# Complex conjugate of a wavefunction

Let ψ be a wavefunction describes the quantum state of a particle at any (x,t), What does ψ* i.e, the complex conjugate of a wavefunction means? I only know probability of finding a particle is given by ∫|ψ|^2 dx= ∫ ψ*ψ dx But what does ψ*ψ really means? I started learning QM with Griffiths without enough mathematical background, Help me.

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As you said, supposing that you have a wave function ψ(x) describing your particle in space, then $\int_a^b \psi^*(x)\psi(x)dx$ is the probability of finding your particle between the points a and b.

It follows that $\psi^*(x)\psi(x)$ is the so called probability density. It is nothing but the function that, when integrated, gives you the probability itself.

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If you measure the position of the particle, at time t, ψ*ψ(x,t) represents the likelihood that you will find it to be at position x... or more accurately within between x and x+dx.

As for what ψ* means, it's the complex conjugate of the wavefunction... but of course you already know that. If you're looking for a nice intuitive way to interpret the physical meaning of the phase of a wavefunction, then I'm afraid I don't think that there is one.

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As for what ψ* means, it's the complex conjugate of the wavefunction... but of course you already know that. If you're looking for a nice intuitive way to interpret the physical meaning of the phase of a wavefunction, then I'm afraid I don't think that there is one.
Exactly, this is just how the mathematical formalism of Quantum Mechanics works.

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ChrisVer
Gold Member
you only need $\psi^{*}$ because you need to have a real thing to integrate over to get probability.... $\psi$ alone is complex, and also it's arbitrary up to an overall phase.... that means that either you take $\psi$ or $e^{i \theta} \psi$ they both are solutions of the same shrodinger equation. As such you must make it real, or you'd have arbitrary probabilities depending on a phase...
One quantity that is for sure real is $\psi^{*} \psi$ and nothing is "lost" from $\psi$'s initial parameters (if for example you write $\psi=a+ib$ the result will contain both a and b...

Another example in which you can see that, is when you have $\psi$ real, where you don't need the complex conjugate and just take the square....

In fact the complex conjugate's "meaning" (mathematical) becomes clearer when you start working on the states in vector spaces... The conjugate appears as a relation between the bras and kets, and thus between the vector and its dual space.

bhobba
Mentor
Let ψ be a wavefunction describes the quantum state of a particle at any (x,t), What does ψ* i.e, the complex conjugate of a wavefunction means?
Its got to do with the bra-ket notation:
http://en.wikipedia.org/wiki/Bra–ket_notation

The wavefunction is looked at as the Ket - its conjugate as the Bra - apply a Bra to a Ket and square it to get probabilities - that's the Born rule.

At a deeper level states are not actually Bra's or Kets - they are really operators - what are normally referred to as states have a special name called pure.

At an even deeper level again it is seen states are really what's required so we can work out probabilities of the outcomes of observables - this is the important Gleason's Theorem (the key assumption is what is known as non contextuality you may have heard about):
http://en.wikipedia.org/wiki/Gleason's_theorem

It's a devil of a thing to prove however so its not usually talked about much at the beginning level despite its importance.

There is a variant that is much easier to prove and some people use it as the foundation of QM eg:
http://arxiv.org/pdf/quant-ph/0205039v1.pdf

Anyway that will take you to a pretty advanced level. For now what I would do is get a hold of Ballentine:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20&tag=pfamazon01-20

Its mathematically quite a bit more advanced than Griffiths. But hold onto your nerves and give the first 3 chapters a read. You wont likely understand too much of the detail, but, hopefully, fingers crossed, you will get the gist so that things are a lot clearer, and Griffiths will be easier. Once you have finished Griffiths you can move onto Ballentine.

Thanks
Bill

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