of calculating the square of the magnitude of the wave function, which is the probability density.

4. May 16, 2015

### vanhees71

If you have a wave function $\psi(t,\vec{x})$ as the solution of the time-dependent Schrödinger equation, you can easily show that the conjugate function solves the conjugate Schrödinger equation, but the conjugate Schrödinger equation looks pretty much like the Schrödinger equation, except that the sign of the term with the time derivative changes, because of the $\mathrm{i}$ in front of it. The Schrödinger equation reads
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=\left [-\frac{\Delta}{2m}+V(\vec{x}) \right ] \psi(t,\vec{x}).$$
Now take the conjugate complex of this equation. The only change is the sign on the left side of the equation and that the conjugate wave function occurs:
$$-\mathrm{i} \partial_t \psi^*(t,\vec{x})=\left [-\frac{\Delta}{2m}+V(\vec{x}) \right ] \psi^*(t,\vec{x}).$$
This shows that complex conjugation has to do with the "time-reversal transformation", because the changed sign in front of the time derivative can be interpreted as following the time evolution backwards. Indeed, it's immediately clear from the equation, that the solution of the complex conjugate equation is not a solution of the original Schrödinger equation, but obviously $\psi^*(-t,\vec{x})$ is.

Indeed, it turns out that the time-reversal transformation is realized as an anti-unitary operator rather than a unitary one. For details, see a good QT textbook, e.g., Sakurai&Napolitano.

5. May 16, 2015

### blue_leaf77

In the relativistic case, the conjugated wavefunction is sometimes related to the antiparticle.

6. Jul 27, 2015

### kaushikquanta

what is wave function actually. i don't get a clear view

7. Jul 28, 2015

### Staff: Mentor

Its the expansion of the state in terms of position eigenstates.

If you don't know the technicalities then the above is likely gibberish - but unfortunately this is one area that cant be explained in lay terms.

At the lay level simply look at it as a way of describing the quantum state that emphasises position measurements.

That still leaves open what is a state. In lay terms its simply a device that allows us to calculate the probabilities of the outcomes of observations.

Thanks
Bill

8. Jul 28, 2015

### Staff: Mentor

A whole lot of very complicated math.

9. Jul 28, 2015

### atyy

1. There is no clear view. It is a meaningless thing that we use to calculate the probabilities of measurement outcomes.

2. However, we make no mistake if we pretend that it is the complete state of a single quantum system.