Quantum Mechanics

QM State Vector

by Matt on Jun.21, 2008, under Physics, Quantum Mechanics

It’s revision time again (this time for a few resits ) so I shall try and put some things I cover up here. To start with something simple!

The quantum mechanical state vector, $\psi$ , is a function of position and time, so in 1 dimension, $\psi (x,t)$ .

The state vector or wave function $\psi$ contains all the information about a system at a given time, $t$ .¹ For a 1-d particle any complex function $\psi (x,t)$ is a possible state of the system, e.g.

$\psi (x,t) = C e^{- a x^2} e^{-i \omega t}$

If this is the state of the system then from it we should be able to find it’s position and it’s velocity. However quantum mechanics in general cannot find these with certainty, we can only say that if we measure it’s position then there is a probability of finding it between $a \leq x \leq b$ In order to find this we need to know the probability density of the state. This is defined as;

$P(x,t) = \psi (x,t) \psi^* (x,t) dx$

Where $\psi^* (x,t)$ is the complex conjugate of $\psi (x,t)$ . We can take this further as a complex function times it’s complex conjugate is the same as takeing the modulus of function squared. So that;

$P(x,t) = | \psi (x,t) |^2 dx$

We now have enough information to find the probability of the functions position lying between a and b, we can do this by integrating.

$\textrm{Prob}(a \leq \textrm{position} \leq b) = \int^b_a P(x,t) dx$

$\textrm{Prob}(a \leq \textrm{position} \leq b) = \int^b_a | \psi (x,t) |^2 dx$

It is important to see that $P(x,t)$ is both real and positive. It’s also important to note that the total probability of finding the particle must be 1, so we can set the constant C by normalizing $\psi$ . This is done as follows

$1 = \int^{\infty}_{- \infty} P(x,t) dx$

Now we’ve normalized the state vector we can look to see where we expect to find the particle. This is the expectation value and is the average value of, in this case, the position.

$< \psi | x | \psi > = \int^{\infty}_{- \infty} P(x,t) x dx$

$<\psi | x | \psi> = \int^{\infty}_{- \infty} \psi^* (x,t) x \psi (x,t) dx$

Note: This is not the most likely place to find the particle, that would be the mode while the expectation value is the mean.

Next time in QM, the Schrodinger Equation!

Quantum Mechanics by F. Mandl Page 2 [↩]

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