# Why we multiply in ψ*

1. Jun 28, 2012

### sciboudy

why in the quantum mechanics we must multiply ψ by ψ*

why we must multiply in the conjugate to find the probability density ?

Last edited: Jun 28, 2012
2. Jun 28, 2012

### Staff: Mentor

using anything else will leave you with a complex result. Using the psi* gets you a real number.

3. Jun 28, 2012

### sciboudy

so why is the probabiblty density = epsi in epsi*

4. Jun 28, 2012

### sciboudy

and why is the probability density equal to multiply OF epsi it self

5. Jun 28, 2012

### Staff: Mentor

perhaps this wikipedia article can explain it better:

http://en.wikipedia.org/wiki/Schrödinger_equation

Part way down in the historical background section, the article mentions that Max Born interpreted the psi function as a probability amplitude which led to interpreting the psi*psi as the probability density something that physicists could measure in a system.

6. Jun 28, 2012

### Khashishi

Multiplying by the complex conjugate gives you the norm squared. If you represent a complex number by the amplitude and phase, then multiplying by the complex conjugate is basically throwing away the phase information. We can't really measure the phase, only differences in phase. The phase is sort of a bookkeeping mechanism for keeping track of quantum interference.

7. Jun 28, 2012

### sciboudy

thank you khashishi and jedishrfu i will read the article

8. Jun 29, 2012

### Many_S_Theory

That isn't a quantum mechanics thing, it's a complex numbers thing. The magnitude of a complex number is given by $$z^* z$$ not $$z^2$$. Think about it, if instead of a complex number I thought of z as a 2-dimensional vector, an x and y coordinate (the y is the complex part). Then the, euclidian, distance from the origin would be $$x^2+y^2$$. So for a complex number $$a+bi$$ you might think the "distance" from the origin to the point in the complex plane is simply $$(a)^2+(bi)^2$$ but that's really $$a^2-b^2$$. In order to get the usual distance we need $$(a+bi)(a-bi)=a^2+b^2+abi-abi = a^2+b^2$$.

The same is equally as true for a complex function. This isn't quantum mechanics it's just the math of complex numbers. In general the use of i in quantum mechanics is a bit of a misdirect for a lot of new students, and this isn't helped by many teachers making it seem mysterious. In reality the role of complex numbers in quantum mechanics is similar to simply saying they there isn't ONE complex equation (the schrodinger equation) which must be followed but in fact two REAL equations. Indeed you can write all of quantum mechanics in such a way that complex numbers don't appear.

9. Jun 29, 2012

### sciboudy

thank you sir but what is the usage of complex in Q m and why we need it ? and what happen if Q. M don't contain complex numbers

10. Jun 29, 2012

### bhobba

You need complex numbers to have infinitesimal displacements of apparatus used to observe systems. Such a displacement does not alter the laws of physics which, without spelling out the detail, means superpositions and lengths are preserved. For definiteness suppose you displace it by an infinitesimal distance dx then the whole quantum vector space is transformed by an operator of the form 1 + Udx where U is a linear operator - it must be linear since superpositions are preserved. Since lengths are preserved we have that 1 + Udx must be unitary which means (1 + Udx)(1 + U(bar)dx) = 1 which implies U = -U(bar) which means U is pure imaginary ie you must use complex numbers.

Why do we multiply by the complex conjugate to get probabilities - its because of a very important theorem called Gleasons Theorem:
http://en.wikipedia.org/wiki/Gleason's_theorem

Basically it follows from some very reasonable assumptions about the probabilities we can define on a complex vector space.

Sorry that the answer is mathematical and requires some linear algebra but its unfortunately in the nature of the beast.

Thanks
Bill

Last edited: Jun 29, 2012
11. Jun 29, 2012

### sciboudy

thank you bill

12. Jun 29, 2012

### grsharma

Because ψ*ψ represents probability density, which is a real quantity. It is know that product of a complex number with its conjugate gives a real quantity.

13. Jun 29, 2012

### akhmeteli

14. Jul 6, 2012

### unchained1978

A general inner product on some $L^{2}$ space between two functions f and g is written as $(f | g ) = \int_{-\infty}^{\infty} f^{*}g\, dx$. The complex conjugation is so that the norm of a function is real valued, i.e. the magnitude of a vector (or function) is real. There are some cases within QM where $\psi (x)$ is real valued, such as the infinite square well wave functions, in which case the complex conjugation is unnecessary.